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stem. For example, point (4, 3), which lies in the region, satisfies the system because its x-value satisfies one of the given inequalities, 4 2, and its y-value satisfies the other inequality, 3 2. EXAMPLE 2 Graph the solution set of 3 x 5 in a coordinate plane. Solution The inequality 3 x 5 means 3 x and x 5. This may be written as x 3 and x 5. (1) Graph x 3 by first graphing the plane divider x 3, which, in the figure at the right, is represented by the dashed line labeled l. The half-plane to the right of line x 3 is the graph of the solution set of x 3. (2) Using the same set of axes, graph x 5 by first graphing the plane divider x 5, which, in the figure at the right, is represented by the dashed line labeled m. The half-plane to the left of line x 5 is the graph of the solution set of x 5. (3) The dark colored region, which is the intersection of the graphs made in steps (1) and (2), is the graph of the solution set of 3 x 5, or x 3 and x 5. All points in this region, and no others, satisfy 3 x 5. For example, point (4, 3), which lies in the region and whose x-value is 4, satisfies 3 x 5 because 3 4 5 is a true statement EXAMPLE 3 Graph the following system of inequalities and label the solution set R: x y 4 y 2x 3 Solution (1) Graph x y 4 by first graphing the plane divider x y 4, which, in the figure at the top of page 433, is represented by the solid line labeled l. The line x y 4 and the half-plane above this line together form the graph of the solution set of x y 4. Graphing the Solution Set of a System of Inequalities 433 (2) Using the same set of axes, graph y 2x 3 by first graphing the plane divider y 2x – 3. In the figure at the right, this is the solid line labeled m. The line y 2x 3 and the half-plane below this line together form the graph of the solution set of y 2x 3. (3) The dark colored region, the intersection of the graphs made in steps (1) and (2), is the solution set of the system x y 4 and y 2x 3. Label it R. Any point in the region R, such as (5, 2), will satisfy x y 4 because 5 2 4, or 7 4, is true, and will satisfy y 2x 3 because 2 2(5) 3, or 2 7, is true < 2x – 3 R x y < 2x – 3 1 –1–1 O 1 m EXAMPLE 4 Sandi boards cats and dogs while their owners are away. Each week she can care for no more than 12 animals. For next week she already has reservations for 4 cats and 5 dogs, but she knows those numbers will probably increase. Draw a graph to show the possible numbers of cats and dogs that Sandi might board next week and list all possible numbers of cats and dogs. Solution (1) Let x the number of cats Sandi will board, and y the number of dogs. (2) Use the information in the problem to write inequalities. Sandi can care for no more than 12 animals: She expects at least 4 cats: She expects at least 5 dogs: (3) Draw the graphs of these three y inequalities. (4) The possible numbers of cats and x y 12 x 4 y 5 dogs are represented by the ordered pairs of positive integers that are included in the solution set of the inequalities. These ordered pairs are shown as points on the graph. The points (4, 5), (4, 6), (4, 7), (4, 8), (5, 5), (5, 6), (5, 7), (6, 5), (6, 6), (7, 5), all satisfy the given conditions1–1 O 1 434 Writing and Solving Systems of Linear Functions Answer If there are 4 cats, there can be 5, 6, 7, or 8 dogs. If there are 5 cats, there can be 5, 6, or 7 dogs. If there are 6 cats, there can be 5 or 6 dogs. If there are 7 cats, there can be 5 dogs. EXERCISES Writing About Mathematics 1. Write a system of inequalities whose solution set is the unshaded region of the graph drawn in Example 3. Explain your answer. 2. What points on the graph drawn in Example 3 are in the solution set of the system of open sentences x y 4 and y 2x 3? Explain your answer. 3. Describe the solution set of the system of inequalities y 4x and 4x y 3. Explain why this occurs. Developing Skills In 4–18, graph each system of inequalities and label the solution set S. Check one solution. 4. x 1 y 2 7. y x y 2x 3 10. y x 5 y 2x 7 13. x y 3 x – y 6 16. 2x 3y 6 x y 4 0 5. x 0 y 0 8. y 2x y x 3 11. y x 1 x y 2 14 2x x 0 17. 6. y 1 y x 1 9. y 2x 3 y x 12. y 3x 6 y 2x 4 15. 2x y 6 18. x y 2 0 x 1 3 1 3y 2 2 4y # 3 3x , 0 In 19–23, in each case, graph the solution set in a coordinate plane. 19. 1 x 4 22. 2 y 3 20. 5 x 1 23. (y 1) and (y 5) 21. 2 y 6 In 24 and 25, write the system of equation whose solution set is labeled S. Graphing the Solution Set of a System of Inequalities 435 25. y 1 x 1 O 1 x S 24. y 1 O S Applying Skills 26. In Ms. Dwyer’s class, the number of boys is more than twice the number of girls. There are at least 2 girls. There are no more than 12 boys. a. Write the three sentences given above as three inequalities, letting x equal the number of girls and y equal the number of boys. b. On one set of axes, graph the three inequalities written in part a. c. Label the solution set of the system of inequalities S. d. Do the coordinates of every point in the region labeled S represent the possible num- ber of girls and number of boys in Ms. Dwyer’s class? Explain your answer. e. Write one ordered pair that could represent the number of girls and boys in Ms. Dwyer’s class. 27. When Mr. Ehmke drives to work, he drives on city streets for part of the trip and on the expressway for the rest of the trip. The total trip is less than 8 miles. He drives at least 1 mile on city streets and at least 2 miles on the expressway. a. Write three inequalities to represent the information given above, letting x equal the number of miles driven on city streets and y equal the number of miles driven on the expressway. b. On one set of axes, graph the three inequalities written in part a. c. Label the solution set of the system of inequalities R. d. Do the coordinates of every point in the region labeled R represent the possible number of miles driven on city streets and driven on the expressway? Explain your answer. e. Write one ordered pair that could represent the number of miles Mr. Ehmke drives on city streets and on the expressway. 28. Mildred bakes cakes and pies for sale. She takes orders for the baked goods but also makes extras to sell in her shop. On any day, she can make a total of no more than 12 cakes and pies. For Monday, she had orders for 4 cakes and 6 pies. Draw a graph and list the possible number of cakes and pies she might make on Monday. 436 Writing and Solving Systems of Linear Functions CHAPTER SUMMARY The equation of a line can be written if one of the following sets of infor- mation is known: 1. The slope and one point 2. The slope and the y-intercept 3. Two points 4. The x-intercept and the y-intercept A system of two linear equations in two variables may be: 1. Consistent: its solution is at least one ordered pair of numbers. (The graphs of the equations are intersecting lines or lines that coincide.) 2. Inconsistent: its solution is the empty set. (The graphs of the equations are parallel with no points in common.) 3. Dependent: its solution is an infinite set of number pairs. (The graphs of the equations are the same line.) 4. Independent: its solution set is exactly one ordered pair of numbers. The solution of an independent linear system in two variables may be found graphically by determining the coordinates of the point of intersection of the graphs or algebraically by using addition or substitution. Systems of linear equations can be used to solve verbal problems. The solution set of a system of inequalities can be shown on a graph as the intersection of the solution sets of the inequalities. VOCABULARY 10-4 System of simultaneous equations • Linear system • System of consistent equations • System of independent equations • System of inconsistent equations • System of dependent equations 10-5 Addition method 10-6 Substitution method • Substitution principle 10-8 System of inequalities REVIEW EXERCISES 1. Solve the following system of equations for x and y and check the solution: 5x 2y 22 x 2y 2 2. a. Graph the following system of inequalities and check one solution: Review Exercises 437 y 2x 3 y 2x b. Describe the solution set of the system in terms of the solution sets of the individual inequalities. In 3–8, write the equation of the line, in the form y mx b, that satisfies each of the given conditions. 3. Through (1, 2) with slope 3 1 4. Through (5, 6) with slope 1 2 5. Through (2, 5) and (2, 3) 6. Through (1, 3) and (5, 3) 2 3 7. With slope 2 and y-intercept 0 8. Graphed to the right y 1 1 O x In 9–14, write the equation of the line, in the form of the given conditions. y x b 5 1 a 1 , that satisfies each 9. With slope 1 and y-intercept 1 10. With slope 3.5 and y-intercept 1.5 11. With x-intercept 5 and y-intercept 2 3 5 12. With x-intercept and y-intercept 7 4 13. Through (1, 3) and (5, 6) 14. Graphed to the right y 1 O 1 x In 15–17, write the equation of the line, in the form Ax By C, that satisfies each of the given conditions. 15. Parallel to the y-axis and 3 units to the left of the y-axis 16. Through (0, 4) and (2, 0) 17. Through (1, 1) and (4, 5) In 18–20, solve each system of equations graphically and check. 19. y x 18. x y 6 y 2x 6 2x y 3 20. 2y x 4 x y 4 0 438 Writing and Solving Systems of Linear Functions In 21–23, solve each system of equations by using addition to eliminate one of the variables. Check. 21. 2x y 10 x y 3 23. 3c d 0 c 4d 52 22. x 4y 1 5x 6y 8 In 24–26, solve each system of equations by using substitution to eliminate one of the variables. Check. 24 x 2y 7 x y 8 25. 3r 2s 20 r 2s 26. x y 7 2x 3y 21 In 27–32, solve each system of equations by using an appropriate algebraic method. Check. 27. x y 0 28. 5a 3b 17 4a 5b 21 31. x y 1,000 0.06x 0.04y t 29. t u 12 1 3u 32. 10t u 24 t u 1 7(10u 1 t) 3x 2y 5 30. 3a 4b 4a 5b 2 33. a. Solve this system of equations algebraically: x y 3 x 3y 9 b. On a set of coordinate axes, graph the system of equations given in part a. In 34–36, graph each system of inequalities and label the solution set A. 34. y 2x 3 y 5 x 35. y 1
2x x 4 36. 2x y 4 x y 2 In 37–41, for each problem, write two equations in two variables and solve algebraically. 37. The sum of two numbers is 7. Their difference is 18. Find the numbers. 38. At a store, 3 notebooks and 2 pencils cost $2.80. At the same prices, 2 notebooks and 5 pencils cost $2.60. Find the cost of a notebook and of a pencil. 39. Two angles are complementary. The larger angle measures 15° less than twice the smaller angle. Find the degree measure of each angle. 40. The measure of the vertex angle of an isosceles triangle is 3 times the measure of each of the base angles. Find the measure of each angle of the triangle. 41. At the beginning of the year, Lourdes’s salary increased by 3%. If her weekly salary is now $254.41, what was her weekly salary last year? Cumulative Review 439 Exploration Jason and his brother Robby live 1.25 miles from school. Jason rides his bicycle to school and Robby walks. The graph below shows their relative positions on their way to school one morning. Use the information provided by the graph to describe Robby’s and Jason’s journeys to school. Who left first? Who arrived at school first? If each boy took the same route to school, when did Jason pass Robby? How fast did Robby walk? How fast did Jason ride his bicycle.25 1.0 0.75 0.50 0.25 0 ROBBY 8:00 8:05 8:10 8:15 8:20 8:25 8:30 TIME JASON CUMULATIVE REVIEW CHAPTERS 1–10 Part I Answer all questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. 1. The multiplicative inverse of 0.4 is (1) 0.4 (2) 1 (3) 2.5 2. Which of the following numbers has the greatest value? (1) 1.5 (2) 21.5 (3) 2 (4) 2.5 (4) 0.5p 2 " 3. Which of the following is an example of the associative property of addi- tion? (1) 6 (x 3) 6 (3 x) (2) 2(x 2) 2x 4 (3) 3(4a) (3 4)a (4) 3 (4 a) (3 4) a 4. Which of the following is a point on the line whose equation is x y 5? (1) (1, 4) (2) (1, 4) (3) (1, 4) (4) (1, 4) 5. The expression 5 2(3x2 8) is equivalent to (1) 9x2 24 (2) 6x2 21 (3) 9x2 8 (4) 6x2 11 6. The solution of the equation 0.2x 3 x 1 is (2) 5 (1) 5 (3) 0.5 (4) 50 440 Writing and Solving Systems of Linear Functions 7. The product of 3.40 103 and 8.50 102 equals (1) 2.89 100 (4) 2.89 101 8. The measure of A is 12° less than twice the measure of its complement. (3) 2.89 102 (2) 2.89 101 What is the measure of A? (1) 51° (3) 39° (2) 34° 9. The y-intercept of the graph of 2x 3y 6 is (2) 2 (3) 3 (1) 6 (4) 56° (4) 3 10. When 2a2 5a is subtracted from 5a2 1, the difference is (1) 3a2 5a 1 (2) 3a2 5a 1 (3) 3a2 6a (4) 3a2 5a 1 Part II Answer all questions in this part. Each correct answer will receive 2 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 11. A straight line has a slope of 3 and contains the point (0, 6). What are the coordinates of the point at which the line intersects the x-axis? 12. Last year the Edwards family spent $6,200 for food. This year, the cost of food for the family was $6,355. What was the percent of increase in the cost of food for the Edwards family? Part III Answer all questions in this part. Each correct answer will receive 3 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 13. Dana paid $3.30 for 2 muffins and a cup of coffee. At the same prices, Damion paid $5.35 for 3 muffins and 2 cups of coffee. What is the cost of a muffin and of a cup of coffee? 14. In trapezoid ABCD, . If AB 54.8 feet, BC ' AB and BC ' DC DC 37.2 feet, and BC 15.8 feet, find to the nearest degree the measure of A. Cumulative Review 441 Part IV Answer all questions in this part. Each correct answer will receive 4 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 15. Tamara wants to develop her own pictures but does not own photographic equipment. The Community Darkroom makes the use of photographic equipment available for a fee. Membership costs $25 a month and members pay $1.50 an hour for the use of the equipment. Non-members may also use the equipment for $4.00 an hour. a. Draw a graph that shows the cost of becoming a member and using the photographic equipment. Let the vertical axis represent cost in dollars and the horizontal axis represent hours of use. b. On the same graph, show the cost of using the photographic equipment for non-members. c. When Tamara became a member of the Community Darkroom in June, she found that her cost for that month would have been the same if she had not been a member. How many hours did Tamara use the facilities of the darkroom in June? 16. ABCD is a rectangle, E is a point on centimeters, and EC 10 centimeters. a. Find the area of nABE . DC , AD 24 centimeters, DE 32 b. Find the perimeter of nABE . CHAPTER 11 CHAPTER TABLE OF CONTENTS 11-1 Factors and Factoring 11-2 Common Monomial Factors 11-3 The Square of a Monomial 11-4 Multiplying the Sum and the Difference of Two Terms 11-5 Factoring the Difference of Two Perfect Squares 11-6 Multiplying Binomials 11-7 Factoring Trinomials 11-8 Factoring a Polynomial Completely Chapter Summary Vocabulary Review Exercises Cumulative Review 442 SPECIAL PRODUCTS AND FACTORS The owners of a fruit farm intend to extend their orchards by planting 100 new apple trees.The trees are to be planted in rows, with the same number of trees in each row. In order to decide how the trees are to be planted, the owners will use whole numbers that are factors of 100 to determine the possible arrangements: 20 rows of 5 trees each 1 row of 100 trees 25 rows of 4 trees each 2 rows of 50 trees each 50 rows of 2 trees each 4 rows of 25 trees each 100 rows of 1 tree each 5 rows of 20 trees each 10 rows of 10 trees each From this list of possibilities, the arrangement that best fits the dimensions of the land to be planted can be chosen. If the owner had intended to plant 90 trees, how many possible arrangements would there be? From earliest times, the study of factors and prime numbers has fascinated mathematicians, leading to the discovery of many important principles. In this chapter you will extend your knowledge of the factors of whole numbers, study the products of special binomials, and learn to write polynomials in factored form. 11-1 FACTORS AND FACTORING Factors and Factoring 443 When two numbers are multiplied, the result is called their product. The numbers that are multiplied are factors of the product. Since 3(5) 15, the numbers 3 and 5 are factors of 15. Factoring a number is the process of finding those numbers whose product is the given number. Usually, when we factor, we are finding the factors of an integer and we find only those factors that are integers. We call this factoring over the set of integers. Factors of a product can be found by using division. Over the set of integers, if the divisor and the quotient are both integers, then they are factors of the dividend. For example, 35 5 7. Thus, 35 5(7), and 5 and 7 are factors of 35. Every positive integer that is the product of two positive integers is also the product of the opposites of those integers. 21 (3)(7) 21 (3)(7) Every negative integer that is the product of a positive integer and a nega- tive integer is also the product of the opposites of those integers. 21 (3)(7) 21 (3)(7) Usually, when we factor a positive integer, we write only the positive inte- gral factors. Two factors of any number are 1 and the number itself. To find other integral factors, if they exist, we use division, as stated above. We let the number being factored be the dividend, and we divide this number in turn by the whole numbers 2, 3, 4, and so on. If the quotient is an integer, then both the divisor and the quotient are factors of the dividend. For example, use a calculator to find the integral factors of 126. We will use integers as divisors and look for quotients that are integers. Pairs of factors of 126 are listed to the right of the quotients. Quotients 126 1 126 126 2 63 126 3 42 126 4 31.5 126 5 25.2 126 6 21 Pairs of Factors of 126 1 126 2 63 3 42 — — 6 21 Quotients 126 7 18 126 8 15.75 126 9 14 126 10 12.6 126 4 11 5 11.45 126 12 10.5 Pairs of Factors of 126 7 18 — 9 14 — — — When the quotient is smaller than the divisor (here, 10.5 12), we have found all possible positive integral factors. The factors of 126 are 1, 2, 3, 6, 7, 9, 14, 18, 21, 42, 63, and 126. Recall that a prime number is an integer greater than 1 that has no positive integral factors other than itself and 1. The first seven prime numbers are 2, 3, 5, 7, 11, 13, and 17. Integers greater than 1 that are not prime are called composite numbers. 444 Special Products and Factors In general, a positive integer greater than 1 is a prime or can be expressed as the product of prime factors. Although the factors may be written in any order, there is one and only one combination of prime factors whose product is a given composite number. As shown below, a prime factor may occur in the product more than once. 21 3 7 20 2 2 5 or 22 5 To express a positive integer, for example 280, as the product of primes, we start with any pair of positive integers, say 28 and 10, whose product is the given number. Then, we factor these factors and continue to factor the factors until all are primes. Finally, we rearrange these factors in numerical order, as shown at the right. Expressing each of two integers as the product of prime factors makes it possible to discover the greatest integer that is a factor of both of them. We call this factor the greatest common factor (GCF) of these integers. 280 280 28 10 280 2 14 2 5
280 2 2 7 2 5 or 280 23 5 7 Let us find the greatest common factor of 180 and 54. 180 2 2 3 3 5 ↓ ↓ ↓ 54 2 3 3 3 ↓ ↓ 3 3 ↓ Greatest common factor 2 or or or 22 32 5 2 33 2 32 or 18 Only the prime numbers 2 and 3 are factors of both 180 and 54. We see that the greatest number of times that 2 appears as a factor of both 180 and 54 is once; the greatest number of times that 3 appears as a factor of both 180 and 54 is twice. Therefore, the greatest common factor of 180 and 54 is 2 3 3, or 2 32, or 18. To find the greatest common factor of two or more monomials, find the product of the numerical and variable factors that are common to the monomials. For example, let us find the greatest common factor of 24a3b2 and 18a2b. 24a3b2 2 2 2 3 ↓ 18a2b 6a2b Greatest common factor 2 The greatest common factor of 24a3b2 and 18a2b is 6a2b. Factors and Factoring 445 When we are expressing an algebraic factor, such as 6a2b, we will agree that: • Numerical coefficients need not be factored. (6 need not be written as 2 3.) • Powers of variables need not be represented as the product of several equal factors. (a2b need not be written as a a b.) EXAMPLE 1 Solution Write all positive integral factors of 72. Quotients 72 1 72 72 2 36 72 3 24 72 4 18 Pairs of Factors of 72 1 72 2 36 3 24 4 18 Quotients 72 5 14.4 72 6 12 72 4 7 5 10.285714 72 8 9 Pairs of Factors of 72 –– 6 12 –– 8 9 It is not necessary to divide by 9, since 9 has just been listed as a factor. Answer The factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, and 72. EXAMPLE 2 Solution Express 700 as a product of prime factors. 700 2 350 700 2 2 175 700 2 2 5 35 700 2 2 5 5 7 or 22 52 7 Answer 22 52 7 EXAMPLE 3 Find the greatest common factor of the monomials 60r2s4 and 36rs2t. Solution To find the greatest common factor of two monomials, write each as the product of primes and variables to the first power and choose all factors that occur in each product. 60r2s4 36rs2t 2 2 3 3 r s s t Greatest common factor 2 2 3 r s s Answer 12rs2 446 Special Products and Factors EXERCISES Writing About Mathematics 1. Ross said that some pairs of number such as 5 and 12 have no greatest common factor. Do you agree with Ross? Explain why or why not. 2. To find the factors of 200, Chaz tried all of the integers from 2 to 14. Should Chaz have used more numbers? Explain why or why not. Developing Skills In 3–12, tell whether each integer is prime, composite, or neither. 3. 5 8. 36 4. 8 9. 41 5. 13 10. 49 6. 18 11. 57 In 13–22, express each integer as a product of prime numbers. 13. 35 18. 400 14. 18 19. 202 15. 144 20. 129 16. 77 21. 590 7. 73 12. 1 17. 128 22. 316 In 23–34, write all the positive integral factors of each given number. 23. 26 29. 37 24. 50 30. 62 25. 36 31. 253 26. 88 32. 102 27. 100 33. 70 28. 242 34. 169 35. The product of two monomials is 36x3y4. Find the second factor if the first factor is: a. 3x2y3 b. 6x3y2 c. 12xy4 d. 9x3y e. 18x3y2 36. The product of two monomials is 81c9d12e. Find the second factor if the first factor is: a. 27e b. c2d11 c. 3cd3e d. 9c4d4 e. 81c5d9e In 37–42, find, in each case, the greatest common factor of the given integers. 37. 10; 15 40. 18; 24; 36 38. 12; 28 41. 75; 50 39. 14; 35 42. 72; 108 In 43–51, find, in each case, the greatest common factor of the given monomials. 43. 4x; 4y 46. 10x2; 15xy2 49. 14a2b; 13ab 44. 4r; 6r2 47. 36xy2z; –27xy2z2 50. 36xyz; 25xyz 45. 8xy; 6xz 48. 24ab2c3; 18ac2 51. 2ab2c; 3x2yz 11-2 COMMON MONOMIAL FACTORS Common Monomial Factors 447 To factor a polynomial over the set of integers means to express the given polynomial as the product of polynomials whose coefficients are integers. For example, since 2(x y) , the polynomial 2x 2y can be written in factored form as 2(x y). The monomial 2 is a factor of each term of the polynomial 2x 2y. Therefore, 2 is called a common monomial factor of the polynomial 2x 2y. 2x 1 2y To factor a polynomial, we look first for the greatest common monomial factor, that is, the greatest monomial that is a factor of each term of the polynomial. For example: 1. Factor 4rs 8st. There are many common factors of 4rs and 8st such as 2, 4, 2s, and 4s. The greatest common monomial factor is 4s. We divide 4rs 8st by 4s to obtain the quotient r 2t, which is the second factor. Therefore, the polynomial 4rs 8st 4s(r 2t). 2. Factor 3x 4y. We notice that 1 is the only common factor of 3x and 4y, so the second factor is 3x 4y. We say that 3x 4y is a prime polynomial. A polynomial with integers as coefficients is a prime polynomial if its only factors are 1 and the polynomial itself. Procedure To factor a polynomial whose terms have a common monomial factor: 1. Find the greatest monomial that is a factor of each term of the polynomial. 2. Divide the polynomial by the factor found in step 1.The quotient is the other factor. 3. Express the polynomial as the product of the two factors. We can check by multiplying the factors to obtain the original polynomial. EXAMPLE 1 Write in factored form: 6c3d 12c2d2 3cd. Solution (1) 3cd is the greatest common factor of 6c3d, –12c2d2, and 3cd. (2) To find the other factor, divide 6c3d 12c2d2 3cd by 3cd. 3cd 1 3cd 3cd (6c3d – 12c2d2 3cd) 3cd 3cd 2 12c2d2 6c3d 2c2 4cd 1 Answer 3cd(2c2 4cd 1) 448 Special Products and Factors EXERCISES Writing About Mathematics 1. Ramón said that the factored form of 3a 6a2 15a3 is 3a(2a 5a2). Do you agree with Ramón? Explain why or why not. 2. a. The binomial 12a2 20ab can be written as 4a(3a 5b). Does that mean that for all positive integral values of a and b, the value of 12a2 20ab has at least two factors other than itself and 1? Justify your answer. b. The binomial 3a 5b is a prime polynomial. Does that mean that for all positive inte- gral values of a and b, 3a 5b is a prime number? Justify your answer. 3. 2a 2b Developing Skills In 3–29, write each expression in factored form. 4. 3x 3y 7. 4x 8y 10. 18c 27d 13. 6 18c 16. ax 5ab 19. 2x 4x3 22. pr2 2prh 25. 3ab2 6a2b 28. c3 c2 2c 6. xc – xd 9. 12x 18y 12. 7y 7 15. 2x2 5x 18. 10x 15x3 21. pr2 prl 24. 12y2 4y 27. 3x2 6x 30 5. bx by 8. 3m 6n 11. 8x 16 14. y2 3y 17. 3y4 3y2 20. p prt 23. 3a2 9 26. 21r3s2 14r2s 29. 9ab2 6ab 3a Applying Skills 30. The perimeter of a rectangle is represented by 2l 2w. Express the perimeter as the prod- uct of two factors. 31. The lengths of the parallel sides of a trapezoid are represented by a and b and its height by . Express this area as the product of 2ah 1 1 1 2bh h. The area of the trapezoid can be written as two factors. 32. A cylinder is cut from a cube whose edge measures 2s. a. Express the volume of the cube in terms of s. b. If the largest possible cylinder is made, express the volume of the cylinder in terms of s. c. Use the answers to parts a and b to express, in terms of s, the volume of the material cut away to form the cylinder. d. Express the volume of the waste material as the product of two factors. 11-3 THE SQUARE OF A MONOMIAL The Square of a Monomial 449 To square a monomial means to multiply the monomial by itself. For example: (3x)2 (3x)(3x) (3)(3)(x)(x) (3)2(x)2 or 9x2 (5y2)2 (5y2)(5y2) (5)(5)(y2)(y2) (5)2(y2)2 or 25y4 (6b4)2 (6b4)(6b4) (6)(6)(b4)(b4) (6)2( b4)2 or 36b8 (4c2d3)2 (4c2d3)(4c2d3) (4)(4)(c2)(c2)(d3)(d3) (4)2(c2)2(d3)2 or 16c4d6 When a monomial is a perfect square, its numerical coefficient is a perfect square and the exponent of each variable is an even number. This statement holds true for each of the results shown above. Procedure To square a monomial: 1. Find the square of the numerical coefficient. 2. Find the square of each literal factor by multiplying its exponent by 2. 3. The square of the monomial is the product of the expressions found in steps 1 and 2. EXAMPLE 1 Square each monomial mentally. a. (4s3)2 2 2 5 ab b. c. (7xy2)2 B A Think (4)2(s3)2 2(a)2(b)2 2 5 A (7)2(x)2(y2)2 B d. (0.3y2)2 (0.3)2(y2)2 EXERCISES Writing About Mathematics Write 16s6 4 25a2b2 49x2y4 0.09y4 1. Explain why the exponent of each variable in the square of a monomial is an even number. 2. When the square of a number is written in scientific notation, is the exponent of 10 always an even number? Explain why or why not and give examples to justify your answer. 450 Special Products and Factors Developing Skills In 3–22, square each monomial. 3. (a2)2 7. (m2n2)2 11. (9ab)2 15. 2 5 7 xy B A 19. (0.8x)2 4. (b3)2 8. (x3y2)2 12. (10x2y2)2 27 8 a2b2 A 20. (0.5y2)2 16. B 2 5. (d5)2 9. (3x2)2 13. (12cd3)2 2 17. x 6 B 21. (0.01xy)2 A 6. (rs)2 10. (5y4)2 2 A 14. 3 4 a B 24x2 18. 5 22. (0.06a2b)2 A B 2 Applying Skills In 23–28, each monomial represents the length of a side of a square. Write the monomial that represents the area of the square. 23. 4x 24. 10y 25. 2 3x 26. 1.5x 27. 3x2 28. 4x2y3 11-4 MULTIPLYING THE SUM AND THE DIFFERENCE OF TWO TERMS Recall that when two binomials are multiplied, the product contains four terms. This fact is illustrated in the diagram below and is also shown by using the distributive property. a ac ad + b bc bd (a b)(c d) a(c d) b(c d) ac ad bc bd If two of the four terms of the product are similar, the similar terms can be combined so that the product is a trinomial. For example: c + d (x 2)(x 3) x(x 3) 2(x 3) x2 3x 2x 6 x2 5x 6 There is, however, a special case in which the sum of two terms is multiplied by the difference of the same two terms. In this case, the sum of the two middle terms of the product is 0, and the product is a binomial, as in the following examples: (a 4)(a 4) a(a 4) 4(a 4) a2 4a 4a 16 a2 16 Multiplying the Sum and the Difference of Two Terms 451 (3x2 5y)(3x2 5y) 3x2(3x2 5y) 5y(3x2 5y) 9x4 15x2y 15x2y 25y2 9x4 25y2 These two examples illustrate the following procedure, which enables us to find the product of the sum and difference of the same two terms mentally. Procedure To multiply the sum of two terms by the difference of the same two terms: 1. Square the first term. 2. From this result, subtract the square of the second term. (a b)(a b) a2 b2 EXAMPLE 1 Find each product. a. (y 7)(y 7) b
. (3a 4b)(3a 4b) Think (y)2 (7)2 (3a)2 (4b)2 Write y2 49 9a2 16b2 EXERCISES Writing About Mathematics 1. Rose said that the product of two binomials is a binomial only when the two binomials are the sum and the difference of the same two terms. Miranda said that that cannot be true because (5a 10)(a 2) 5a2 20, a binomial. Show that Rose is correct by writing the factors in the example that Miranda gave in another way. 2. Ali wrote the product (x 2)2(x 2)2 as (x2 4)2. Do you agree with Ali? Explain why or why not. Developing Skills In 3–17, find each product. 3. (x 8)(x 8) 6. (12 a)(12 a) 9. (8x 3y)(8x 3y) 4. (y 10)(y 10) 7. (c d)(c d) 10. (x2 + 8)(x2 8) 5. (n 9)(n 9) 8. (3x 1)(3x 1) 11. (3 5y3)(3 5y3) 452 Special Products and Factors a 1 1 2 12. 15. (a 5)(a 5)(a2 25) B A A B a 2 1 2 13. (r 0.5)(r 0.5) 16. (x 3)(x 3)(x2 9) 14. (0.3 m)(0.3 m) 17. (a b)(a b)(a2 b2) Applying Skills In 18–21, express the area of each rectangle whose length l and width w are given. 18. l x 7, w x 7 19. l 2x 3, w 2x 3 20. l c d, w c d 21. l 2a 3b, w 2a 3b In 22–25, find each product mentally by thinking of the factors as the sum and difference of the same two numbers. 22. (27)(33) (30 3)(30 3) 23. (52)(48) (50 2)(50 2) 24. (65)(75) 25. (19)(21) 11-5 FACTORING THE DIFFERENCE OF TWO PERFECT SQUARES An expression of the form a2 b2 is called a difference of two perfect squares. Factoring the difference of two perfect squares is the reverse of multiplying the sum of two terms by the difference of the same two terms. Since the product (a b)(a b) is a2 b2, the factors of a2 b2 are (a b) and (a b). Therefore: a2 b2 (a b)(a b) Procedure To factor a binomial that is a difference of two perfect squares: 1. Express each of its terms as the square of a monomial. 2. Apply the rule a2 – b2 (a b)(a – b). Remember that a monomial is a square if and only if its numerical coeffi- cient is a square and the exponent of each of its variables is an even number. EXAMPLE 1 Factor each polynomial. a. r2 1 25x2 2 1 49y2 b. c. 0.04 c6d4 Think (r)2 (1)2 Write (r 1)(r 1) 2 1 7y (5x)2 2 A (0.2)2 (c3d 2)2 B 5x 1 1 7y A (0.2 c3d 2)(0.2 c3d 2) 5x 2 1 7y B A B Factoring the Difference of Two Perfect Squares 453 EXAMPLE 2 Express x2 – 100 as the product of two binomials. Solution Since x2 100 is a difference of two squares, the square of x and the square of 10, the factors of x2 100 are (x 10) and (x 10). Answer (x 10)(x 10) EXERCISES Writing About Mathematics 1. If 391 can be written as 400 9, find two factors of 391 without using a calculator or pencil and paper. Explain how you found your answer. 2. Does 5a2 45 have two binomial factors? Explain your answer. Developing Skills In 3–18, factor each binomial. 3. a2 4 7. 16a2 b2 11. 25 s4 15. 0.04 49r2 4. c2 100 8. 25m2 n2 12. 100x2 81y2 16. 0.16y2 9 5. 9 x2 9. d 2 4c2 w2 2 1 13. 64 17. 0.81 y2 6. 144 c2 10. r4 9 14. x2 0.64 18. 81m4 49 Applying Skills In 19–23, each given polynomial represents the area of a rectangle. Express the area as the product of two binomials. 19. x2 4 23. 4x2 y2 22. t4 64 21. t2 49 20. y2 9 In 24–26, express the area of each shaded region as: a. the difference of the areas shown, and b. the product of two binomials. 24. c d c d d c 25. 2x 26. x d c 2x x y y y y y y 2x 2x 454 Special Products and Factors In 27 and 28, express the area of each shaded region as the product of two binomials. 27. 5a 28. 2b 2b 2b 2b 5a 3x 2y 2y 2y 2y 2y 2y 2y 2y 3x 11-6 MULTIPLYING BINOMIALS We have used the “FOIL” method from Section 5-4 to multiply two binomials of the form ax b and cx d. The pattern of the multiplication in the following example shows us how to find the product of two binomials mentally. ➤ (2x 3)(4x 5) 2x(4x) 2x(5) 3(4x) 3(5) ➤ ➤ ➤ 8x2 10x 12x 15 8x2 2x 15 Examine the trinomial 8x2 2x 15 and note the following: 1. The first term of the trinomial is the product of the first terms of the binomials: 2. The middle term of the trinomial is the product of the outer terms plus the product of the inner terms of the binomials: 3. The last term of the trinomial is the product of the last terms of the binomials: 8x2 \|________\| (2x 3)(4x 5) → 2x(4x) 8x2 12x Think \|____\| (2x 3)(4x 5) → (12x) (10x) 2x \|______________\| 10x (2x 3)(4x 5) → (3)(5) 15 \|_________\| 15 Multiplying Binomials 455 Procedure To find the product of two binomials (ax b) and (cx d) where a, b, c, and d have numerical values: 1. Multiply the first terms of the binomials. 2. Multiply the first term of each binomial by the last term of the other bino- mial (the outer and inner terms), and add these products. 3. Multiply the last terms of the binomials. 4. Write the results of steps 1, 2, and 3 as a sum. Write the product (x 5)(x 7) as a trinomial. ➤ (x 5)(x 7) ➤ ➤ ➤ 1. (x)(x) x2 2. (x)(7) (5)(x) (7x) (5x) –12x 3. (5)(7) 35 4. x2 12x 35 Answer Write the product (3y 8)(4y 3) as a trinomial. ➤ ➤ (3y 8)(4y 3) ➤ ➤ 1. (3y)(4y) 12y2 2. (3y)(3) (8)(4y) 9y (32y) –23y 3. (8)(3) –24 4. 12y2 23y 24 Answer EXAMPLE 1 Solution EXAMPLE 2 Solution EXAMPLE 3 Write the product (z 4)2 as a trinomial. Solution (z 4)2 (z 4)(z 4) z2 4z 4z 16 z2 8z 16 Answer A trinomial, such as z2 8z 16, that is the square of a binomial is called a perfect square trinomial. Note that the first and last terms are the squares of the terms of the binomial, and the middle term is twice the product of the terms of the binomial. 456 Special Products and Factors EXERCISES Writing About Mathematics 1. What is a shorter procedure for finding the product of binomials of the form (ax b) and (ax b)? 2. Which part of the procedure for multiplying two binomials of the form (ax b) and (cx d) does not apply to the product of binomials such as (3x 5) and (2y 7)? Explain your answer. Developing Skills In 3–29, perform each indicated operation mentally. 4. (6 d)(3 d) 7. (x 7)(x 2) 10. (3x 2)(x 5) 13. (a 4)2 16. (7x 3)(2x 1) 19. (2x 5)2 22. (2t 3)(5t 1) 25. (2c 3d)(5c 2d) 28. (6t 1)(4t z) 3. (x 5)(x 3) 6. (8 c)(3 c) 9. (5 t)(9 t) 12. (y 8)2 15. (3x 2)2 18. (3x 4)2 21. (5y 4)(5y 4) 24. (5x 7y)(3x 4y) 27. (a b)(2a 3) Applying Skills 5. (x 10)(x 5) 8. (n 20)(n 3) 11. (c 5)(3c 1) 14. (2x 1)2 17. (2y 3)(3y 2) 20. (3t 2)(4t 7) 23. (2a b)(2a 3b) 26. (a b)2 29. (9y w)(9y 3w) 30. Write the trinomial that represents the area of a rectangle whose length and width are: a. (x 5) and (x 4) b. (2x 3) and (x 1) 31. Write the perfect square trinomial that represents the area of a square whose sides are: a. (x 6) b. (x 2) c. (2x 1) d. (3x 2) 32. The length of a garden is 3 feet longer than twice the width. a. If the width of the garden is represented by x, represent the length of the garden in terms of x. b. If the width of the garden is increased by 5 feet, represent the new width in terms of x. c. If the length of the enlarged garden is also increased so that it is 3 feet longer than twice the new width, represent the length of the enlarged garden in terms of x. d. Express the area of the enlarged garden in terms of x. 11-7 FACTORING TRINOMIALS Factoring Trinomials 457 (x 1 3) We have learned that (x 3)(x 5) x2 8x 15. Therefore, factors of x2 8x 15 are and (x + 5). Factoring a trinomial of the form ax2 + bx c is the reverse of multiplying binomials of the form (dx e) and (fx g). When we factor a trinomial of this form, we use combinations of factors of the first and last terms. We list the possible pairs of factors, then test the pairs, one by one, until we find the pair that gives the correct middle term. For example, let us factor x2 7x 10: 1. The product of the first terms of the binomials must be x2. Therefore, for each first term, we use x. We write: x2 7x 10 (x )(x ) 2. Since the product of the last terms of the binomials must be +10, these last terms must be either both positive or both negative. The pairs of integers whose products is +10 are: (1)(10) (5)(2) (1)(10) (2)(5) 3. From the products obtained in steps 1 and 2, we see that the possible pairs of factors are: (x 10)(x 1) (x 10)(x 1) (x 5)(x 2) (x 5)(x 2) 4. Now, we test each pair of factors: (x 10)(x 1) is not correct because the middle term is 11x, not 7x. (x 5)(x 2) is correct because the middle term is 7x. 10x \|____\| (x 10)(x 1) → 10x 1x 11x \|_____________\| ✘ 1x 5x \|___\| (x 5)(x 2) → 2x 5x 7x \|____________\| ✔ 2x Neither of the remaining pairs of factors is correct because each would result in a negative middle term. 5. The factors of x2 7x 10 are (x 5) and (x 2). Observe that in this trinomial, the first term, x2, is positive and must have factors with the same signs. We usually choose positive factors of the first terms. The last term, +10, is positive and must have factors with the same signs. Since the middle term of the trinomial is also positive, the last terms of both binomial factors must be positive (5 and 2). In this example, if we had chosen x times x as the factors of x2, the factors of x2 7x 10 would have been written as (x 5) and (x 2). Every trinomial that has two binomial factors also has the opposites of these binomials as factors. Usually, however, we write only the pair of factors whose first terms have positive coefficients as the factors of a trinomial. 458 Special Products and Factors Procedure To factor a trinomial of the form ax2 bx c, find two binomials that have the following characteristics: 1. The product of the first terms of the binomials is equal to the first term of the trinomial (ax2). 2. The product of the last terms of the binomials is equal to the last term of the trinomial (c). 3. When the first term of each binomial is multiplied by the last term of the other binomial and the sum of these products is found, the result is equal to the middle term of the trinomial. EXAMPLE 1 Factor y2 8y 12. Solution (1) The product of the first terms of the binomials must be y2. Therefore, for each first term, we use y. We write: y2 8y 12 (y )(y ) (2) Since the product of the last terms of the binomials must be +12, these last terms must be either both positive or both negative. The pairs of integers whose product is +12 are: (1)(12
) (1)(12) (2)(6) (2)(6) (3)(4) (3)(4) (3) The possible factors are: (y l)(y 12) (y l)(y 12) (y 2)(y 6) (y 2)(y 6) (y 3)(y 4) (y 3)(y 4) (4) When we find the middle term in each of the trinomial products, we see that only for the factors (y 6)(y 2) is the middle term 8y: 6y \|____\| (y 6)(y 2) → 2y 6y 8y \|____________\| 2y Answer y2 8y 12 (y 6)(y 2) When the first and last terms are both positive (y2 and 12 in this example) and the middle term of the trinomial is negative, the last terms of both binomial factors must be negative (6 and –2 in this example). Factoring Trinomials 459 EXAMPLE 2 Factor: a. c2 5c 6 b. c2 5c 6 Solution a. (1) The product of the first terms of the binomials must be c2. Therefore, for each first term, we use c. We write: c2 5c 6 (c )(c ) (2) Since the product of the last terms of the binomials must be –6, one of these last terms must be positive and the other negative. The pairs of integers whose product is –6 are: (1)(6) (1)(6) (3)(2) (3)(2) (3) The possible factors are: (c 1)(c 6) (c 1)(c 6) (c 3)(c 2) (c 3)(c 2) (4) When we find the middle term in each of the trinomial products, we see that only for the factors (c 1)(c 6) is the middle term 5c: 1c \|____\| (c 1)(c 6) → 6c 1c 5c \|____________\| 6c b. (1) As in the solution to part a, we write: c2 5c 6 (c )(c ) (2) Since the product of the last terms of the binomials must be 6, these last terms must be either both positive or both negative. The pairs of integers whose product is 6 are: (1)(6) (1)(6) (3)(2) (3)(2) (3) The possible factors are: (c 1)(c 6) (c 1)(c 6) (c 3)(c 2) (c 3)(c 2) (4) When we find the middle term in each of the trinomial products, we see that only for the factors (c 3)(c 2) is the middle term 5c. 3c \|____\| (c 3)(c 2) → 2c 3c 5c \|____________\| 2c Answer a. c2 5c 6 (c 1)(c 6) b. c2 5c 6 (c 3)(c 2) 460 Special Products and Factors EXAMPLE 3 Factor 2x2 7x 15. Solution (1) Since the product of the first terms of the binomials must be 2x2, we use 2x and x as the first terms. We write: 2x2 7x 15 (2x )(x ) (2) Since the product of the last terms of the binomials must be 15, one of these last terms must be positive and the other negative. The pairs of integers whose product is –15 are: (1)(15) (1)(15) (3)(5) (3)(5) (3) These four pairs of integers will form eight pairs of binomial factors since there are two ways in which the first terms can be arranged. Note how (2x 1)(x 15) is not the same product as (2x 15)(x 1). The possible pairs of factors are: (2x 1)(x 15) (2x 15)(x 1) (2x 1)(x 15) (2x 15)(x 1) (2x 3)(x 5) (2x 5)(x 3) (2x 3)(x 5) (2x 5)(x 3) (4) When we find the middle term in each of the trinomial products, we see that only for the factors (2x 3)(x 5) is the middle term 7x: 3x \|____\| (2x 3)(x 5) → 3x 10x 7x \|_____________\| 10x Answer 2x2 7x 15 (2x 3)(x 5) In factoring a trinomial of the form ax2 bx c, when a is a positive inte- ger (a 0): 1. The coefficients of the first terms of the binomial factors are usually writ- ten as positive integers. 2. If the last term, c, is positive, the last terms of the binomial factors must be either both positive (when the middle term, b, is positive), or both negative (when the middle term, b, is negative). 3. If the last term, c, is negative, one of the last terms of the binomial factors must be positive and the other negative. Factoring a Polynomial Completely 461 EXERCISES Writing About Mathematics 1. Alicia said that the factors of x2 bx c are (x d)(x e) if c de and b d e. Do you agree with Alicia? Explain why or why not. 2. Explain why there is no positive integral value of c for which x2 x c has two binomial factors. Developing Skills In 3–33, factor each trinomial. 3. a2 3a 2 7. y2 6y 8 11. y2 2y 8 15. x2 11x 24 19. x2 10x 24 23. c2 2c 35 27. 3x2 10x 8 31. 10a2 9a 2 4. c2 6c 5 8. y2 6y 8 12. y2 2y 8 16. a2 11a 18 20. x2 x 2 24. x2 7x 18 28. 16x2 8x 1 32. 3a2 7ab 2b2 5. x2 8x 7 9. y2 9y 8 13. y2 7y – 8 17. z2 10z 25 21. x2 6x 7 25. z2 9z 36 29. 2x2 x 3 33. 4x2 5xy 6y2 6. x2 11x 10 10. y2 9y 8 14. y2 7y 8 18. x2 5x 6 22. y2 4y 5 26. 2x2 5x 2 30. 4x2 12x 5 Applying Skills In 34–36, each trinomial represents the area of a rectangle. In each case, find two binomials that could be expressions for the dimensions of the rectangle. 34. x2 10x 9 35. x2 9x 20 36. 3x2 14x 15 In 37–39, each trinomial represents the area of a square. In each case, find a binomial that could be an expression for the measure of each side of the square. 37. x2 10x 25 38. 81x2 18x 1 39. 4x2 + 12x 9 11-8 FACTORING A POLYNOMIAL COMPLETELY Some polynomials, such as x2 + 4 and x2 + x 1, cannot be factored into other polynomials with integral coefficients. We say that these polynomials are prime over the set of integers. Factoring a polynomial completely means finding the prime factors of the polynomial over a designated set of numbers. In this book, we will consider a polynomial factored when it is written as a product of monomials or prime polynomials over the set of integers. 462 Special Products and Factors Procedure To factor a polynomial completely: 1. Look for the greatest common factor. If it is greater than 1, factor the given polynomial into a monomial times a polynomial. 2. Examine the polynomial factor or the given polynomial if it has no common factor (the greatest common factor is 1).Then: • Factor any trinomial into the product of two binomials if possible. • Factor any binomials that are the difference of two perfect squares as such. 3. Write the given polynomial as the product of all of the factors. Make certain that all factors except the monomial factor are prime polynomials. EXAMPLE 1 Factor by2 4b. Solution How to Proceed (1) Find the greatest common factor of the terms: by2 4b b(y2 4) (2) Factor the difference of by2 4b b(y 2)(y 2) Answer two squares: EXAMPLE 2 Factor 3x2 – 6x 24. Solution How to Proceed (1) Find the greatest common 3x2 – 6x 24 3(x2 – 2x 8) factor of the terms: (2) Factor the trinomial: 3x2 – 6x 24 3(x 4)(x 2) Answer EXAMPLE 3 Factor 4d2 6d 2. Solution How to Proceed (1) Find the greatest common 4d2 6d 2 2(2d2 3d 1) factor of the terms: (2) Factor the trinomial: 4d2 6d 2 2(2d 1)(d 1) Answer Factoring a Polynomial Completely 463 EXAMPLE 4 Factor x4 16. Solution How to Proceed (1) Find the greatest common The greatest common factor is 1. factor of the terms: (2) Factor the binomial as the difference of two squares: x4 16 (x2 4)(x2 4) (3) Factor the difference of x4 16 (x2 4)(x 2)(x 2) Answer two squares: EXERCISES Writing About Mathematics 1. Greta said that since 4a2 a2b2 is the difference of two squares, the factors are . Has Greta factored 4a2 a2b2 into prime polynomial factors? Explain (2a 1 ab)(2a 2 ab) why or why not. 2. Raul said that the factors of x3 1 are (x 1)(x 1)(x 1). Do you agree with Raul? Explain why or why not. Developing Skills In 3–29, factor each polynomial completely. 4. 4x2 4 7. 2x2 32 10. 63c2 7 13. 4a2 36 16. pc2 pd2 19. x3 7x2 10x 22. 2ax2 2ax 12a 25. y4 13y2 36 28. 16x2 16x 4 3. 2a2 2b2 6. st2 9s 9. 18m2 8 12. z3 z 15. y4 81 18. 4r2 4r 48 21. d3 8d2 16d 24. a4 10a2 9 27. 2a2b 7ab 3b 5. ax2 ay2 8. 3x2 27y2 11. x3 4x 14. x4 1 17. 3x2 6x 3 20. 4x2 6x 4 23. 16x2 x2y4 26. 5x4 10x2 5 29. 25x2 100xy 100y2 464 Special Products and Factors Applying Skills 30. The volume of a rectangular solid is represented by 12a3 5a2b 2ab2. Find the algebraic expressions that could represent the dimensions of the solid. 31. A rectangular solid has a square base. The volume of the solid is represented by 3m2 1 12m 1 12 . a. Find the algebraic expression that could represent the height of the solid. b. Find the algebraic expression that could represent the length of each side of the base. 32. A rectangular solid has a square base. The volume of the solid is represented by 10a3 1 20a2 1 10a . a. Find the algebraic expression that could represent the height of the solid. b. Find the algebraic expression that could represent the length of each side of the base CHAPTER SUMMARY Factoring a number or a polynomial is the process of finding those numbers or polynomials whose product is the given number or polynomial. A perfect square trinomial is the square of a binomial, whereas the difference of two perfect squares is the product of binomials that are the sum and difference of the same two. A prime polynomial, like a prime number, has only two factors, 1 and itself. To factor a polynomial completely: 1. Factor out the greatest common monomial factor if it is greater than 1. 2. Write any factor of the form a2 – b2 as (a b)(a – b). 3. Write any factor of the form ax2 + bx c as the product of two binomial factors if possible. 4. Write the given polynomial as the product of these factors. VOCABULARY 11-1 Product • Factoring a number • Factoring over the set of integers • Greatest common factor 11-2 Factoring a polynomial • Common monomial factor • Greatest common monomial factor • Prime polynomial 11-3 Square of a monomial 11-5 Difference of two perfect squares 11-6 Perfect square trinomial 11-8 Prime over the set of integers • Factoring a polynomial completely REVIEW EXERCISES Review Exercises 465 1. Leroy said that 4x2 16x 12 (2x 2)(2x 6) 2(x 1)(x 3). Do you agree with Leroy? Explain why or why not. 2. Express 250 as a product of prime numbers. 3. What is the greatest common factor of 8ax and 4ay? 4. What is the greatest common factor of 16a3bc2 and 24a2bc4? In 5–8, square each monomial. 5. (3g3)2 6. (4x4)2 7. (0.2c2y)2 In 9–14, find each product. 8. A 1 2 a3b5 2 B 9. (x 5)(x 9) 12. (3d 1)(d 2) 10. (y 8)(y 6) 13. (2w 1)2 11. (ab 4)(ab 4) 14. (2x 3c)(x 4c) In 15–29, in each case factor completely. 15. 6x 27b 18. x2 16h2 21. 64b2 9 24. a2 7a 30 27. 2x2 12bx 32b2 30. Express the product (k 15)(k 15) as a binomial. 31. Express 4ez2(4e z) as a binomial. 32. Factor completely: 60a2 37a 6. 16. 3y2 + 10y 19. x2 4x 5 22. 121 k2 25. x2 16x 60 28. x4 1 17. m2 81 20. y2 9y 14 23. x2 8x 16 26. 16y2 16 29. 3x3 6x2 24x 33. Which of the following polynomials has a factor that is a perfect square trinomia
l and a factor that is a perfect square? (1) a2y 10ay 25y (2) 2ax2 2ax 12a (3) 18m2 24m 8 (4) c2z2 18cz2 81z2 34. Of the four polynomials given below, explain how each is different from the others. x2 9 x3 5x2 6x 35. If the length and width of a rectangle are represented by 2x 3 and 3x 2, respectively, express the area of the rectangle as a trinomial. x2 2x 1 x2 2x 1 36. Find the trinomial that represents the area of a square if the measure of a side is 8m 1 1 . 466 Special Products and Factors 37. If 9x2 30x 25 represents the area of a square, find the binomial that represents the length of a side of the square. 38. A group of people wants to form several committees, each of which will have the same number of persons. Everyone in the group is to serve on one and only one committee. When the group tries to make 2, 3, 4, 5, or 6 committees, there is always one extra person. However, they are able to make more than 6 but fewer than 12 committees of equal size. a. What is the smallest possible number of persons in the group? b. Using the group size found in a, how many persons are on each com- mittee that is formed? Exploration a. Explain how the expression (a b)(a b) a2 b2 can be used to find the product of two consecutive even or two consecutive odd integers. b. The following diagrams illustrate the formula: 1 3 5 . . . (2n 3) (2n 1) n2 1 + 3 = 4 = 22 1 + 3 + 5 = 9 = 32 1 + 3 + 5 + 7 = 16 = 42 (1) Explain how 3, 5, and 7 are the difference of two squares using the diagrams. (2) Use the formula to explain how any odd number, 2n 1, can be written as the difference of squares. [Hint: 1 3 5 . . . (2n 3) (n 1)2] CUMULATIVE REVIEW CHAPTERS 1–11 Part I Answer all questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. 1. If apples cost $3.75 for 3 pounds, what is the cost, at the same rate, of 7 pounds of apples? (1) $7.25 (2) $8.25 (3) $8.50 (4) $8.75 Cumulative Review 467 (1) a2 4a 4 2. When 3a2 4 is subtracted from 2a2 4a, the difference is (2) a2 4a 4 (3) a2 8a 3. The solution of the equation x 3(2x 4) 7x is (4) a2 a (1) 6 (2) 6 (3) 1 (4) 1 4. The area of a circle is 16p square centimeters. The circumference of the circle is (1) 16p centimeters (2) 8p centimeters (3) 4p centimeters (4) 2p centimeters 5. The x-intercept of the line whose equation is 2x y 5 is (1) 5 (2) 5 6. The factors of x2 3x 10 are (1) (x 5)(x 2) (2) (x 5)(x 2) (3) 2 5 (4) 5 2 (3) (x 5)(x 2) (4) (x 5)(x 2) 7. The fraction (1) 5 10–3 2.5 3 1025 5.0 3 1028 is equal to (2) 5 10–2 (3) 5 102 (4) 5 103 8. When factored completely, 5a3 45a is equal to (1) 5a(a 3)2 (2) 5(a2 3)(a 3) (3) a(5a 9)(a 5) (4) 5a(a 3)(a 3) 9. When a 3, a2 5a is equal to (1) 6 (2) 6 (3) 24 (4) 24 10. The graph of the equation 2x y 7 is parallel to the graph of (1) y 2x 3 (2) y –2x 5 (3) x 2y 4 (4) y 2x 2 Part II Answer all questions in this part. Each correct answer will receive 2 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 11. The measures of the sides of a right triangle are 32, 60, and 68. Find, to the nearest degree, the measure of the smallest angle of the triangle. 12. Solve and check: x x 1 8 5 2 3 . 468 Special Products and Factors Part III Answer all questions in this part. Each correct answer will receive 3 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 13. The width of Mattie’s rectangular garden is 10 feet less than the length. She bought 25 yards of fencing and used all but 7 feet to enclose the garden. a. Write an equation or a system of equations that can be used to find the dimensions, in feet, of the garden. b. What are the dimensions, in feet, of the garden? 14. The population of a small town at the beginning of the year was 7,000. Records show that during the year there were 5 births, 7 deaths, 28 new people moved into town, and 12 residents moved out. What was the percent of increase or decrease in the town population? Part IV Answer all questions in this part. Each correct answer will receive 4 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 15. Admission to a museum is $5.00 for adults and $3.00 for children. In one day $2,400 was collected for 620 paid admissions. How many adults and how many children paid to visit the museum that day? 16. a. Write an equation of a line whose slope is –2 and whose y-intercept is 5. b. Sketch the graph of the equation written in a. c. Are (1, 3) the coordinates of a point on the graph of the equation written in a? d. If (k, 1) are the coordinates of a point on the graph of the equation written in a, what is the value of k? OPERATIONS WITH RADICALS Whenever a satellite is sent into space, or astronauts are sent to the moon, technicians at earthbound space centers monitor activities closely.They continually make small corrections to help the spacecraft stay on course. The distance from the earth to the moon varies, from 221,460 miles to 252,700 miles, and both the earth and the moon are constantly rotating in space.A tiny error can send the craft thousands of miles off course.Why do such errors occur? Space centers rely heavily on sophisticated computers, but computers and calculators alike work with approximations of numbers, not necessarily with exact values. We have learned that irrational numbers, such as 2 and 5 " , are shown on a calculator as decimal " approximations of their true values.All irrational numbers, which include radicals, are nonrepeating decimals that never end. How can we work with them? In this chapter, we will learn techniques to compute with radicals to find exact answers. We will also look at methods for working with radicals on a calculator to understand how to minimize errors when using these devices. CHAPTER 12 CHAPTER TABLE OF CONTENTS 12-1 Radicals and the Rational Numbers 12-2 Radicals and the Irrational Numbers 12-3 Finding the Principal Square Root of a Monomial 12-4 Simplifying a Square-Root Radical 12-5 Addition and Subtraction of Radicals 12-6 Multiplication of Square-Root Radicals 12-7 Division of Square-Root Radicals Chapter Summary Vocabulary Review Exercises Cumulative Review 469 470 Operations With Radicals 12-1 RADICALS AND THE RATIONAL NUMBERS Squares and Square Roots Recall from Section 1-3 that to square a number means to multiply the number by itself. To square 8, we write: 82 8 8 64 On a calculator: ENTER: 8 x2 ENTER DISPLAY: 8 2 6 4 To find the square root of a number means to find the value that, when multiplied by itself, is equal to the given number. To express the square root of 64, we write: 64 5 8 " On a calculator: ENTER: 2nd ¯ 64 ENTER DISPLAY: √ ( 6 4 8 The symbol " sign is the radicand. For example, in 64,” the radicand is 64. " is called the radical sign, and the quantity under the radical , which we read as “the square root of 64 A radical, which is any term containing both a radical sign and a radicand, is a root of a quantity. For example, 64 is a radical. In general, the square root of b is x (written as " and x2 b. b 5 x ) if and only if x 0 " Some radicals, such as 2 " 3 , are irrational numbers. We begin this study of radicals by examining " 9 , are rational numbers; others, such as " and radicals that are rational numbers. 4 " and Radicals and the Rational Numbers 471 Perfect Squares Any number that is the square of a rational number is called a perfect square. For example, 3 3 9 0 0 0 1.4 1.4 1.96 2 7 3 2 7 5 4 49 Therefore, perfect squares include 9, 0, 1.96, and Then, by applying the inverse operation, we know that: 4 49 . The square root of every perfect square is a rational number. 49 5 2 4 7 1.4 3 0 1.96 0 " " # 9 " Radicals That Are Square Roots Certain generalizations can be made for all radicals that are square roots, whether they are rational numbers or irrational numbers: 1. Since the square root of 36 is a number whose square is 36, we can write the statement also true that 2 36 B (6)2 5 A " " 36 5 6 . " 5 36 . We notice that In general, for every nonnegative real number n: 36 A " 2 B 5 (6)2 5 36 . It is n A " B 2 5 n and n2 5 n . " 2. Since (6)(6) 36 and (6)(6) 36, both 6 and 6 are square roots of 36. This example illustrates the following statement: Every positive number has two square roots: one root is a positive number called the principal square root, and the other root is a negative number. These two roots have the same absolute value. To indicate the positive or principal square root only, place a radical sign over the number: 25 5 5 and 0.49 5 0.7 " " To indicate the negative square root only, place a negative sign before the radical: 2 25 5 25 and 2 0.49 5 20.7 To indicate both square roots, place both a positive and a negative sign before the radical: 6 25 5 65 " and 6 0.49 5 60.7 " " " 472 Operations With Radicals 3. Every real number, when squared, is either positive or 0. Therefore: The square root of a negative number does not exist in the set of real numbers. For example, real number that, when multiplied by itself, equals 25. 225 " does not exist in the set of real numbers because there is no Calculators and Square Roots A calculator will return only the principal square root of a positive number. To display the negative square root of a number, the negative sign must be entered before the radical. ENTER: (-) 2nd ¯ 25 ENTER DISPLAY: - √ ( 2 5 - 5 The calculator will display an error message if it is set in “real” mode and the square root of a negative number is entered. ENTER: 2nd ¯ (-) 25 ENTER DISPLAY Cube Roots and Other Roots A cube root of a
number is one of the three equal factors of the number. For example, 2 is a cube root of 8 because 2 2 2 8, or 23 8. A cube root of 8 is written as 3 8 . " Finding a cube root of a number is the inverse operation of cubing a ) if and only number. In general, the cube root of b is x (written as if x3 b. 3 b 5 x " 3 28 " then 225 We have said that does not exist in the set of real numbers. However, does exist in the set of real numbers. Since (2)3 (2)(2)(2) 8, 3 28 5 22 " " . Radicals and the Rational Numbers 473 In the set of real numbers, every number has one cube root. The cube root of a positive number is positive, and the cube root of a negative number is negative. In the expression , the integer n that indicates the root to be taken, is called the index of the radical. Here are two examples: n b " • In • In 3 8 " 4 16 " , read as “the cube root of 8,” the index is 3. , read as “the fourth root of 16,” the index is 4. Since 24 16, 2 is one of the four equal factors of 16, and 4 16 5 2 . " When no index appears, the index is understood to be 2. Thus, 25 5 2 25 5 5 " . " When the index of the root is even and the radicand is positive, the value is a real number. That real number is positive if a plus sign or no sign precedes the radical, and negative if a minus sign precedes the radical. When the index of the root is even and the radicand is negative, the root has no real number value. When the index of the root is odd and the radicand either positive or negative, the value is a real number. That real number is positive if the radicand is positive and negative if the radicand is negative. Calculators and Roots A radical that has an index of 3 or larger can be evaluated on most graphing calkey to display a list of choices. Cube root culators. To do so, first press the MATH is choice 4. To show that ENTER: MATH 4 64 DISPLAY: 3 √ ( 6 4 3 64 5 4 : " ENTER 4 Any root with an index greater than 3 can be found using choice 5 of the MATH menu. The index, indicated by x in the menu, must be entered first. To show that ENTER: 4 MATH 4 625 5 5 : " 5 625 ENTER DISPLAY: 4 x √ 6 2 5 5 474 Operations With Radicals EXAMPLE 1 Find the principal square root of 361. Solution Since 19 19 361, then 361 5 19 . " ENTER Calculator Solution ENTER: 2nd ¯ 361 DISPLAY: √ ( 3 6 1 1 9 Answer 19 EXAMPLE 2 Find the value of " 0.0016 . Solution Since (0.04)(0.04) 0.0016, then 2 0.0016 5 20.04 . Calculator Solution ENTER: (-) 2nd ¯ .0016 DISPLAY " ENTER Answer 0.04 EXAMPLE 3 Is 13 2 a rational or an irrational number? Explain your answer. A " Solution Since B B Answer The quantity A " 2 n n for n 0, then 2 13. 13 A " B 2 13 is a rational number since its value, 13, can be written as A " the quotient of two integers, B 13 1 , where the denominator is not 0. Note: 213 does not exist in the set of real numbers since, by the order of 2 must be evaluated first. There is no real number that, when A " operations, squared, equals 13. B 213 " Radicals and the Rational Numbers 475 EXAMPLE 4 Solve for x: x2 36. Solution If x2 a, then x 5 6 when a is a positive number. a " Check x2 36 36 x 5 6 " x 6 x2 36 (6)2 36 36 36 ✔ x2 36 (6)2 36 36 36 ✔ Answer x 6 or x 6; the solution set is {6, 6}. EXERCISES Writing About Mathematics 1. Explain the difference between 2 9 " and 29 . " 2. We know that 53 125 and 54 625. Explain why 3 2125 5 25 " but 4 2625 2 25 . " Developing Skills In 3–22, express each radical as a rational number with the appropriate signs. 16 3. 8. " 2 " 625 4. 9. 64 2 " 1 4 $ 6 5. 6 2 10. " 100 9 16 $ 13. 2 1.44 14. 0.09 15. 6 0.0004 18. " 5 32 " " 3 28 " In 23–32, evaluate each radical by using a calculator. " 3 2125 " 19. 20. 2 6. 6 169 11. 16. 21. " 6 25 81 $ 3 1 " 4 0.1296 " 7. 12. 17. 22. 400 " 0.64 " 4 81 " 2 36 4 $ 2 23. 24. 10.24 " 5 21,024 " 46.24 " 3 21,000 " In 33–38, find the value of each expression in simplest form. 32.49 3 2.197 " 2 25. 28. 29. 30. " 2 26. 31. 3 23,375 " 3 20.125 " 27. 32. 4 4,096 " 6 5.76 " 33. 36. 8 A " $A 2 2 B 34. 37. B 9 3 $A 2 1 2 B 97 A " B A " 35. 38. B 2 0.7 A " B (29)2 1 " 83 A " 2 B 97 476 Operations With Radicals In 39–47, replace each ? with , , or to make a true statement. 39. 42 $ 45. m ? m " 2 40. 1 ? 12 , 0 m 1 43. 4 25 ? 46. m ? 4 25 $ " m , m 1 41. 3 2 ? 44. 1 ? 2 3 2 A B 1 " 47. m ? m " , m 1 In 48–55, solve each equation for the variable when the replacement set is the set of real numbers. 48. x2 4 52. y2 30 6 4 81 49. y2 53. 2x2 50 50. x2 0.49 54. 3y2 27 0 51. x2 16 0 55. x3 8 Applying Skills In 56–59, in each case, find the length of the hypotenuse of a right triangle when the lengths of the legs have the given values. 56. 6 inches and 8 inches 58. 15 meters and 20 meters 57. 5 centimeters and 12 centimeters 59. 15 feet and 36 feet In 60–63, find, in each case: a. the length of each side of a square that has the given area b. the perimeter of the square. 60. 36 square feet 61. 196 square yards 62. 121 square centimeters 64. Express in terms of x, (x 0), the perimeter of a square whose area is represented by x2. 63. 225 square meters 65. Write each of the integers from 101 to 110 as the sum of the smallest possible number of perfect squares. (For example, 99 72 72 12.) Use positive integers only. 12-2 RADICALS AND THE IRRATIONAL NUMBERS We have learned that type of number is examine n is a rational number when n is a perfect square. What n" when n is not a perfect square? As an example, let us using a calculator. " 5 " 2nd ENTER: DISPLAY: ¯ 5 ENTER √ ( imation of Can we state that 5 " 2.236067977? Or is 2.236067977 a rational approx? To answer this question, we will find the square of 2.236067977. 5 " Radicals and the Irrational Numbers 477 ENTER: 2.236067977 x2 ENTER DISPLAY We know that if , then x2 b. The calculator displays shown above b 5 x 2.236067977 because (2.236067977)2 5. The rational 5" demonstrate that number 2.236067977 is an approximate value for " Recall that a number such as is called an irrational number. Irrational , where a and b are integers and numbers cannot be expressed in the form b 0. Furthermore, irrational numbers cannot be expressed as terminating or repeating decimals. The example above illustrates the truth of the following statement: " a b 5 5 . " If n is any positive number that is not a perfect square, then tional number. is an irra- n " Radicals and Estimation 5 " represents the exact value of the irrational number whose The radical is a rational approximation of the irrasquare is 5. The calculator display for tional number. It is a number that is close to, but not equal to . There are other values correctly rounded from the calculator display that are also approximations of 5 " 5 " 5 : " 2.236067977 2.236068 2.23607 2.2361 2.236 2.24 calculator display, to nine decimal places rounded to six decimal places (nearest millionth) rounded to five decimal places (nearest hundred-thousandth) rounded to four decimal places (nearest ten-thousandth) rounded to three decimal places (nearest thousandth) rounded to two decimal places (nearest hundredth) is greater Each rational approximation of than 2 but less than 3. This fact can be further demonstrated by placing the square 5 , which is 5, between the squares of two consecutive integers, one less than of " 5 and one greater than 5, and then finding the square root of each number. 5 , as seen above, indicates that 5 " " Since then or " 4 5 9 . " " 478 Operations With Radicals In the same way, to get a quick estimate of any square-root radical, we place its square between the squares of two consecutive integers. Then we take the square root of each term to show between which two consecutive integers the 73 : radical lies. For example, to estimate Since 64 73 81, 73 81 , then 73 , 9" . or 64 8 , " " " " Basic Rules for Radicals That Are Irrational Numbers There are general rules to follow when working with radicals, especially those that are irrational numbers: 1. If the degree of accuracy is not specified in a question, it is best to give the exact answer in radical form. In other words, if the answer involves a radical, leave the answer in radical form. For example, the sum of 2 and 5 " 2. If the degree of accuracy is not specified in a question and a rational 5 , an exact value. is written as 2 1 " approximation is to be given, the approximation should be correct to two or more decimal places. For example, an exact answer is 5 . By using a calculator, a student " is approximately 2 2.236067977 4.236067977. discovers that Acceptable answers would include the calculator display and correctly rounded approximations of the display to two or more decimal places: 5 " 2 1 2 1 4.2360680 4.236068 4.23607 (seven places) (six places) (five places) 4.2361 4.236 4.24 (four places) (three places) (two places) Unacceptable answers would include values that are not rounded correctly, as well as values with fewer than two decimal places such as 4.2 (one decimal place) and 4 (no decimal place). 3. When the solution to a problem involving radicals has two or more steps, no radical should be rounded until the very last step. For example, to find the value of 3 3 first multiply the calculator approximation for product to two decimal places. 5 " 5 , rounded to the nearest hundredth, " by 3 and then round the Correct Solution: Incorrect Solution(2.236067977) 6.708203931 6.71 3(2.236067977) 3(2.24) 6.72 5 The solution, 6.72, is incorrect because the rational approximation of rounded too soon. To the nearest hundredth, the correct answer is 6.71." was Radicals and the Irrational Numbers 479 EXAMPLE 1 Between which two consecutive integers is Solution How to Proceed (1) Place 42 between the squares of consecutive integers: (2) Take the square root of each number: (3) Simplify terms: (4) Multiply each term of the inequality by 1: Recall that when an inequality is multiplied by a negative number, the order of the inequality is reversed. Answer EXAMPLE 2 2 42 is between 7 and 6. " 2 42 ? " 36 42 49 " 49 36 , 6 6 " " 2 42 , " 7 7 42 42 7 2 42 6
" " Is 56 " a rational or an irrational number? Solution Since 56 is a positive integer that is not a perfect square, there is no rational number that, when squared, equals 56. Therefore, is irrational. 56 " Calculator Solution STEP 1. Evaluate 56 . ENTER: ¯ 56 ENTER " 2nd DISPLAY STEP 2. To show that the calculator displays a rational approximation, not an exact value, show that the square of 7.483314774 does not equal 56. ENTER: 7.483314774 x2 ENTER DISPLAY Since (7.483314774)2 56, then 7.483314774 is a rational approximation for , not an exact value. 56 " Answer 56 " is an irrational number. 480 Operations With Radicals EXAMPLE 3 Calculator Solution Is 8.0656 " rational or irrational? STEP 1. Evaluate 8.0656 . " 2nd ENTER: DISPLAY: ¯ 8.0656 ENTER √ ( STEP 2. It appears that the rational number in the calculator display is an exact value of does equal 8.0656. " 8.0656 . To verify this, show that the square of 2.84 ENTER: 2.84 x2 ENTER DISPLAY Since (2.84)2 8.0656, then 8.0656 2.84. " Answer 8.0656 " is a rational number. If n is a positive rational number written as a terminating decimal, then n2 has twice as many decimal places as n. For example, the square of 2.84 has four decimal places. Also, since the last digit of 2.84 is 4, note that the last digit of 2.842 must be 6 because 42 16. EXAMPLE 4 Calculator Solution Approximate the value of the expression 82 1 13 2 4 a. to the nearest thousandth b. to the nearest hundredth " ENTER: 2nd ¯ 8 x2 13 ) 4 ENTER DISPLAY. To round to the nearest thousandth, look at the digit in the ten-thousandth (4th) decimal place. Since this digit (9) is greater than 5, add 1 to the digit in Radicals and the Irrational Numbers 481 the thousandth (3rd) decimal place. When rounded to the nearest thousandth, 4.774964387 is approximately equal to 4.775 b. To round to the nearest hundredth, look at the digit in the thousandth (3rd) decimal place. Since this digit (4) is not greater than or equal to 5, drop this digit and all digits to the right. When rounded to the nearest hundredth, 4.774964387 is approximately equal to 4.77 Answers a. 4.775 b. 4.77 Note: It is not correct to round 4.774964387 to the nearest hundredth by rounding 4.775, the approximation to the nearest thousandth. EXERCISES Writing About Mathematics 1. a. Use a calculator to evaluate 999,999 . " b. Enter your answer to part a and square the answer. Is the result 999,999? c. Is 999,999 " rational or irrational? Explain your answer. 2. Ursuline said that is an irrational number because it is the ratio of 18 which is irra- tional and 50 which is irrational. Do you agree with Ursuline? Explain why or why not. " 18 50 $ " Developing Skills In 3–12, between which consecutive integers is each given number? 3. 8 " 4. 13 " 5. 8. 52 9. 73 40 " 2 10. 125 6. 2 2 " 143 11. 7. 2 14 " 9 1 36 12. " " In 13–18, in each case, write the given numbers in order, starting with the smallest. " " " 13. 2, 16. 0 " 14. 4, 17. 5, 17 , 3 " 21 , " 30 " 15. 18. In 19–33, state whether each number is rational or irrational. 19. 24. 29. 25 " 400 1,156 " " 20. 25. 30. 40 " 1 2 $ 951 " 21. 26. 31. 2 2 36 " 4 9 $ 6.1504 " 22. 27. 32. 2 2 " " 15 , 3, 4 11 , 2 23 , 2 19 " " 2 54 " 0.36 " " 2,672.89 23. 28. 33. 2 150 " 0.1 " " 5.8044 482 Operations With Radicals In 34–48, for each irrational number, write a rational approximation: a. as shown on a calculator display b. rounded to four decimal places 34. 39. 44. 2 " 90 " " 28.56 35. 40. 45. 3 " 108 " " 67.25 36. 41. 46. 21 23.5 4,389 " " " 37. 42. 47. 39 88.2 123.7 " " " 38. 43. 48. " 2 " 80 115.2 " 134.53 In 49–56, use a calculator to approximate each expression: a. to the nearest hundredth b. to the nearest thousandth. 49. 7 1 7 " 2 1 53. 3 " 57. Both 50. 54. 2 1 " 19 2 6 " 3 " 8 1 8 " 130 2 4 51. 55. " " 50 1 17 27 1 4.0038 52. 56. " " and 58.01 are irrational numbers. Find a rational number n such that 58 " n 58 " 58.01 . " " Applying Skills In 58–63, in each case, find to the nearest tenth of a centimeter the length of a side of a square whose area is the given measure. 58. 18 square centimeters 59. 29 square centimeters 60. 96 square centimeters 61. 140 square centimeters 62. 202 square centimeters 63. 288 square centimeters In 64–67, find the perimeter of each figure, rounded to the nearest hundredth. 64. 3 √34 5 65. √3 √7 66. √5 √5 67. √19 10 20 √19 12-3 FINDING THE PRINCIPAL SQUARE ROOT OF A MONOMIAL Just as we can find the principal square root of a number that is a perfect square, we can find the principal square roots of variables and monomials that represent perfect squares. • Since (6)(6) 36, • Since (x)(x) x2, • Since (a2)(a2) a4, • Since (6a2)(6a2) 36a4, then then then then 36 x2 a4 6. x (x 0). a2. 36a4 6a2. " " " " Finding the Principal Square Root of a Monomial 483 In the last case, where the square root contains both numerical and variable factors, we can determine the square root by finding the square roots of its factors and multiplying: 36a4 5 36 ? " " a4 5 6 ? a2 5 6a2 " Procedure To find the square root of a monomial that has two or more factors, write the indicated product of the square roots of its factors. Note: In our work, we limit the domain of all variables under a radical sign to nonnegative numbers. EXAMPLE 1 Find the principal square root of each monomial. Assume that all variables represent positive numbers. a. 25y2 b. 1.44a6b2 c. 1,369m10 d. 81 4 g6 Solution a. b. c. d. 25 25y2 5 A " 1.44a6b2 5 B A " 1.44 " " y2 (5)(y) 5y Answer B a6 b2 (1.2)(a3)(b) 1.2a3b Answer A " B A " B A " m10 B (37)(m5) 37m5 Answer 1,369m10 5 " 81 4 g6 5 $ 81 4 A " 1,369 A " (g6g3) 5 9 2g3 A B Answer EXERCISES Writing About Mathematics 1. Is it true that for x 0, x2 5 2x ? Explain your answer. " 2. Melanie said that when a is an even integer and x 0, Melanie? Explain why or why not. xa 5 x a 2 " . Do you agree with 484 Operations With Radicals Developing Skills In 3–14, find the principal square root of each monomial. Assume that all variables represent positive numbers. 3. 4a2 7. 9c2 4. 49z2 8. 36y4 5. 16 25r2 9. c2d2 6. 0.81w2 10. 4x2y2 11. 144a4b2 12. 0.36m2 13. 0.49a2b2 14. 70.56b2x10 Applying Skills In 15–18, where all variables represent positive numbers: a. Represent each side of the square whose area is given. b. Represent the perimeter of that square. 15. 49c2 16. 64x2 17. 100x2y2 18. 144a2b2 19. The length of the legs of a right triangle are represented by 9x and 40x. Represent the length of the hypotenuse of the right triangle. 12-4 SIMPLIFYING A SQUARE-ROOT RADICAL , can often be simplified. To A radical that is an irrational number, such as understand this procedure, let us first consider some radicals that are rational and that numbers. We know that 400 5 20 . 36 5 6 " 12 " 36 5 6 Since and then " " " . " " " Since and then " " " 16 ? 25 5 16 ? " 16 ? 25 5 400 5 20 . 25 5 4 ? 5 5 20 , 16 ? 25 . " " " These examples illustrate the following property of square-root radicals: The square root of a product of nonnegative numbers is equal to the prod- uct of the square roots of these numbers. In general, if a and b are nonnegative numbers: a ? b 5 " a ? b " " and a ? " " b 5 a ? b " Now we will apply this rule to a square-root radical with a radicand that is not a perfect square. Simplifying a Square Root Radical 485 1. Express 50 as the product of 25 and 2, where 25 is the greatest perfect-square factor of 50: 2. Write the product of the two square roots: 3. Replace 25 with 5 to obtain the simplified expression with the smallest possible radicand: " 50 5 " 25 ? 2 " 5 25 ? " 2 " 5 5 2 " The expression . The simplest form is called the simplest form of of a square-root radical is one in which the radicand is an integer that has no perfect-square factor other than 1. " " 50 2 5 If the radicand is a fraction, change it to an equivalent fraction that has a denominator that is a perfect square. Write the radicand as the product of a fraction that is a perfect square and an integer that has no perfect-square factor other than 1. For example: 8 3 5 $ Procedure 3 3 3 8 3 5 $ 24 " To simplify the square root of a product: 1. If the radicand is a fraction, write it as an equivalent fraction with a denomi- nator that is a perfect square. 2. Find, if possible, two factors of the radicand such that one of the factors is a perfect square and the other is an integer that has no factor that is a perfect square. 3. Express the square root of the product as the product of the square roots of the factors. 4. Find the square root of the factor with the perfect-square radicand. EXAMPLE 1 Simplify each expression. Assume that y 0. Answers a. b. c. d. 150 18 " 4 " 9 2 $ 4y3 " " 25 ? 6 5 4 " 25 ? 4y3 4y2 ? y 5 " $ " 18 4 5 4y2 ? 5 5 6 5 20 2y " y " " 486 Operations With Radicals EXERCISES Writing About Mathematics 1. Does 1 3 27 5 " 9 ? Explain why or why not. " 2. Abba simplified the expression 192 by writing 192 5 a. Explain why 4 12 is not the simplest form of 192 . " " " " b. Show how it is possible to find the simplest form of 16 ? 12 5 4 12 . " " 192 " by starting with 4 12 . " c. What is the simplest form of 192 ? " Developing Skills In 3–22, write each expression in simplest form. Assume that all variables represent positive numbers. 3. 8. 13. 18. 20 12 " 2 " 1 2 $ 8 " 9x 27 " 5 " 24 36 5 $ 3x3 " 48 4. 9. 14. 19. " 23. The expression 2 (1) 3 " 24. The expression (1) 8 " 25. The expression 3 (1) 54 " 26. The expression 3 (1) 9 " 4 2 " " 3 " is equivalent to (2) 4 12 " is equivalent to (2) 42 " 18 is equivalent to (2) 3 2 " is equivalent to (2) 6 " 5. 10. 15. 20. 63 " 1 4 " 48 32 3 $ 49x5 " 6. 11. 16. 21. 98 96 " 3 4 " 2 1 8 $ 36r2s " 7. 4 12 " 80 3 2 " 15 12. 17. 22. 2 5 $ 243xy2 " (3) 4 3 " (3) 32 " (3) 9 2 " (3) 12 " (4) 16 3 " (4) 64 " (4) 3 6 " (4) 27 " In 27–30, for each expression: a. Use a calculator to find a rational approximation of the expression. b. Write the original expression in simplest radical form. c. Use a calculator to find a rational approximation of the answer to part b. d. Are t
he approximations in parts a and c equal? 27. 300 " 28. 180 " 29. 2 288 " 30. 1 3 252 " Addition and Subtraction of Radicals 487 ? Explain your answer. 31. a. Does 9 1 16 5 9 1 16 b. Is finding a square root distributive over addition? " " 32. a. Does 169 2 25 5 169 2 25 ? Explain your answer. b. Is finding a square root distributive over subtraction? " " " " 12-5 ADDITION AND SUBTRACTION OF RADICALS Radicals are exact values. Computations sometimes involve many radicals, as in the following example: 12 1 75 1 3 3 " " Rather than use approximations obtained with a calculator to find this sum, it may be important to express the answer as an exact value in radical form. To learn how to perform computations with radicals, we need to recall some algebraic concepts. " Adding and Subtracting Like Radicals We have learned how to add like terms in algebra: 2x 5x 3x = 10x If we replace the variable x by an irrational number, ment must be true: 3 , the following state 10 3 " " Like radicals are radicals that have the same index and the same radicand. are like are like radicals and and " " 6 5 3 2 For example, radicals. 3 , " 3 , and " 3 " 3 7 " 3 7 " To demonstrate that the sum of like radicals can be written as a single term, we can apply the concept used to add like terms, namely, the distributive property: 2 Similarly(2 1 5 1 3) 5 " " 3(10) 5 10 (6 2 1) 5 " 3 7(5) 5 5 " 3 7 " Procedure To add or subtract like terms that contain like radicals: 1. Add or subtract the coefficients of the radicals. 2. Multiply the sum or difference obtained by the common radical. 488 Operations With Radicals Adding and Subtracting Unlike Radicals Unlike radicals are radicals that have different radicands or different indexes, or both. For example: • • • and and 5 3 " 3 2 " 9 10 " different. 2 2 are unlike radicals because their radicands are different. " 2 are unlike radicals because their indexes are different. " 3 4 and " are unlike radicals because their radicands and indexes are The sum or difference of unlike radicals cannot always be expressed as a single term. For instance: • The sum of 3 and 2 2 is 3 5 1 2 5 " • The difference of 5 " and " 11 is 5 7 " 2 . " 7 2 11 . " However, it is sometimes possible to transform unlike radicals into equivalent like radicals. These like radicals can then be combined into a single term. Let us return to the problem posed at the start of this section: " " 12 1 75 25 ? 10 " " 3 Procedure To combine unlike radicals: 1. Simplify each radical if possible. 2. Combine like radicals by using the distributive property. 3. Write the indicated sum or difference of the unlike radicals. EXAMPLE 1 Combine: a. 6 " " n 2 2 49n " Solution a(8 1 1 2 2) 5 5(7) 5 7 " " b. 6 n 2 2 49n 5 6 " n 2 2(7(6 2 14) 5 28 n " Answers a. 7 5 " b. 28 n " Addition and Subtraction of Radicals 489 EXAMPLE 2 Alexis drew the figure at the right. Triangles ABC and CDE are isosceles right triangles. AB 5.00 centimeters and DE 3.00 centimeters. B 5.00 cm C D 3.00 cm a. Find AC and CE. b. Find, in simplest radical form, AC CE. c. Alexis measured AC CE and found the measure to be 11.25 centimeters. Find the percent of error of the measurement that Alexis made. Express the answer to the nearest hundredth of a percent. A E Solution a. Use the Pythagorean Theorem. AC2 AB2 BC2 52 52 25 25 50 CE2 CD2 DE2 32 32 9 9 18 CE 18 " 2 5 5 18 5 25 ? " " " AC 50 " 50 1 " " b. AC CE c. Error 8 2 2 11.25 " Percent of error 8 " 2 2 11.25 2 8 " ¯ 2 8 2nd 8 2nd ¯ 2 ENTER: ( ( ) ) ) 11.25 ) ENTER DISPLAY Multiply the number in the display by 100 to change to percent. Percent of error 0.5631089% 0.56% Answers a. AC 50 cm, CE " cm 18 " b. 8 2 " cm c. 0.56% 490 Operations With Radicals EXERCISES Writing About Mathematics 1. Compare adding fractions with adding radicals. How are the two operations alike and how are they different? 2. Marc said that 3 5 2 5 5 3 . Do you agree with Marc? Explain why or why not. " " Developing Skills In 3–23, in each case, combine the radicals. Assume that all variables represent positive numbers. 8 5. 7 7. 11. 13. 15. 17. 19. 21. 23. " 3 " 15 " " " $ " " " y 2 7 y " 27 1 " 75 " 12 2 48 1 3 " " 0.2 1 0.45 3 4 1 " 1 3 $ 100b 2 64b 1 9b 3a2 1 " 12a2 " " 4. 6. 8. 10. 12. 14. 16. 18. 20. 22. 24 " 50 2 1 " 80 2 5 " 0.98 2 4 " 72 8 9 2 $ " 7a 1 28a " 3 " 3x 2 12x " x " " " a2 1 6 a 2 3 a " 0.08 1 3 1.28 " In 25–27, in each case, select the numeral preceding the correct choice. 25. The difference 5 2 2 32 is equivalent to (1) 2 " 26. The sum 3 (1) 9 10 " " " 9 (2) 2 " 8 1 6 2 " " is equivalent to (2) 72 " 27. The sum of (1) 39 " and 12 " " (2) 5 6 " 27 is equivalent to (3) 4 30 " (3) 18 10 " (3) 13 3 " (4) 5 30 " (4) 12 2 " (4) 5 3 " Applying Skills In 28 and 29: a. Express the perimeter of the figure in simplest radical form. b. Using a calculator, approximate the expression obtained in part a to the nearest thousandth. Multiplication of Square-Root Radicals 491 28. 29. 5√5 4√5 3√5 √27 2√3 30. On the way to softball practice, Maggie walks diagonally through a square field and a rec- tangular field. The square field has a length of 60 yards. The rectangular field has a length of 70 yards and a width of 10 yards. What is the total distance Maggie walks through the fields? 12-6 MULTIPLICATION OF SQUARE-ROOT RADICALS To find the area of the rectangle pictured at the 3 5 2 right, we multiply . We have learned by " " ab a ? when a and b are nonnegative that " 2 4 numbers. To multiply , we use the com" mutative and associative laws of multiplication as follows: 3 " b 5 by " " 4 5 5 3 5 (4)(5 In general, if a and b are nonnegative numbers: 3 5 (4 ? 5) 4√2 5√3 2 ? 3 A " 5 20 xy ab " Procedure To multiply two monomial square roots: 1. Multiply the coefficients to find the coefficient of the product. 2. Multiply the radicands to find the radicand of the product. 3. If possible, simplify the result. 492 Operations With Radicals EXAMPLE 1 a. Multiply 3 6 5 and write the product in simplest radical form. " A B A 2 " B b. Check the work performed in part a using a calculator. Solution a(5) 6(2) " 12 5 15 5 15 " 4 A " 5 15(2) 5 30 Answer 5 b. To check, evaluate ENTER: 3 2nd 2nd ¯ 2 ) ENTER DISPLAY . " ¯ Then evaluate 30 ENTER: 30 2nd DISPLAY: 3 0 √ ( 3 3 ENTER Therefore 30 B 3 " appears to be true. EXAMPLE 2 Solution Alternative Solution A Answer 12 Find the value of 2)(2) 5 4(3) 5 12 (3) 5 12 " EXAMPLE 3 Find the indicated product: Solution 3x ? " 6x 5 " " 3x ? 6x 5 3x ? " 18x2 5 " " 6x , (x 0). 9x2 ? " " 2 5 3x Answer 2 " Multiplication of Square-Root Radicals 493 EXERCISES Writing About Mathematics 1. When a and b are unequal prime numbers, is answer. rational or irrational? Explain your ab " 2. Is 4a2 a rational number for all values of a? Explain your answer. " Developing Skills In 3–26: in each case multiply, or raise to the power, as indicated. Then simplify the result. Assume that all variables represent positive numbers. 3. 6. 9. 12. 15. 18. 21. 24. " " " 5 3 ? 3 " 2x ? 14 ? 2x " 2 " 3 " 8 ? 7 " 24 25x 9a 4x B B A " ab . 7. 10. 13. 16. 19. 22. 25. 7 ? 7 " 12 ? 12 60 ? " 5 " 24 ? " 21 2 B 27a " A 2 t A " A " A " 5x B A " 2 B In 27–33: a. Perform each indicated operation. ber or a rational number. B 5. 8. a ? a " 18 ? 3 " 8 11. 14 " 15 " " " 2 B y 26 3a " 17. 20. State whether the product is an irrational num- B 15x 2 B A " 26. 23. 3x B B B 27. 5 12 31. B A B A 35. Two square-root radicals that are irrational numbers are multiplied. B A 3 B 33. 32. B B " " 18 28. 3 2 32 29 " 30. 34. A A 5 8 " 11 " 1 2 B A 38 10 " 1 11 B A B 45 " B a. Give two examples where the product of these radicals is also an irrational number. b. Give two examples where the product of these radicals is a rational number. Applying Skills In 36–39, in each case, find the area of the square in which the length of each side is given. 36. 2 " 37. 2 3 " 38. 6 2 " 39. 5 3 " In 40 and 41: a. Express the area of the figure in simplest radical form. formed in part a by using a calculator. b. Check the work per- 40. √2 2√3 41. √3 2√12 494 Operations With Radicals 12-7 DIVISION OF SQUARE-ROOT RADICALS Since and then " " $ , Since and then . 4 9 , 16 25 5 4 5 16 5 4 5 25 16 25 5 " " $ " " $ 16 25 . These examples illustrate the following property of square-root radicals: The square root of a fraction that is the quotient of two positive numbers is equal to the square root of the numerator divided by the square root of the denominator. In general, if a and b are positive numbers: a b 5 " " $ a b and 5 a b a b $ " " 8 . In this example, notice that the " We use this principle to divide 72 by quotient of two irrational numbers is a rational number. " 5 72 8 " " 72 8 5 " 9 5 3 $ We can also divide radical terms by using the property of fractions: bd 5 a ac b ? Note, in the following example, that the quotient of two irrational numbers c d is irrational: 10 2 In general, if a and b are positive, and y 0: 10 ? # 10 $ Procedure To divide two monomial square roots: 1. Divide the coefficients to find the coefficient of the quotient. 2. Divide the radicands to find the radicand of the quotient. 3. If possible, simplify the result. Division of Square-Root Radicals 495 EXAMPLE 1 Divide 8 by 4 48 " " 2 , and simplify the quotient. Solution 8 48 4 4 " 2 5 8 4 " 48 2 5 2 $ 24 (2) B 6 A " B 5 4 6 " Answer EXAMPLE 2 Find the indicated division: , (x 0, y 0, z 0). 2x2y3z " 6y " Solution " 2x2y3z 6y " 5 1 3 ? $ " x2y2z 5 3 9 ? $ " x2y2z 1 9x2y2 ? $ " 3z 5 1 3xy " 3z Answer EXERCISES Writing About Mathematics 1. Ross simplified 16 81 $ by writing 5 4 9 5 2 3 16 81 " " . Do you agree with Ross? Explain why or why not. 2. Is 1 16 rational or irrational? Explain your answer. $ Developing Skills In 3–18, divide. Write the quotient in simplest form. Assume that all variables represent positive numbers. 3. 7. 11. 15. 72 4 2 " 48 " " 12 3 " " 20 . 8. 12. 16. 75 4 " 24 4 " 7 " 20 4 50 " 2 " 5. 9. 13. 17. 70 4 " 10 " 150 " b3c4 a2 a " " 6. 10. 14. 18. 14 4 2 " 21 " 40 4 " 9y 4 " 3 x3y " z 6 " 5 " y " In 19–26, state
whether each quotient is a rational number or an irrational number. 19. 5 " 7 23. 9 16 " " 20. 24. 50 2 18 25 " " " " 21. 25. 18 3 " " 25 " 5 " 24 2 22. 49 " 7 " 26. 3 " 6 " 54 3 496 Operations With Radicals In 27–34, simplify each expression. Assume that all variables represent positive numbers. 27. 31. 36 49 $ 10 8 25 $ 28. 32. 3 4 $ 9 18 xy2 y $ 29. 4 33. 5 16 $ a6b5c4 a4b3c2 $ 30. 34. 8 49 $ 15 3 a5 a2b2 $ Applying Skills In 35–38, in each case, the area A of a parallelogram and the measure of its base b are given. Find the height h of the parallelogram, expressed in simplest form. 35. A 7 37. A 8 12 , b 7 45 , b 2 3 15 " " " " 36. A 38. A CHAPTER SUMMARY 640 , b 98 , b " 2 " 32 32 " " A radical, which is the root of a quantity, is written in its general form as , with an . A radical consists of a radicand, b, placed under a radical sign, " n b " index, n. 2 49 5 " " Finding the square root of a quantity reverses the result of the operation of squaring. A square-root radical has an index of 2, which is generally not written. because 72 49. In general, for nonnegative numbers b Thus, and x, 49 5 7 if and only if x2 b. " b 5 x If k is a positive number that is a perfect square, then k is a rational num- ber. If k is positive but not a perfect square, then Every positive number has two square roots: a positive root called the principal square root, and a negative root. These roots have the same absolute value: " k " is an irrational number. Principal Square Root Negative Square Root Both Square Roots x2 5 ZxZ " 2 x2 5 2ZxZ " 6 x2 5 6x " Finding the cube root of a number is the inverse of the operation of cubing. if and only if because 43 = 4 4 4 64. In general, 3 b 5 x " 3 64 5 4 Thus, " x3 b. and 7 " Like radicals have the same radicand and the same index. For example, 7 3 . Unlike radicals can differ in their radicands ( and " indexes ( and A square root of a positive integer is simplified by factoring out the square root of its greatest perfect square. The radicand of a simplified radical, then, has no perfect square factor other than 1. When a radical is irrational, the radical expresses its exact value. Most calculators display only rational approximations of radicals that are irrational numbers. 7 2 " 2 ), in their " ), or in both ( 3 6 ). " 3 7 " 4 7 " 7 " and Review Exercises 497 Operations with radicals include: 1. Addition and subtraction: Combine like radicals by adding or subtracting their coefficients and then multiplying this result by their common radical. The sum or difference of unlike radicals, unless transformed to equivalent like radicals, cannot be expressed as a single term. 2. Multiplication and division: For all radicals whose denominators are not equal to 0, multiply (or divide) coefficients, multiply (or divide) radicands, and simplify. The general rules for these operations are as follows xy B " ab and $ VOCABULARY 12-1 Radicand • Radical • Perfect square • Principal square root • Cube root • Index 12-4 Simplest form of a square-root radical 12-5 Like radicals • Unlike radicals REVIEW EXERCISES 1. When a b $ is an integer, what is the relationship between a and b? 2. When a is a positive perfect square and b is a positive number that is not a perfect square, is ab rational or irrational? Explain your answer. " 3. Write the principal square root of 1,225. 4. Write the following numbers in order starting with the smallest: 18 , " 2 2 , 3. " In 5–14, write each number in simplest form. 9 25 $ 400y4 5. 9. 13. " " 6. 10. 2 49 " 180 " 0.01m16 " 7. 11. 3 227 " 3 18 " 8. 6 1.21 " 12. 1 2 28 " 16. 48b3 , b 0 14. 15. 9 27x3y5, x . 0, y . 0 $ In 17–20, in each case, solve for the variable, using the set of real numbers as the replacement set. 17. y2 81 20. 2k2 144 0 18. m2 0.09 19. 3x2 600 0.25a8b10 " 498 Operations With Radicals 21. a. Use a calculator to evaluate 315.4176 . " b. Is 315.4176 " a rational or an irrational number? Explain your answer. In 22–33, in each case, perform the indicated operation and simplify the result. 18 1 8 2 32 " 50 2 " 98 1 1 2 72 " 22. 24. 26. 28. 30. 32. " 2 8 2 " 2 " 7 " 16 " 2 " 3 " " 2 " 70 2 A A " 21 7 24 A " B B B 2 5 " 10 A " B 23. 3 20 2 2 45 25. 27. 29. 31. 33. " 75 2 3 " 12 " 3 A 5 " B 98 4 " 2 2 " 162 50 " 5 " 9 " 80 2 " " 1 6 " 60 A " B In 34 and 35, in each case, select the numeral preceding the correct choice. 34. The expression 108 2 (1) 105 " 35. The sum of (1) 9 34 " " (2) 35 and 9 2 " " (2) 13 is equivalent to (3) 5 3 " 3 " 3 " is 32 2 " (3) 10 34 " (4) 6 (4) 15 In 36–39, for each irrational number given, write a rational approximation: a. as shown on a calculator display b. rounded to four significant digits. 36. 194 3 16 " 40. The area of a square is 28.00 square meters. 38. 37. " " 2 0.7 39. 5 227 " a. Find, to the nearest thousandth of a meter, the length of one side of the square. b. Find, to the nearest thousandth of a meter, the perimeter of the square. c. Explain why the answer to part b is not equal to 4 times the answer to part a. 41. Write as a polynomial in simplest form: 42. What is the product of 3.5x2 and 6.2x3? A 2x Exploration STEP 1. On a sheet of graph paper, draw the positive ray of the real number line. Draw a square, using the interval from 0 to 1 as one side of the square. Draw the diagonal of this square from 0 to its opposite vertex. Cumulative Review 499 2 . Why? Place the point of a compass at The length of the diagonal is " 0 and the pencil of a compass at the opposite vertex of the square so 2 . Keep that the measure of the opening of the pair of compasses is " the point of the compass at 0 and use the pencil to mark the endpoint of a segment of length on the number line. 2 " STEP 2. Using the same number line, draw a rectangle whose dimensions are 2 as one by 1, using the interval on the number line from 0 to " side. Draw the diagonal of this rectangle from 0 to the opposite vertex. 3 . Place the point of a pair of compasses The length of the diagonal is at 0 and the pencil at the opposite vertex of the rectangle so that the 3 . Keep the point measure of the opening of the pair of compasses is " of the compasses at 0 and use the pencil to mark the endpoint of a segment of length on the number line. " " 2 STEP 3. Repeat step 2, drawing a rectangle whose dimensions are by 1 to on the number line. This point should coincide with 2 on the " 4 3 3 " locate number line. " STEP 4. Explain how these steps can be used to locate integer n. n for any positive " CUMULATIVE REVIEW CHAPTERS 1–12 Part I Answer all questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. 1. A flagpole casts a shadow five feet long at the same time that a man who is six feet tall casts a shadow that is two feet long. How tall is the flagpole? (1) 12 feet (2) 15 feet (4) 24 feet (3) 18 feet 2. Which of the following is an irrational number? 6 " (1) 12 3 (2) 24 4 (3. Parallelogram ABCD is drawn on the coordinate plane with the vertices A(0, 0), B(8, 0), C(10, 5), and D(2, 5). The number of square units in the area of ABCD is (1) 40 (2) 20 4. The product (3a 2)(2a 3) can be written as (3) 16 (4) 10 " " 5 2 5 (4) 3 " (1) 6a2 6 (2) 6a2 a 6 (3) 6a2 5a 6 (4) 6a2 5a 6 500 Operations With Radicals 5. If 0.2x 8 x 4, then x equals (1) 120 (2) 12 (3) 15 (4) 15 6. If the height of a right circular cylinder is 12 centimeters and the measure of the diameter of a base is 8 centimeters, then the volume of the cylinder is (1) 768p (2) 192p 7. The identity 3(a 7) 3a 21 is an example of (4) 48p (3) 96p (1) the additive inverse property (2) the associative property for addition (3) the commutative property for addition (4) the distributive property of multiplication over addition 8. The value of a share of stock decreased from $24.50 to $22.05. The percent of decrease was (1) 1% (2) 10% (3) 11% (4) 90% 9. When written in scientific notation, 384.5 is equal to (1) 38.45 101 (2) 3.845 102 (3) 3.845 103 (4) 3.845 10–2 10. In right triangle ABC, C is the right angle. The cosine of B is AC BC AC AB BC AB BC AC (3) (2) (1) (4) Part II Answer all questions in this part. Each correct answer will receive 2 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 11. A car uses of a tank of gasoline to travel 600 kilometers. The tank holds 48 liters of gasoline. How far can the car go on one liter of gasoline? 3 4 12. A ramp that is 20.0 feet long makes an angle of 12.5° with the ground. What is the perpendicular distance from the top of the ramp to the ground? Part III Answer all questions in this part. Each correct answer will receive 3 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 13. ABCD is a trapezoid with BC ' AB and BC ' CD and CD = 8. A line segment is drawn from A to E, the midpoint of , AB 13, BC 12, .CD Cumulative Review 501 a. Find the area of AED. b. Find the perimeter of AED. 14. Maria’s garden is in the shape of a rectangle that is twice as long as it is wide. Maria increases the width by 2 feet, making the garden 1.5 times as long as it is wide. What are the dimensions of the original garden? Part IV Answer all questions in this part. Each correct answer will receive 4 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 15. The length of the base of an isosceles triangle is and the length of the . Express the perimeter as an exact value in simplest form. 2 " 10 altitude is 12 2 " 16. In the coordinate plane, O is the origin, A is a point on the y-axis, and B is g AB is 2 and its y-intercept i
s 8. a point on the x-axis. The slope of a. Write the equation of g AB . b. Draw g AB on graph paper. c. What are the coordinates of B? d. What is the x-intercept of g ?AB CHAPTER 13 CHAPTER TABLE OF CONTENTS 13-1 Solving Quadratic Equations 13-2 The Graph of a Quadratic Function 13-3 Finding Roots from a Graph 13-4 Graphic Solution of a Quadratic-Linear System 13-5 Algebraic Solution of a Quadratic-Linear System Chapter Summary Vocabulary Review Exercises Cumulative Review 502 QUADRATIC RELATIONS AND FUNCTIONS When a baseball is hit, its path is not a straight line. The baseball rises to a maximum height and then falls, following a curved path throughout its flight.The maximum height to which it rises is determined by the force with which the ball was hit and the angle at which it was hit.The height of the ball at any time can be found by using an equation, as can the maximum height to which the ball rises and the distance between the batter and the point where the ball hits the ground. In this chapter we will study the quadratic equation that models the path of a baseball as well as functions and relations that are not linear. Solving Quadratic Equations 503 13-1 SOLVING QUADRATIC EQUATIONS The equation x2 3x 10 0 is a polynomial equation in one variable. This equation is of degree two, or second degree, because the greatest exponent of the variable x is 2. The equation is in standard form because all terms are collected in descending order of exponents in one side, and the other side is 0. A polynomial equation of degree two is also called a quadratic equation. The standard form of a quadratic equation in one variable is ax2 bx c 0 where a, b, and c are real numbers and a 0. To write an equation such as x(x 4) 5 in standard form, rewrite the left side without parentheses and add 5 to both sides to make the right side 0. x(x 4) 5 x2 4x 5 x2 4x 5 0 Solving a Quadratic Equation by Factoring When 0 is the product of two or more factors, at least one of the factors must be equal to 0. This is illustrated by the following examples: 5 0 0 0 7 0 (2) 0 0 0 (3 In general: When a and b are real numbers, ab 0 if and only if a 0 or b 0. This principle is used to solve quadratic equations. For example, to solve the quadratic equation x2 3x 2 0, we can write the left side as (x 2)(x 1). and (x 1) represent real numbers whose product is 0. The The factors equation will be true if the first factor is 0, that is, if (x 2) 0 or if the second factor is 0, that is, if (x 1) 0. (x 2 2) x2 3x 2 0 (x 2)(x 1 check will show that both 2 and 1 are values of x for which the equation is true. 504 Quadratic Relations and Functions Check for x 2: x2 3x 2 0 Check for x 1: x2 3x 2 0 (2)2 3(2) 2 4 6 2 0 5? 5? 0 0 0 ✔ (1)2 3(1) 2 1 3 2 0 5? 5? 0 0 0 ✔ Since both 2 and 1 satisfy the equation x2 3x 2 0, the solution set of this equation is {2, 1}. The roots of the equation, that is, the values of the variable that make the equation true, are 2 and 1. Note that the factors of the trinomial x2 3x 2 are (x 2) and (x 1). If the trinomial x2 3x 2 is set equal to zero, then an equation is formed and this equation has a solution set, {2, 1}. Procedure To solve a quadratic equation by factoring: 1. If necessary, transform the equation into standard form. 2. Factor the quadratic expression. 3. Set each factor containing the variable equal to 0. 4. Solve each of the resulting equations. 5. Check by substituting each value of the variable in the original equation. The real number k is a root of ax2 bx c 0 if and only if (x k) is a factor of ax2 bx c. EXAMPLE 1 Solve and check: x2 7x 10 Solution How to Proceed (1) Write the equation in standard form: (2) Factor the quadratic expression: (3) Let each factor equal 0: (4) Solve each equation: x2 7x 10 x2 7x 10 0 (x 2)(x 55) Check both values in the original equation: Solving Quadratic Equations 505 Check for x 2: x2 7x 10 210 (2)2 7(2) 4 14 5? 5? 10 10 ✔ 210 Check for x 5: x2 7x 10 210 (5)2 7(5) 25 35 5? 5? 10 10 ✔ 210 Answer x 2 or x 5; the solution set is {2, 5}. EXAMPLE 2 Solution List the members of the solution set of 2x2 3x. How to Proceed (1) Write the equation in standard form: (2) Factor the quadratic expression: (3) Let each factor equal 0: (4) Solve each equation: (5) Check both values in the original equation. 2x2 3x 2x2 3x 0 x(2x 3) 0 x 0 2x 3 0 2x 3 x 5 3 2 Check for x 0: 2x2 3x 2(0)2 5? 3(0) 0 0 ✔ Answer: The solution set is 0, 3 2 . V U 3 Check for x : 2 2x2 3x 2 5? 3 3 2 B 9 5 Note: We never divide both sides of an equation by an expression containing a variable. If we had divided 2x2 = 3x by x, we would have obtained the equation 2x 3, whose solution is but would have lost the solution x 0. 3 2 506 Quadratic Relations and Functions EXAMPLE 3 Find the solution set of the equation x(x 8) 2x 25. Solution How to Proceed (1) Use the distributive property on the left side of the equation: (2) Write the equation in standard form: (3) Factor the quadratic expression: x(x 8) 2x 25 x2 8x 2x 25 x2 10x 25 0 (x 5)(x 5) 0 (4) Let each factor equal 0: (5) Solve each equation: (6) Check the value in the original equation(x 8) 2x 25 5? 2(5) 5(5 8) 5? 10 2 25 15 15 15 ✔ 25 Answer x 5; the solution set is {5}. Every quadratic equation has two roots, but as Example 3 shows, the two roots are sometimes the same number. Such a root, called a double root, is written only once in the solution set. EXAMPLE 4 The height h of a ball thrown into the air with an initial vertical velocity of 24 feet per second from a height of 6 feet above the ground is given by the equation h l6t2 24t 6 where t is the time, in seconds, that the ball has been in the air. After how many seconds is the ball at a height of 14 feet? Solution (1) In the equation, let h 14: (2) Write the equation in standard form: (3) Factor the quadratic expression: (4) Solve for t: h l6t2 24t 6 14 16t2 24t 6 0 16t2 24t 8 0 8(2t2 3t 1) 0 8(2t 1)(t 1) 2t 1 0 2t Answer The ball is at a height of 14 feet after second as it rises and after 1 second 1 2 as it falls. EXAMPLE 5 The area of a circle is equal to 3 times its circumference. What is the radius of the circle? Solving Quadratic Equations 507 Solution How to Proceed (1) Write an equation from the given information: (2) Set the equation in standard form: (3) Factor the quadratic expression: (4) Solve for r: (5) Reject the zero value. Use the positive value to write the answer. Answer The radius of the circle is 6 units. pr2 3(2pr) pr2 6pr 0 pr(r 6) 0 pr 0 r 0 r 6 0 r 6 EXERCISES Writing About Mathematics 1. Can the equation x2 9 be solved by factoring? Explain your answer. 2. In Example 4, the trinomial was written as the product of three factors. Only two of these factors were set equal to 0. Explain why the third factor was not used to find a solution of the equation. Developing Skills In 3–38, solve each equation and check. 3. x2 3x 2 0 6. r2 12r 35 0 9. x2 2x 1 0 12. x2 x 6 0 15. x2 x 12 0 18. m2 64 0 21. s2 s 0 24. x2 x 6 27. r2 4 30. s2 4s 33. 30 x x2 36. x(x 2) 35 4. z2 5z 4 0 7. c2 6c 5 0 10. y2 11y 24 0 13. x2 2x 15 0 16. x2 49 0 19. 3x2 12 0 22. x2 3x 0 25. y2 3y 28 28. x2 121 31. y2 8y 20 34. x2 3x 4 50 37. y(y 3) 4 5. x2 8x 16 0 8. m2 10m 9 0 11. x2 4x 5 0 14. r2 r 72 0 17. z2 4 0 20. d2 2d 0 23. z2 8z 0 26. c2 8c 15 29. y2 6y 32. x2 9x 20 35. 2x2 7 5 5x 38. x(x 3) 40 508 Quadratic Relations and Functions In 39–44, solve each equation and check. y 1 3 3 5 6 y 24x 12 x x 1 4 21 5 4 x 40. 42. 43. 39. 41. 44. x 3 5 12 x 2x Applying Skills 45. The height h of a ball thrown into the air with an initial vertical velocity of 48 feet per sec- ond from a height of 5 feet above the ground is given by the equation h 16t2 48t 5 where t is the time in seconds that the ball has been in the air. After how many seconds is the ball at a height of 37 feet? 46. A batter hit a baseball at a height of 4 feet with a force that gave the ball an initial vertical velocity of 64 feet per second. The equation h 16t2 64t 4 gives the height h of the baseball t seconds after the ball was hit. If the ball was caught at a height of 4 feet, how long after the batter hit the ball was the ball caught? 47. The length of a rectangle is 12 feet more than twice the width. The area of the rectangle is 320 square feet. a. Write an equation that can be used to find the length and width of the rectangle. b. What are the dimensions of the rectangle? 48. A small park is enclosed by four streets, two of which are parallel. The park is in the shape of a trapezoid. The perpendicular distance between the parallel streets is the height of the trapezoid. The portions of the parallel streets that border the park are the bases of the trapezoid. The height of the trapezoid is equal to the length of one of the bases and 20 feet longer than the other base. The area of the park is 9,000 square feet. a. Write an equation that can be used to find the height of the trapezoid. b. What is the perpendicular distance between the two parallel streets? 49. At a kennel, each dog run is a rectangle whose length is 4 feet more than twice the width. Each run encloses 240 square feet. What are the dimensions of the runs? 50. One leg of a right triangle is 14 centimeters longer than the other leg. The length of the hypotenuse is 26 centimeters. What are the lengths of the legs? 13-2 THE GRAPH OF A QUADRATIC FUNCTION A batter hits a baseball at a height of 3 feet off the ground, with an initial vertical velocity of 72 feet per second. The height, y, of the baseball can be found using the equation when x represents the number of seconds from the time the ball was hit. The graph of the equation––and, inci- y 5 216x2 1 72x 1 3 The Graph of a Quadratic Function 509 dentally, the actual path of the ball––is a curve called a parabola. The special properties of parabolas are discussed in this section. An equation of the form y ax2 bx c (a 0) is called a second-degree polynomial function or a quadratic function. It is a function because for every ordered pair in its solutio
n set, each different value of x is paired with one and only one value of y. The graph of any quadratic function is a parabola. Because the graph of a quadratic function is nonlinear, a larger number of points are needed to draw the graph than are needed to draw the graph of a linear function. The graphs of two equations of the form y ax2 bx c, one that has a positive coefficient of x2 (a 0) and the other a negative coefficient of x2 (a 0), are shown below. CASE 1 The graph of ax2 bx c where a 0. Graph the quadratic function y x2 4x 1 for integral values of x from 1 to 5 inclusive: (1) Make a table using integral values of x from 1 to 5. (2) Plot the points associated with each ordered pair (x, y). (3) Draw a smooth curve through the points. x 1 0 1 2 3 4 5 x2 4x 12 1 16 16 1 25 20 –1 – The values of x that were chosen to draw this graph are not a random set of numbers. These numbers were chosen to produce the pattern of y-values shown in the table. Notice that as x increases from 1 to 2, y decreases from 6 to 3. Then the graph reverses and as x continues to increase from 2 to 5, y increases from 3 to 6. The smallest value of y occurs at the point (2, 3). The point is called the minimum because its y-value, 3, is the smallest value of y for the equation. The minimum point is also called the turning point or vertex of the parabola. 510 Quadratic Relations and Functions The graph is symmetric with respect to the vertical line, called the axis of symmetry of the parabola. The axis of symmetry of the parabola is determined by the formula x 5 2b 2a where a and b are the coefficients x2 and x, respectively, from the standard form of the quadratic equation. For the function y x2 4x 1, the equation of the vertical line of symor x 2. Every point on the parabola to the left of x 2 metry is matches a point on the parabola to the right of x 2, and vice versa. 2(24) 2(1) x 5 This example illustrates the following properties of the graph of the qua- dratic equation y x2 4x 1: 1. The graph of the equation is a parabola. 2. The parabola is symmetric with respect to the vertical line x 2. 3. The parabola opens upward and has a minimum point at (2, 3). 4. The equation defines a function. For every x-value there is one and only one y-value. 5. The constant term, 1, is the y-intercept. The y-intercept is the value of y when x is 0. CASE 2 The graph of ax2 bx c where a 0. Graph the quadratic function y x2 2x 5 using integral values of x from 4 to 2 inclusive: (1) Make a table using integral values of x from 4 to 2. (2) Plot the points associated with each ordered pair (x, y). (3) Draw a smooth curve through the points. x 4 3 2 1 0 1 2 x2 2x 5 16 1 – Again, the values of x that were chosen to produce the pattern of y-values shown in the chart. Notice that as x increases from 4 to 1, y increases from The Graph of a Quadratic Function 511 3 to 6. Then the graph reverses, and as x continues to increase from 1 to 2, y decreases from 6 to 3. The largest value of y occurs at the point (1, 6). This point is called the maximum because its y-value, 6, is the largest value of y for the equation. In this case, the maximum point is the turning point or vertex of the parabola. The graph is symmetric with respect to the vertical line whose equation is x 1. As shown in Case 1, this value of x is again given by the formula x 5 2b 2a where a and b are the coefficients of x2 and x, respectively. For the function y x2 2x 5, the equation of the axis of symmetry is x or x 1. This example illustrates the following properties of the graph of the qua- 2(22) 2(21) dratic equation y 5 2x2 2 2x 1 5 : 1. The graph of the equation is a parabola. 2. The parabola is symmetric with respect to the vertical line x 1. 3. The parabola opens downward and has a maximum point at (1, 6) 4. The equation defines a function. 5. The constant term, 5, is the y-intercept. When the equation of a parabola is written in the form y ax2 bx c 0, the equation of the axis of symmetry is x and the x-coordinate of the turn. This can be used to find a convenient set of values of x to be used ing point is when drawing a parabola. Use as a middle value of x with three values that are smaller and three that are larger. 2b 2a 2b 2a 2b 2a KEEP IN MIND 1. The graph of y ax2 bx c, with a 0, is a parabola. 2. The axis of symmetry of the parabola is a vertical line whose equation is 2b x . 2a 3. A parabola has a turning point on the axis of symmetry. The x-coordinate of the turning point is by substituting the vertex of the parabola. 2b 2a 2b 2a . The y-coordinate of the turning point is found into the equation of the parabola. The turning point is 4. If a is positive, the parabola opens upward and the turning point is a minimum. The minimum value of y for the parabola is the y-coordinate of the turning point. 5. If a is negative, the parabola opens downward and the turning point is a maximum. The maximum value of y for the parabola is the y-coordinate of the turning point. 512 Quadratic Relations and Functions EXAMPLE 1 a. Write the equation of the axis of symmetry of y x2 3. b. Graph the function. c. Does the function have a maximum or a minimum? d. What is the maximum or minimum value of the function? e. Write the coordinates of the vertex. Solution a. In this equation, a 1. Since there is no x term in the equation, the equation can be written as y x2 0x 3 with b 0. The equation of the axis of symmetry is x 2b 2a 5 2(0) 2(1) 0 or x 0. b. (1) Since the vertex of the parabola is on the axis of symmetry, the x-coordinate of the vertex is 0. Use three values of x that are less than 0 and three values of x that are greater than 0. Make a table using integral values of x from 3 to 3. (2) Plot the points associated with each ordered pair (x, y). (3) Draw a smooth curve through the points to draw a parabola. x –3 –2 –1 0 1 2 3 x2 3 (3)2 3 (2)2 3 (1)2 3 (0)2 3 (1)2 3 (2)2 3 (3)1 –1 O 1 x c. Since a 1 0, the function has a minimum. d. The minimum value of the function is the y-coordinate of the vertex, 3, which can be read from the table of values. e. Since the vertex is the turning point of this parabola, the coordinates of the vertex are (0, 3). The Graph of a Quadratic Function 513 Calculator Solution a. Determine the equation of the axis of symmetry as before. b. Enter the equation in the Y list of functions and graph the function. Clear any equations already in the list. ENTER: Y X,T,,n x2 3 ZOOM 6 DISPLAY: 1 2 Plot1 Plot2 Plot3 = = X 2 –3 \Y = \Y = \Y = \Y = \Y = \Y = \Y 3 4 5 6 7 c. The graph shows that the function has a minimum. d. Since the minimum of the function occurs at the vertex, use value 2nd CALC 1 from the CALC menu to evaluate the Q function at x 0, the x-coordinate of the vertex: R ENTER: 2nd CALC 1 0 ENTER DISPLAY: Calculate 1: value 2: zero 3: minimum 4: minimum 5: intersect 6: dy/dx 7: f [x] dx Y1 = X2 –3 * X = 0 Y = –3 The calculator displays the minimum value, 3. e. The coordinates of the vertex are (0, 3). Answers a. The axis of symmetry is the y-axis. The equation is x 0. b. Graph c. The function has a minimum. d. The minimum value is 3. e. The vertex is (0, 3). 514 Quadratic Relations and Functions When the coordinates of the turning point are rational numbers, we can use from the or maximum minimum CALC menu to find the vertex. In Example 1, since the turning point is a minimum, use minimum: CALC CALC 2nd 2nd 3 4 R Q Q R ENTER: 2nd CALC 3 When the calculator asks “LeftBound?” move the cursor to any point to the left of the vertex using the left or right arrow keys, and then press enter. When the calculator asks “RightBound?” move the cursor to the right of the vertex, and then press enter. When the calculator asks “Guess?” move the cursor near the vertex and then press enter. The calculator displays the coordinates of the vertex at the bottom of the screen. Y1=X^2–3 Y1=X^2–3 * * LeftBound? X = -.8510638 Y = –2.27569 RightBound? X = 1.0638298 Y = –1.868266 Y1=X^2–3 * Y = -3 Guess? X = 0 * Y = -3 Minimum X = 0 EXAMPLE 2 Sketch the graph of the function y x2 7x 10. Solution (1) Find the equation of the axis of symmetry and the x-coordinate of the vertex: 2b x 2a 2(27) 2(1) 3.5 (2) Make a table of values using three integral values smaller than and three larger than 3.5. (3) Plot the points whose coordinates are given in the table and draw a smooth curve through them. x x2 7x 10 y 1 2 3 3.5 4 5 6 4 1 7 10 4 14 10 0 9 21 10 2 12.25 24.5 10 2.25 16 28 10 2 25 35 10 0 36 42 10 4 The Graph of a Quadratic Function 515 y 1 –1 –1 O 1 x y = x2 – 7x + 10 Note: The table can also be displayed on the calculator. First enter the equation into Y1. ENTER: Y X,T,,n x2 7 X,T,,n 10 Then enter the starting value and the interval between the x values. We will use 1 as the starting value and 0.5 as the interval in order to include 3.5, the x-value of the vertex. ENTER: 2nd TBLSET 1 ENTER .5 ENTER Before creating the table, make sure that “Indpnt:” and “Depend:” are set to “auto.” If . they are not, press ENTER ENTER 2nd Finally, press to create the table. Scroll up and down to view the values of x and y. TABLE X Y1 4 1.75 0 -1.25 -2 -2.25 -2 1 1.5 2 2.5 3 3.5 4 X=1 Calculator Solution Enter the equation in the Y= menu and sketch the graph of the function in the standard window. ENTER: Y X,T,,n x2 7 X,T,,n 10 ZOOM 6 DISPLAY: 516 Quadratic Relations and Functions EXAMPLE 3 The perimeter of a rectangle is 12. Let x represent the measure of one side of the rectangle and y represent the area. a. Write an equation for the area of the rectangle in terms of x. b. Draw the graph of the equation written in a. c. What is the maximum area of the rectangle? Solution a. Let x be the measure of the length of the rectangle. Use the formula for perimeter to express the measure of the width in terms of x: Write the formula for area in terms of x and y: P 2l 2w 12 2x 2w 6 x w w 6 x A lw y x(6 x) y 6x x2 y x2 6x b. The equation of the axis of symmetry is x values using values of x on each side of 3. 26 2(21) or x 3.
Make a table of y x 0 1 2 3 4 5 6 x2 6x 0 0 1 6 4 12 9 18 16 24 25 30 36 36 1 –1 1 x c. The maximum value of the area, y, is 9. Note: The graph shows all possible values of x and y. Since both the measure of a side of the rectangle, x, and the area of the rectangle, y, must be positive, 0 x 6 and 0 y 9. Since (2, 8) is a point on the graph, one possible rectangle has dimensions 2 by (6 2) or 2 by 4 and an area of 8. The rectangle with maximum area, 9, has dimensions 3 by (6 3) or 3 by 3, a square. The Graph of a Quadratic Function 517 Translating, Reflecting, and Scaling Graphs of Quadratic Functions Just as linear and absolute value functions can be translated, reflected, or scaled, graphs of quadratic functions can also be manipulated by working with the graph of the quadratic function y x2. y 5 x2 1 2.5 For instance, the graph of shifted 2.5 units up. The graph of y x2 is the graph of y x2 reflected in the x-axis. The y 5 3x2 stretched vertically by a factor of 3, graph of y 5 x2 while the graph of compressed vertically by a fac1 tor of . 3 is the graph of y 5 1 3x2 y 5 x2 is the graph of is the graph of y 5 x2 2 + 2.5 y = x y 1 y = x2 –1 1 x y 1 –2 –1 y = x2 x 2 y = – x2 1 1 x Translation Rules for Quadratic Functions If c is positive: The graph of y 5 x2 1 c is the graph of y 5 x2 shifted c units up. The graph of y 5 x2 2 c is the graph of y 5 x2 shifted c units down. The graph of y 5 (x 1 c)2 is the graph of y 5 x2 shifted c units to the left. The graph of y 5 (x 2 c)2 is the graph of y 5 x2 shifted c units to the right. Reflection Rule for Quadratic Functions The graph of y 5 2x2 is the graph of y 5 x2 reflected in the x-axis. Scaling Rules for Quadratic Functions When c 1, the graph of y 5 cx2 is the graph of y 5 x2 stretched vertically by a factor of c. When 0 c 1, the graph of y 5 cx2 is the graph of y 5 x2 compressed vertically by a factor of c. 518 Quadratic Relations and Functions EXAMPLE 4 In a–e, write an equation for the resulting function if the graph of y 5 x2 is: a. shifted 5 units down and 1.5 units to the left b. stretched vertically by a factor of 4 and shifted 2 units down c. compressed vertically by a factor of and reflected in the x-axis 1 6 d. reflected in the x-axis, shifted 2 units up, and shifted 2 units to the right Solution a. y 5 (x 1 1.5)2 2 5 Answer b. First, stretch vertically by a factor of 4: y 5 4x2 Then, translate the resulting function 2 units down: 1 c. First, compress vertically by a factor of : 6 Then, reflect in the x-axis: d. First, reflect in the x-axis: Then, translate the resulting function 2 units up: Finally, translate the resulting function 2 units to the right: y 5 4x2 2 2 Answer 1 y 6x2 21 y 6x2 y 5 2x2 Answer y 5 2x2 1 2 y 5 2(x 2 2)2 1 2 Answer EXERCISES Writing About Mathematics h(x) 5 1 1. Penny drew the graph of 4x2 2 2x x 4 to x 4. Her graph is shown to the right. Explain why Penny’s graph does not look like a parabola. from y 1 1O x The Graph of a Quadratic Function 519 2. What values of x would you choose to draw the graph of h(x) 5 1 4x2 2 2x so that points on both sides of the turning point would be shown on the graph? Explain your answer. Developing Skills In 3–14: a. Graph each quadratic function on graph paper using the integral values for x indicated in parentheses to prepare the necessary table of values. b. Write the equation of the axis of symmetry of the graph. c. Write the coordinates of the turning point of the graph. 3. y x2 (3 x 3) 5. y x2 1 (3 x 3) 7. y x2 4 (3 x 3) 9. y x2 2x (2 x 4) 11. y x2 4x 3 (1 x 5) 13. y x2 2x 3 (4 x 2) 4. y x2 (3 x 3) 6. y x2 1 (3 x 3) 8. y x2 2x (2 x 4) 10. y x2 6x 8 (0 x 6) 12. y x2 2x 1 (2 x 4) 14. y x2 4x 3 (1 x 5) In 15–20: a. Write the equation of the axis of symmetry of the graph of the function. coordinates of the vertex. three points with integral coefficients on each side of the vertex. b. Find the c. Draw the graph on graph paper or on a calculator, showing at least 15. y x2 6x 1 18. y x2 4x 3 21. Write an equation for the resulting function if the graph of y x2 is: 16. y x2 2x 8 19. y x2 3x 7 17. y x2 8x 12 20. y x2 x 5 a. reflected in the x-axis and shifted 3 units left. b. compressed vertically by a factor of and shifted 9 units up. 2 7 c. reflected in the x-axis, stretched vertically by a factor of 6, shifted 1 unit down, and shifted 4 units to the right. In 22–25, each graph is a translation and/or a reflection of the graph of y x2. For each graph, a. determine the vertex and the axis of symmetry, and b. write the equation of each graph. 22. y 23. y x x 520 Quadratic Relations and Functions 24. y 25. y x x 26. Of the graphs below, which is the graph of a quadratic function and the graph of an absolute value function? (1) y (3) y (2) y x x (4) y x x Applying Skills 27. The length of a rectangle is 4 more than its width. a. If x represents the width of the rectangle, represent the length of the rectangle in terms of x. b. If y represents the area of the rectangle, write an equation for y in terms of x. c. Draw the graph of the equation that you wrote in part b. d. Do all of the points on the graph that you drew represent pairs of values for the width and area of the rectangle? Explain your answer. The Graph of a Quadratic Function 521 28. The height of a triangle is 6 less than twice the length of the base. a. If x represents the length of the base of the triangle, represent the height in terms of x. b. If y represents the area of the triangle, write an equation for y in terms of x. c. Draw the graph of the equation that you wrote in part b. d. Do all of the points on the graph that you drew represent pairs of values for the length of the base and area of the triangle? Explain your answer. 29. The perimeter of a rectangle is 20 centimeters. Let x represent the measure of one side of the rectangle and y represent the area of the rectangle. a. Use the formula for perimeter to express the measure of a second side of the rectangle. b. Write an equation for the area of the rectangle in terms of x. c. Draw the graph of the equation written in b. d. What are the dimensions of the rectangle with maximum area? e. What is the maximum area of the rectangle? f. List four other possible dimensions and areas for the rectangle. 30. A batter hit a baseball at a height 3 feet off the ground, with an initial vertical velocity of 64 feet per second. Let x represent the time in seconds, and y represent the height of the baseball. The height of the ball can be determined over a limited period of time by using the equation y 16x2 64x 3. a. Make a table using integral values of x from 0 to 4 to find values of y. b. Graph the equation. Let one horizontal unit second, and one vertical unit 10 1 4 feet. (See suggested coordinate grid below.) h 100 80 60 40 20 . If the ball was caught after 4 seconds, what was its height when it was caught? d. From the table and the graph, determine: (1) the maximum height reached by the baseball; (2) the time required for the ball to reach this height. 522 Quadratic Relations and Functions 13-3 FINDING ROOTS FROM A GRAPH In Section 1, you learned to find the solution of an equation of the form ax2 bx c 0 by factoring. In Section 2, you learned to draw the graph of a function of the form . How are these similar expressions related? To answer this question, we will consider three possible cases. y 5 ax2 1 bx 1 c CASE 1 A quadratic equation can have two distinct real roots. The equation x2 7x 10 0 has exactly two roots or solutions, 5 and 2, that make the equation true. The function, y x2 7x 10 has infinitely many pairs of numbers that make the equation true. The graph of this function shown at the right intersects the x-axis in two points, (5, 0) and (2, 0). Since the y-coordinates of these points are 0, the x-coordinates of these points are the roots of the equation x2 7x 10 0. y 1 O –1 – CASE 2 A quadratic equation can have only one distinct real root. The two roots of the equation x2 6x 9 0 are equal.There is only one number, 3, that makes the equation true. The function y x2 6x 9 has infinitely many pairs of numbers that make the equation true. The graph of this function shown at the right intersects the x-axis in only one point, (3, 0). Since the y-coordinate of this point is 0, the x-coordinate of this point is the root of the equation x2 6x 9 0. y 1 –1 – Recall from Section 1 that when a quadratic equation has only one distinct root, the root is said to be a double root. In other words, when the root of a quadratic equation is a double root, the graph of the corresponding quadratic function intersects the x-axis exactly once. CASE 3 A quadratic equation can have no real roots. Finding Roots from a Graph 523 The equation x2 2x 3 0 has no real roots. There is no real number that makes the equation true. The function, y x2 2x 3 has infinitely many pairs of numbers that make the equation true. The graph of this function shown at the right does not intersect the x-axis. There is no point on the graph whose y-coordinate is 0. Since there is no point whose y-coordinate is 0, there are no real numbers that are roots of the equation x2 2x 3 01 –1 O 1 x The equation of the x-axis is y 0. The x-coordinates of the points at which the graph of y ax2 bx c intersects the x-axis are the roots of the equation ax2 bx c 0. The graph of the function y ax2 bx c can intersect the x-axis in 0, 1, or 2 points, and the equation ax2 bx c 0 can have 0, 1, or 2 real roots. A real number k is a root of the quadratic equation ax2 bx c 0 if and only if the graph of y ax2 bx c intersects the x-axis at (k, 0). EXAMPLE 1 Use the graph of y x2 5x 6 to find the roots of x2 5x 6 0. y Solution The graph intersects the x-axis at (2, 0) and (3, 0). The x-coordinates of these points are the roots of x2 5x 6 0. Answer The roots are 2 and 3. The solution set is {2, 3}. 1 –1 1 Note that in Example 1 the quadratic expression x2 5x 6 can be factored into (x 2)(x 3), from which we can obtain the solution set 2, 3 . 6 5 524 Quad
ratic Relations and Functions EXAMPLE 2 Use the graph of y x2 3x 4 to find the linear factors of x2 1 3x 2 4 . y Solution The graph intersects the x-axis at (4, 0) and (1, 0). Therefore, the roots are 4 and 1. If x 4, then x (4) (x 4) is a factor. If x 1, then (x 1) is a factor. O 1 –1 –1 x Answer The linear factors of x2 3x 4 are (x 4) and (x 1). y = x2 + 3x – 4 EXERCISES Writing About Mathematics 1. The coordinates of the vertex of a parabola y x2 2x 5 are (1, 4). Does the equation x2 2x 5 0 have real roots? Explain your answer. 2. The coordinates of the vertex of a parabola y x2 2x 3 are (1, 4). Does the equa- tion x2 2x 3 0 have real roots? Explain your answer. Developing Skills In 3–10: a. Draw the graph of the parabola. b. Using the graph, find the real numbers that are elements of the solution set of the equation. c. Using the graph, factor the corresponding quadratic expression if possible. 3. y x2 6x 5; 0 x2 6x 5 5. y x2 2x 3; 0 x2 2x 3 7. y x2 2x 1; 0 x2 2x 1 9. y x2 4x 5; 0 x2 4x 5 4. y x2 2x 1; 0 x2 2x 1 6. y x2 x 2; 0 x2 x 2 8. y x2 3x 2; 0 x2 3x 2 10. y x2 5x 6; 0 x2 5x 6 11. If the graph of a quadratic function, f(x), crosses the x-axis at x 6 and x 8, what are two factors of f(x)? 12. If the factors of a quadratic function, h(x), are (x 6) and (x 3), what is the solution set for the equation h(x) 0? Graphic Solution of a Quadratic-Linear System 525 In 13–15, refer to the graph of the parabola shown below. 13. Which of the following is the equation of the parabola? (1) y (x 2)(x 3) (2) y (x 2)(x 3) (3) y (x 2)(x 3) (4) y (x 2)(x 3) y 1 14. Set the equation of the parabola equal to 0. What are the 1O x roots of this quadratic equation? 15. If the graph of the parabola is reflected in the x-axis, how many roots will its corresponding quadratic equation have? 13-4 GRAPHIC SOLUTION OF A QUADRATIC-LINEAR SYSTEM In Chapter 10 you learned how to solve a system of linear equations by graphing. For example, the graphic solution of the given system of linear equations is shown below. y 5 21 2x 1 4 y 2x 1 Since the point of intersection, (2, 3), is a solution of both equations, the common solution of the system is x 2 and y 3. A quadratic-linear system consists of a quadratic equation and a linear equation. The solution of a quadratic-linear system is the set of ordered pairs of numbers that make both equations true. As shown below, the line may intersect the curve in two, one, or no points. Thus the solution set may contain two ordered pairs, one ordered pair, or no ordered pairs2, 3) y = ––x + 4 2 1 O 1 x Two points of intersection One point of intersection No point of intersection 526 Quadratic Relations and Functions EXAMPLE 1 Solve the quadratic-linear system graphically: y x2 6x 6 y x 4 Solution (1) Draw the graph of y x2 6x 6. The axis of symmetry of a parabola is x . Therefore, the axis of symme- 2b 2a try for the graph of y x2 6x 6 is x or x 3. Make a table of values using integral values of x less than 3 and greater than 3. Plot the points associated with each pair (x, y) and join them with a smooth curve. 2(26) 2(1) x 0 1 2 3 4 5 6 x2 6x 6 0 0 6 1 6 6 4 12 6 9 18 6 16 24 6 25 30 6 36 36 2) On the same set of axes, draw the graph of y x 4. Make a table of val- ues and plot the points2, –2) y = x – 4 (5, 1) x Graphic Solution of a Quadratic-Linear System 527 The line could also have been graphed by using the y-intercept, 4. Starting at the point (0, 4) move 1 unit up and 1 unit to the right to locate a second point. Then again, move 1 unit up and 1 unit to the right to locate a third point. Draw a line through these points. , and the slope 5 1 or 1 1 (3) Find the coordinates of the points at which the graphs intersect. The graphs intersect at (2, 2) and at (5, 1). Check each solution in each equation. Four checks are required in all. The checks are left for you. Calculator Solution (1) Enter the equations into the Y= menu. ENTER: Y X,T,,n x2 6 DISPLAY: X,T,,n X,T,,2) Calculate the first intersection by choosing intersect from the CALC menu. Accept Y1 X2 6X 6 as the first curve and Y2 X 4 as the second curve. Then press enter when the screen prompts “Guess?”. 2nd CALC 5 ENTER ENTER ENTER ENTER: DISPLAY: * Y = –2 intersection X = 2 The calculator displays the coordinates of the intersection point at the bottom of the screen. (3) To calculate the second intersection point, repeat the process from (2), but when the screen prompts “Guess?” move the cursor with the left and right arrow keys to the approximate position of the second intersection point. * Answer The solution set is {(2, 2), (5, 1)}. intersection X = 5 Y = 1 528 Quadratic Relations and Functions EXERCISES Writing About Mathematics 1. What is the solution set of a system of equations when the graphs of the equations do not intersect? Explain your answer. 2. Melody said that the equations y x2 and y 2 do not have a common solution even before she drew their graphs. Explain how Melody was able to justify her conclusion. Developing Skills In 3–6, use the graph on the right to find the common solution of the system. 3. y x2 2x 3 y 0 4. y x2 2x 3 y 3 5. y x2 2x 3 y 4 6. y x2 2x 1 –1 x 7. For what values of c do the equations y x2 2x 3 and y c have no points in common? 8. a. Draw the graph of y x2 4x 2, in the interval 1 x 5. b. On the same set of axes, draw the graph of y x 2. c. Write the coordinates of the points of intersection of the graphs made in parts a and b. d. Check the common solutions found in part c in both equations. In 9–16, find graphically and check the solution set of each system of equations. 9. y x2 y x 2 11. y x2 2x 1 y 2x 5 13. y x2 8x 15 x y 5 15. y x2 4x 1 y 2x 1 10. y x2 2x 4 y x 12. y 4x x2 y x 4 14. y x2 6x 5 y 3 16. y x2 x 4 2x y 2 Algebraic Solution of a Quadratic-Linear System 529 Applying Skills 17. When a stone is thrown upward from a height of 5 feet with an initial velocity of 48 feet per second, the height of the stone y after x seconds is given by the function y 5 216x2 1 48x 1 5 . a. Draw a graph of the given function. Let each horizontal unit equal second and each vertical unit equal 5 feet. Plot points for 0, , 2, b. On the same set of axes, draw the graph of y 25. c. From the graphs drawn in parts a and b, determine when the stone is at a height of , 1, 1 2 3 2 5 2 1 2 , and 3 seconds. 25 feet. 18. If you drop a baseball on Mars, the gravity accelerates the baseball at 12 feet per second squared. Let’s suppose you drop a baseball from a height of 100 feet. A formula for the height, y, of the baseball after x seconds is given by y 6x2 100. a. On a calculator, graph the given function. Set Xmin to –10, Xmax to 10, Xscl to 1, Ymin to 5, Ymax to 110, and Yscl to 10. b. Graph the functions y 46 and y 0 as Y2 and Y3. c. Using a calculator method, determine, to the nearest tenth of a second, when the baseball has a height of 46 feet and when the baseball hits the ground (reaches a height of 0 feet). 13-5 ALGEBRAIC SOLUTION OF A QUADRATIC-LINEAR SYSTEM = x2 – 4 x In the last section, we learned to solve a quadratic-linear system by finding the points of intersection of the graphs. The solutions of most of the systems that we solved were integers that were easy to read from the graphs. However, not all solutions are integers. For example, the graphs of y x2 4x 3 and 2x 1 1 are shown at the right. They intersect in two points. One of those points, (4, 3), has integral coordinates and can be read easily from the graph. However, the coordinates of the other point are not integers, and we are not able to identify the exact values of x and y from the graph. y 5 1 530 Quadratic Relations and Functions In Chapter 10 we learned that a system of linear equations can be solved by an algebraic method of substitution. This method can also be used for a quadratic-linear system. The algebraic solution of the system graphed on the previous page is shown in Example 1. EXAMPLE 1 Solve algebraically and check: y x2 4x 3 y 5 1 2x 1 1 Solution (1) Since y is expressed in terms of x in the linear equation, substitute 1 2x 1 1 the expression for y in the quadratic equation to form an equation in one variable: y x2 4x 3 1 2x 1 1 x2 4x 3 (2) To eliminate fractions as coefficients, multiply both sides of the equation by 2: 2 A 1 2x 1 1 B (3) Write the quadratic equation in standard form: (4) Solve the quadratic equation by factoring: (5) Set each factor equal to 0 and solve for x. (6) Substitute each value of x in the linear equation to find the corresponding value of y: (7) Write each solution as coordinates: 2(x2 4x 3) x 2 2x2 8x 6 0 2x2 9x 4 0 (2x 1)(x 4) 2x 1 0 2x 1 x 5 1 2 1 2x x, y) y y y y 1 2 2x 1 1 1 2(4x, y) (4, 3) (8) Check each ordered pair in each of the given equations: Check for x2 4x 3 5 1 4 5? 2 A A 5 4 5 2x 1 1 5 1 4 5? 1 2 2 B A 5 4 5 Algebraic Solution of a Quadratic-Linear System 531 Check for x 4, y 3 y x2 4x 3 3 5? (4)2 2 4(4) 1 3 3 5? 16 2 16 1 3 ✔ 3 5 3 y 3 5? 1 1 2x 1 1 2(4) 1 1 3 5? 2 1 1 ✔ 3 5 3 Answer EXAMPLE 2 2, 5 1 4 B U A , (4, 3) V The length of the longer leg of a right triangle is 2 units more than twice the length of the shorter leg. The length of the hypotenuse is 13 units. Find the lengths of the legs of the triangle. Solution Let a the length of the longer leg and b the length of the shorter leg. (1) Use the Pythagorean Theorem to write an equation: (2) Use the information in the first sentence of the problem to write another equation: (3) Substitute the expression for a from step 2 in the equation in step 1: (4) Square the binomial and write the equation in standard form: (5) Factor the left member of the equation: a2 b2 (13)2 a2 b2 169 a 2b 2 a2 b2 169 (2b 2)2 b2 169 4b2 8b 4 b2 169 5b2 8b 165 0 (5b 33)(b 5) 0 (6) Set each factor equal to 0 and solve 5b 33 0 for b: (7) For each value of b, find the value of a. Use the linear equation in step 2: (8) Reject the negative values. Use the pair of positive values to write the answer. a Answer The lengths of the legs are 12 units and 5 units. 5b 33 233 b
5 a 2b 2 b 5 0 b 5 b 5 a 2b 2 a 2 233 5 A 256 5 1 2 B a 2(5) 2 a 12 532 Quadratic Relations and Functions EXERCISES Writing About Mathematics 1. Explain why the equations y x2 and y 4 have no common solution in the set of real numbers. 2. Explain why the equations x2 y2 49 and x 8 have no common solution in the set of real numbers. 3. y x2 2x Developing Skills In 3–20, solve each system of equations algebraically and check all solutions. 4. y x2 5 y x 5 7. x2 y 9 y x 9 y x 1 8. x2 y 2 y x 6. y x2 2x 1 y x 5. y x2 4x 3 y x 3 9. x2 2y 5 y x 1 12. y 3x2 8x 5 x y 3 15. x2 y2 25 x 2y 5 18. x2 y2 40 y 2x 2 Applying Skills 10. y 2x2 6x 5 11. y 2x2 2x 3 y x 2 13. y x2 3x 1 y 5 1 3x 1 2 16. x2 y2 100 y x 2 19. x2 y2 20 y x 2 y x 3 14. y x2 6x 8 y 5 21 2x 1 2 17. x2 y2 50 x y 20. x2 y2 2 y x 2 21. A rectangular tile pattern consists of a large square, two small squares, and a rectangle arranged as shown in the diagram. The height of the small rectangle is 1 unit and the area of the tile is 70 square units. x a. Write a linear equation that expresses the relationship between the length of the sides of the quadrilaterals that make up the tile. b. Write a second-degree equation using the sum of the areas of the quadrilaterals that make up the tile. c. Solve algebraically the system of equations written in parts a and b. d. What are the dimensions of each quadrilateral in the tile? y 1 y " " Chapter Summary 533 22. A doorway to a store is in the shape of an arch whose equation is h 5 23 , where x represents the horizontal distance, in feet, from the left end of the base of the doorway and h is the height, in feet, of the doorway x feet from the left end of the base. 4x2 1 6x h a. How wide is the doorway at its base? b. What is the maximum height of the doorway? (0, 0) x c. Can a box that is 6 feet wide, 6 feet long, and 5 feet high be moved through the door- way? Explain your answer. 23. The length of the diagonal of a rectangle is 85 meters. The length of the rectangle is 1 meter longer than the width. Find the dimensions of the rectangle. 24. The length of one leg of an isosceles triangle is 29 feet. The length of the altitude to the base of the triangle is 1 foot more than the length of the base a. Let a the length of the altitude to the base and b the distance from the vertex of a base angle to the vertex of the right angle that the altitude makes with the base. Use the Pythagorean Theorem to write an equation in terms of a and b. b. Represent the length of the base in terms of b. c. Represent the length of the altitude, a, in terms of b. d. Solve the system of equations from parts a and c. e. Find the length of the base and the length of the altitude to the base. f. Find the perimeter of the triangle. g. Find the area of the triangle. CHAPTER SUMMARY The equation y ax2 bx c, where a 0, is a quadratic function whose domain is the set of real numbers and whose graph is a parabola. The axis of symmetry of the parabola is the vertical line x . The vertex or turning point of the parabola is on the axis of symmetry. If a 0, the parabola opens upward and the y-value of the vertex is a minimum value for the range of the function. If a 0, the parabola opens downward and the y-value of the vertex is a maximum value for the range of the function. 2b 2a A quadratic-linear system consists of two equations one of which is an equation of degree two and the other a linear equation (an equation of degree one). The common solution of the system may be found by graphing the equations on the same set of axes or by using the algebraic method of substitution. A quadratic-linear system of two equations may have two, one, or no common solutions. 534 Quadratic Relations and Functions The roots of the equation ax2 bx c 0 are the x-coordinates of the points at which the function y ax2 bx c intersects the x-axis. The real number k is a root of ax2 bx c 0 if and only if (x k) is a factor of ax2 bx c. When the graph of y x2 is translated by k units in the vertical direction, the equation of the image is y x2 k. When the graph of y x2 is translated k units in the horizontal direction, the equation of the image is y (x k)2. When the graph of y x2 is reflected over the x-axis, the equation of the image is y x2. The graph of y kx2 is the result of stretching the graph of y x2 in the vertical direction when k 1 or of compressing the graph of y x2 when 0 k 1. VOCABULARY 13-1 Standard form • Polynomial equation of degree two • Quadratic equation • Roots of an equation • Double root 13-2 Parabola • Second-degree polynomial function • Quadratic function • Minimum • Turning point • Vertex • Axis of symmetry of a parabola • Maximum 13-4 Quadratic-linear system REVIEW EXERCISES 1. Explain why x y2 is not a function when the domain and range are the set of real numbers. 2. Explain why x y2 is a function when the domain and range are the set of positive real numbers. In 3–6, the set of ordered pairs of a relation is given. For each relation, a. list the c. determine if the elements of the domain, b. list the elements of the range, relation is a function. 3. {(1, 3), (2, 2), (3, 1), (4, 0), (5, 1)} 4. {(1, 3), (1, 2), (1, 1), (1, 0), (1, 1)} 5. {(3, 9), (2, 4), (1, 1), (0, 0), (1, 1), (2, 4), (3, 9)} 6. {(0, 1), (1, 1), (2, 1), (3, 1), (4, 1)} In 7–10, for each of the given functions: a. Write the equation of the axis of symmetry. b. Draw the graph. c. Write the coordinates of the turning point. Review Exercises 535 d. Does the function have a maximum or a minimum? e. What is the range of the function? 7. y x2 6x 6 9. f(x) x2 2x 6 8. y x2 4x 1 10. f(x) x2 6x 1 In 11–16, solve each system of equations graphically and check the solution(s) if they exist. 11. y x2 6 x y 6 14. y x2 4x x y 4 y x 16. y 2x x2 x y 2 y x 5 15. y 5 x2 y 4 12. y x2 2x 1 13. y x2 x 3 In 17–22, solve each system of equations algebraically and check the solutions. 18. y x2 4x 9 17. x2 y 5 y 1 2x y 3x 1 21. x2 y2 40 20. y x2 6x 5 y 3x y x 1 19. x2 2y 11 y x 4 22. x2 y2 5 y 5 1 2x y 5 x2 23. Write an equation for the resulting function if the graph of is shifted 3 units up, 2.5 units to the left, and is reflected over the x-axis. 24. The sum of the areas of two squares is 85. The length of a side of the larger square minus the length of a side of the smaller square is 1. Find the length of a side of each square. 25. Two square pieces are cut from a rectangular piece of carpet as shown in the diagram. The area of the original piece is 144 square feet, and the width of the small rectangle that is left is 2 feet. Find the dimensions of the original piece of carpet. Exploration Write the square of each integer from 2 to 20. Write the prime factorization of each of these squares. What do you observe about the prime factorization of each of these squares? Let n be a positive integer and a, b, and c be prime numbers. If the prime rational or irra- , is n a perfect square? Is n 5 a3 3 b2 3 c4 n factorization of tional? Express in terms of a, b, and c. n " " 536 Quadratic Relations and Functions CUMULATIVE REVIEW CHAPTERS 1–13 Part I Answer all questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. 1. When a 7, 5 2a2 equals (1) 147 (2) 93 (3) 103 (4) 147 2. Which expression is rational? p (1) 1 3 $ 3. When factored completely, 3x2 75 can be expressed as 2 8 $ (2) (3) (4) 0.4 " (1) (3x 15)(x 5) (2) (x 5)(3x 15) (3) 3x(x 5)(x 5) (4) 3(x 5)(x 5) 4. When b2 4b is subtracted from 3b2 3b the difference is (1) 3 7b (2) 2b2 7b 5. The solution set of the equation 0.5x 4 2x 0.5 is (3) 2b2 7b (4) 2b2 b (1) {3} (2) {1.4} (3) {30} (4) {14} 6. Which of these represents the quadratic function y x2 5 shifted 2 units down and 4 units to the right? (1) y (x 4)2 2 (2) y (x 4)2 3 (3) y (x 4)2 7 (4) y (x 2)2 9 7. The slope of the line whose equation is x 2y 4 is (1) 2 (2) 2 (3) 1 2 (4) 21 2 8. The solution set of the equation x2 7x 10 0 is (1) {–2, 5} (2) {2, 5} (3) (x 2)(x 5) (4) (x 2)(x 5) 9. Which of these shows the graph of a linear function intersecting the graph of an absolute value function? (1) y (2) y x x (3) y (4) y Cumulative Review 537 x x 10. In ABC, mA 72 and mB 83. What is the measure of C? (1) 155° (2) 108° (3) 97° (4) 25° Part II Answer all questions in this part. Each correct answer will receive 2 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 11. In a class of 300 students, 242 take math, 208 take science, and 183 take both math and science. How many students take neither math nor science? 12. Each of the equal sides of an isosceles triangle is 3 centimeters longer than the base. The perimeter of the triangle is 54 centimeters, what is the measure of each side of the triangle? Part III Answer all questions in this part. Each correct answer will receive 3 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 13. The base of a right circular cylinder has a diameter of 5.00 inches. Sally measured the circumference of the base of the cylinder and recorded it to be 15.5 inches. What is the percent of error in her measurement? Express your answer to the nearest tenth of a percent. 14. Solve the following system of equations and check your solutions. y x2 3x 1 y x 1 538 Quadratic Relations and Functions Part IV Answer all questions in this part. Each correct answer will receive 4 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 15. Find to the nearest degree the measure of the acute angle that the graph of y 2x 4 makes with the x-axis. 1
6. Jean Forester has a small business making pies and cakes. Today, she must make at least 4 cakes to fill her orders and at least 3 pies. She has time to make a total of no more than 10 pies and cakes. a. Let x represent the number of cakes that Jean makes and y represent the number of pies. Write three inequalities that can be used to represent the number of pies and cakes that she can make. b. In the coordinate plane, graph the inequalities that you wrote and indi- cate the region that represents their common solution. c. Write at least three ordered pairs that represent the number of pies and cakes that Jean can make. ALGEBRAIC FRACTIONS,AND EQUATIONS AND INEQUALITIES INVOLVING FRACTIONS Although people today are making greater use of decimal fractions as they work with calculators, computers, and the metric system, common fractions still surround us. 1 4 1 2 11 3 -yard gain in football, pint of cream, We use common fractions in everyday measures: 1 21 2 2 1 -inch 4 cups of flour. nail, We buy dozen eggs, not 0.5 dozen eggs. We describe 15 minutes as hour, not 0.25 hour. Items are sold at a third off, or at a fraction of the original price. B Fractions are also used when sharing. For example, Andrea designed some beautiful Ukrainian eggs this year. She gave onefifth of the eggs to her grandparents.Then she gave one-fourth of the eggs she had left to her parents. Next, she presented her aunt with one-third of the eggs that remained. Finally, she gave one-half of the eggs she had left to her brother, and she kept six eggs. Can you use some problem-solving skills to discover how many Ukrainian eggs Andrea designed? 1 3 A CHAPTER 14 CHAPTER TABLE OF CONTENTS 14-1 The Meaning of an Algebraic Fraction 14-2 Reducing Fractions to Lowest Terms 14-3 Multiplying Fractions 14-4 Dividing Fractions 14-5 Adding or Subtracting Algebraic Fractions 14-6 Solving Equations with Fractional Coefficients 14-7 Solving Inequalities with Fractional Coefficients 14-8 Solving Fractional Equations Chapter Summary Vocabulary Review Exercises Cumulative Review In this chapter, you will learn operations with algebraic fractions and methods to solve equations and inequalities that involve fractions. 539 540 Algebraic Fractions, and Equations and Inequalities Involving Fractions 14-1 THE MEANING OF AN ALGEBRAIC FRACTION A fraction is a quotient of any number divided by any nonzero number. For example, the arithmetic fraction indicates the quotient of 3 divided by 4. An algebraic fraction is a quotient of two algebraic expressions. An algebraic fraction that is the quotient of two polynomials is called a fractional expression or a rational expression. Here are some examples of algebraic fractions that are rational expressions: 3 4 x 2 2 x 4c 3d x 1 5 x 2 2 x2 1 4x 1 3 x 1 1 a b The fraction means that the number represented by a, the numerator, is to be divided by the number represented by b, the denominator. Since division by 0 is not possible, the value of the denominator, b, cannot be 0. An algebraic fraction is defined or has meaning only for values of the variables for which the denominator is not 0. EXAMPLE 1 Find the value of x for which 12 x 2 9 is not defined. Solution The fraction 12 x 2 9 is not defined when the denominator, x 9, is equal to 0. x 9 0 x 9 Answer EXERCISES Writing About Mathematics 1. Since any number divided by itself equals 1, the solution set for 1 is the set of all real x x numbers. Do you agree with this statement? Explain why or why not. 2. Aaron multiplied by (equal to 1) to obtain the fraction b2 2 b b 1 1 . Is the fraction b 1 1 1 b equal to the fraction for all values of b? Explain your answer. b b b 1 1 1 b b2 2 b b 1 1 Developing Skills In 3–12, find, in each case, the value of the variable for which the fraction is not defined. 3. 2 x 8. y 1 5 y 1 2 4. 9. 25 6x 10 2x 2 1 5. 10. 12 y2 2y 1 3 4y 1 2 6. 1 x 2 5 11. 1 x2 2 4 7. 7 2 2 x 12. 3 x2 2 5x 2 14 Reducing Fractions to Lowest Terms 541 Applying Skills In 13–17, represent the answer to each problem as a fraction. 13. What is the cost of one piece of candy if five pieces cost c cents? 14. What is the cost of 1 meter of lumber if p meters cost 980 cents? 15. If a piece of lumber 10x 20 centimeters in length is cut into y pieces of equal length, what is the length of each of the pieces? 16. What fractional part of an hour is m minutes? 17. If the perimeter of a square is 3x 2y inches, what is the length of each side of the square? 14-2 REDUCING FRACTIONS TO LOWEST TERMS A fraction is said to be reduced to lowest terms or is a lowest terms fraction when its numerator and denominator have no common factor other than 1 or 1. Each of the fractions 5 10 The arithmetic fraction and 5 10 a 2a 1 can be expressed in lowest terms as . 2 is reduced to lowest terms when both its numer- ator and denominator are divided by 5: The algebraic fraction a 2a tor and denominator are divided by a, where a 0: 5 10 5 5 4 5 10 4 5 5 1 2 is reduced to lowest terms when both its numera- a 2a 5 a 4 a Fractions that are equal in value are called equivalent fractions. Thus, 2a 4 a 5 1 2 5 10 and 1 2 are equivalent fractions, and both are equivalent to a 2a , when a 0. The examples shown above illustrate the division property of a fraction: if the numerator and the denominator of a fraction are divided by the same nonzero number, the resulting fraction is equal to the original fraction. In general, for any numbers a, b, and x, where b 0 and x 0: bx 5 ax 4 x ax bx 4 x 5 a b When a fraction is reduced to lowest terms, we list the values of the variables that must be excluded so that the original fraction is equivalent to the reduced form and also has meaning. For example: 5x 4 x 5 4 5 5x 5 4x 4 x 4x cy 4 y dy 4 y 5 c d cy dy 5 (where x 0) (where y 0, d 0) 542 Algebraic Fractions, and Equations and Inequalities Involving Fractions When reducing a fraction, the division of the numerator and the denomina- tor by a common factor may be indicated by a cancellation. Here, we use cancellation to divide the numerator and the denominator by 3: 3(x 1 5) 18 5 1 3(x 1 5) 18 5 x 1 5 6 6 Here, we use cancellation to divide the numerator and the denominator by (a 2 3) : 1 a2 2 9 3a 2 9 5 (a 2 3)(a 1 3) 3(a 2 3) 5 a 1 3 3 1 (where a 3) By re-examining one of the examples just seen, we can show that the multi- plication property of one is used whenever a fraction is reduced: 3(x 1 5) 18 5 3 ? (x 1 5) 3 ? 6 5 3 3 ? (x 1 5) 6 5 1 ? (x 1 5) 6 5 x 1 5 6 However, when the multiplication property of one is applied to fractions, it is referred to as the multiplication property of a fraction. In general, for any numbers a, b, and x, where b 0 and x 0 Procedure To reduce a fraction to lowest terms: METHOD 1 1. Factor completely both the numerator and the denominator. 2. Determine the greatest common factor of the numerator and the denominator. 3. Express the given fraction as the product of two fractions, one of which has as its numerator and its denominator the greatest common factor determined in step 2. 4. Use the multiplication property of a fraction. METHOD 2 1. Factor both the numerator and the denominator. 2. Divide both the numerator and the denominator by their greatest common factor. Reducing Fractions to Lowest Terms 543 EXAMPLE 1 Reduce 15x2 35x4 to lowest terms. Solution METHOD 1 15x2 5x2 35x4 5 3 7x2 ? 5x2 5 3 7x2 ? 1 5 3 7x2 Answer 3 7x2 (x 0) EXAMPLE 2 METHOD 2 15x2 35x4 5 3 ? 5x2 ? 5x2 7x2 1 5 3 ? 5x2 7x2 1 ? 5x2 5 3 7x2 Express 2x2 2 6x 10x as a lowest terms fraction. Solution METHOD 1 2x2 2 6x 10x 5 2x(x 2 3) 2x ? 5 METHOD 2 2x2 2 6x 10x 5 2x(x 2 3) 10x 5 2x 2x ? 5 1 ? (x 2 3) 5 (x 2 3) 5 1 2x(x 2 3) 10x Answer x 2 3 5 (x 0) EXAMPLE 3 Reduce each fraction to lowest terms. a. x2 2 16 x2 2 5x 1 4 Solution a. Use Method 1: x2 2 16 x2 2 5x 1 4 5 (x 1 4)(x 2 4) (x 2 1)(x 2 4. 2 2 x 4x 2 8 b. Use Method 2: 2 2 x 4x 2 8 5 21(x 2 2) 4(x 2 2) 1 5 21(x 2 2) 4(x 2 2) 1 5 21 4 Answers a. x 1 4 x 2 1 (x 1, x 4) b. 21 4 (x 2) 544 Algebraic Fractions, and Equations and Inequalities Involving Fractions EXERCISES Writing About Mathematics 1. Kevin used cancellation to reduce a 1 4 a 1 8 Kevin’s work? to lowest terms as shown below. What is the error in . Kevin let a 4 to prove that when reduced to lowest terms . Explain to Kevin why his reasoning is incorrect. 1 2 Developing Skills In 3–54, reduce each fraction to lowest terms. In each case, list the values of the variables for which the fractions are not defined. 3. 4x 12x 7. ab cb 11. 15x2 5x 15. 112a2b 28ac 19. 3x 1 6 4 23. 2ax 1 2bx 6x2 27. 18b2 1 30b 9b3 31. 2a2 6a2 2 2ab 35. x2 2 1 5x 2 5 39. 16 2 a2 2a 2 8 43. x2 1 7x 1 12 x2 2 16 47. x2 2 25 x2 2 2x 2 15 51. r2 2 4r 2 5 r2 2 2r 2 15 4. 8. 12. 16. 20. 24. 27y2 36y 3ay2 6by2 5x2 25x4 220x2y2 290xy2 8y 2 12 6 5a2 2 10a 5a2 28. 4x 4x 1 8 32. 14 7r 2 21s 36. 1 2 x x 2 1 40. 44. 48. 52. x2 2 y2 3y 2 3x x2 1 x 2 2 x2 1 4x 1 4 a2 2 a 2 6 a2 2 9 48 1 8x 2 x2 x2 1 x 2 12 5. 9. 13. 17. 24c 36d 5xy 9xy 27a 36a2 232a3b3 148a3b3 21. 5x 2 35 5x 25. 12ab 2 3b2 3ab 29. 7d 7d 1 14 33. 12a 1 12b 3a 1 3b 37. 3 2 b b2 2 9 41. 45. 49. 53. 2b(3 2 b) b2 2 9 3y 2 3 y2 2 2y 1 1 a2 2 6a a2 2 7a 1 6 2x2 2 7x 1 3 (x 2 3)2 6. 9r 10r 10. 2abc 4abc 14. 18. 8xy2 24x2y 5xy 45x2y2 22. 8m2 1 40m 8m 26. 30. 6x2y 1 9xy2 12xy 5y 5y 1 5x 34. x2 2 9 3x 1 9 38. 2s 2 2r s2 2 r2 42. r2 2 r 2 6 3r 2 9 46. x2 2 3x x2 2 4x 1 3 50. 2x2 2 50 x2 1 8x 1 15 54. x2 2 7xy 1 12y2 x2 1 xy 2 20y2 Multiplying Fractions 545 55. a. Use substitution to find the numerical value of x2 2 5x x 2 5 , then reduce each numerical frac- tion to lowest terms when: (1) x 7 (4) x 2 (2) x 10 (5) x 4 (3) x 20 (6) x 10 b. What pattern, if any, do you observe for the answers to part a? c. Can substitution be used to evaluate x2 2 5x x 2 5 d. Reduce the algebraic fraction x2 2 5x x 2 5 to lowest terms. when x 5? Explain your answer. e. Using the answer to part d, find the value of x 38,756. x2 2 5x x 2 5 , reduced to lowest terms, when f. If the fraction is multiplied by x2 2
5x x 2 5 ? Explain your answer. x2 2 5x x 2 5 x x to become x(x2 2 5x) x(x 2 5) , will it be equivalent to 14-3 MULTIPLYING FRACTIONS The product of two fractions is a fraction with the following properties: 1. The numerator is the product of the numerators of the given fractions. 2. The denominator is the product of the denominators of the given fractions. In general, for any numbers a, b, x, and y, when b 0 and y 0: a b ? y 5 ax x by We can find the product of 7 27 and 9 4 in lowest terms by using either of two methods. METHOD 1. 7 27 3 4 5 7 3 9 9 27 3 4 5 63 108 5 7 3 9 12 3 9 5 7 12 3 9 9 5 7 12 3 1 5 7 12 METHOD 2. 27 3 9 7 4 5 7 27 3 1 3 9 4 5 7 12 Notice that Method 2 requires less computation than Method 1 since the reduced form of the product was obtained by dividing the numerator and the denominator by a common factor before the product was found. This method may be called the cancellation method. The properties that apply to the multiplication of arithmetic fractions also apply to the multiplication of algebraic fractions. 546 Algebraic Fractions, and Equations and Inequalities Involving Fractions To multiply and express the product in lowest terms, we may use 5x2 7y by 14y2 15x3 either of the two methods. In this example, x 0 and y 0. METHOD 1. 5x2 7y ? 14y2 15x3 5 5x2 ? 14y2 7y ? 15x3 5 70x2y2 105x3y 5 2y 3x ? 35x2y 35x2y 5 2y 3x ? 1 5 2y 3x METHOD 2. 5x2 7x ? 14y2 15x3 5 5x2 1 7y 1 ? 2y 14y2 15x3 3x 5 2y 3x (the cancellation method) While Method 1 is longer, it has the advantage of displaying each step as a property of fractions. This can be helpful for checking work. Procedure To multiply fractions: METHOD 1 1. Multiply the numerators of the given fractions. 2. Multiply the denominators of the given fractions. 3. Reduce the resulting fraction, if possible, to lowest terms. METHOD 2 1. Factor any polynomial that is not a monomial. 2. Use cancellation to divide a numerator and a denominator by each common factor. 3. Multiply the resulting numerators and the resulting denominators to write the product in lowest terms. EXAMPLE 1 Multiply and express the product in reduced form: 5a3 9bx ? 6bx a2 Solution METHOD 1 (1) Multiply the numerators and denominators of the given fractions: (2) Reduce the resulting fraction to lowest terms: 5a3 9bx ? 6bx a2 5 5a3 ? 6bx 9bx ? a2 5 30a3bx 9a2bx 5 10a 3 ? 3a2bx 3a2bx 5 10a 3 ? 1 5 10a 3 Multiplying Fractions 547 METHOD 2 (1) Divide the numerators and denominators by the common factors 3bx and a2: 5a3 9bx ? a 6bx a2 5 5a3 9bx 3 ? 2 6bx a2 1 (2) Multiply the resulting numerators and the resulting denominators: 5 10a 3 Answer 10a 3 (a 0, b 0, x 0) EXAMPLE 2 Multiply and express the product in simplest form: 12a ? 3 8a Solution Think of 12a as 12a 1 . 12a ? 8a 5 36a 8a 5 12a 3 3 1 ? 8a 9 9 2 5 9 5 4a 4a ? 2 5 1 ? 2 Answer 9 2 (a 0) EXAMPLE 3 Multiply and simplify the product: x2 2 5x 1 6 3x ? 2 4x 2 12 Solution x2 2 5x 1 6 3x ? 2 4x 2 12 5 1 (x23) (x 2 2) 3x 1 2 (x 2 3) 1 ? 4 2 5 x 2 2 6x Answer x 2 2 6x (x 0, 3) EXERCISES Writing About Mathematics 1. When reduced to lowest terms, a fraction whose numerator is x2 3x 2 equals 1. What is the denominator of the fraction? Explain your answer. 2. Does x2 xz 1 z2 ? x2 2 z2 x2 2 xz 5 x z for all values of x and z? Explain your answer. 548 Algebraic Fractions, and Equations and Inequalities Involving Fractions Developing Skills In 3–41, find each product in lowest terms. In each case, list any values of the variable for which the fractions are not defined. 3. 8. 13. 18. 30a 36 8 12 ? mn ? 30m2 18n ? 12a 2 4 b 8 m2n2 6n 5m ? b3 12 21. 2r r 2 1 ? r 2 1 10 24. 1 x2 2 1 ? 2x 1 2 6 27. 30. 33. 36. 39. a(a 2 b)2 4b ? 4b a(a2 2 b2) ? 3y x 1 1 ? x2 1 6x 1 5 9y2 x2 2 25 4x2 2 9 8x 2x2 2 8 b2 1 81 b2 2 81 ? ? 2x 1 3 x 2 5 8x 1 16 32x2 81 2 b2 81 1 b2 4. 9. 14. 36 ? 5y 9y 14y 24x 35y ? 8x 24a3b2 7c3 ? 21c2 12ab 5. 10. 15. 1 2 ? 20x 12x 5y ? 7 8 ? 15y2 36x2 2x 1 4 21 b3 2 b2 a ? ab 2 a b2 7s s 1 2 ? a2 2 9 3 2s 1 4 21 ? 12 2a 2 6 (a 2 2)2 4b ? 16b3 4 2 a2 y2 2 2y 2 3 2c3 ? 4c2 2y 1 2 ? 5x 1 15 x2 2 4 3x 3x 2 6 4x 1 8 6x 1 18 2 2 x 2x ? d2 2 25 4 2 d2 ? 5d2 2 20 d 1 5 when x 65,908? 19. 22. 25. 28. 31. 34. 37. 40. 6. 11. 16. 5 d ? d2 m2 32 8 ? 3m 3a 1 9 15a ? 20. a3 18 7. 12. 17. x2 36 ? 20 6r2 10rs 5s2 ? 6r3 5x 2 5y x2y ? xy2 25 x2 2 1 x2 ? 3x2 2 3x 15 8x 2x 1 6 ? x 1 3 x2 x2 2 x 2 2 3 ? 21 x2 2 4 23. 26. 29. a2 2 7a 2 8 2a 1 2 ? 5 a 2 8 32. 35. 38. 41. 10y 1 90 5y 2 45 6a 1 12 5a 2 15 4a 2 6 4a 1 8 ? y2 2 81 (y 1 9)2 ? x2 2 3x 1 2 2x2 2 2 a2 1 12a 1 36 a2 2 36 ? 2x x 2 2 ? 36 2 a2 36 1 a2 42. What is the value of x2 2 4 6x 1 12 ? 4x 2 12 x2 2 5x 1 6 14-4 DIVIDING FRACTIONS We know that the operation of division may be defined in terms of the multiplicative inverse, the reciprocal. A quotient can be expressed as the dividend times the reciprocal of the divisor. Thus 24 35 and We use the same rule to divide algebraic fractions. In general, for any num- bers a, b, c, and d, when b 0, c 0, and d 0 ad d bc Procedure To divide by a fraction, multiply the dividend by the reciprocal of the divisor. EXAMPLE 1 Divide: 16c3 21d2 4 24c4 14d3 Solution How to Proceed (1) Multiply the dividend by the reciprocal of the divisor: (2) Divide the numerators and denominators by the common factors: (3) Multiply the resulting numerators and the resulting denominators: Dividing Fractions 549 16c3 21d2 4 24c4 14d3 5 16c3 21d2 ? 14d3 24c4 2 5 16c3 21d2 3 2d ? 14d3 24c4 3c 5 4d 9c Answer 4d 9c (c 0, d 0) EXAMPLE 2 Divide: 8x 1 24 x2 2 25 4 4x x2 1 8x 1 15 Solution How to Proceed (1) Multiply the dividend by the reciprocal of the divisor: (2) Factor the numerators and denominators, and divide by the common factors: (3) Multiply the resulting numerators and the resulting denominators: Answer 2(x 1 3)2 x(x 2 5) (x 0, 5, 5, 3) 8x 1 24 x2 2 25 4 4x x2 1 8x 1 15 5 8x 1 24 x2 2 25 ? x2 1 8x 1 15 4x 5 2 8 (x 1 5) 1 (x 1 3) (x 2 5) ? 1 (x 1 5) 4 1 (x 1 3) x 5 2(x 1 3)2 x(x 2 5) Note: If x 5, 5, or 3, the dividend and the divisor will not be defined. If x 0, the reciprocal of the divisor will not be defined. 550 Algebraic Fractions, and Equations and Inequalities Involving Fractions EXERCISES Writing About Mathematics 1. Explain why the quotient 2. To find the quotient 3 2 4 1 nator and wrote 3 2(x 2 4 15 , Ruth canceled (x 4) in the numerator and denomi- is undefined for x 2 and for x 3. . Is Ruth’s answer correct? Explain why or why not. Developing Skills In 3–27, find each quotient in lowest terms. In each case, list any values of the variables for which the quotient is not defined. 3. 7. 11. 15. 18. 21. 24. 27. 4. 8. 12. 7a 10 4 21 5 3x 5y 4 21x 2y 4x 1 4 9 4 3 8x x2 2 5x 1 4 2x 4 2x 2 2 8x2 4a 4 (a2 2 b2) a2 2 ab 35 4 4b 12 7 7ab2 10cd 3y2 1 9y 4 14b3 5c2d2 18 4 16. 19. 5. 9. 8 4 x 2y xy2 x2y a3 2 a b 4 x y3 4 a3 4b3 6. 10. 14. 9 4 x x 3 6a2b2 8c 4 3ab x2 2 1 5 4 x 2 1 10 4 b2 2 4 b2 4 21x 3x 1 6 17. 20. b2 2 b 2 6 2b (x 2 2)2 4x2 2 16 13. 5y2 27 10 4 10a 1 15 25 4a2 2 9 12y 2 6 8 4 (2y2 2 3y 1 1) 5x 1 10y 3x 1 12y x2 2 2xy 2 8y2 x2 2 16y2 4 2x 1 2 x 1 2 4 4x a 1 b)2 a2 2 b2 4 a 1 b b2 2 a2 ? a 2 b (a 2 b)2 22. x2 2 4x 1 4 3x 2 6 4 (2 2 x) 25. x 1 y x2 1 y2 ? x x 2 y 4 (x 1 y)2 x4 2 y4 23. (9 2 y2) 4 y2 1 8y 1 15 2y 1 10 26. 2a 1 6 a2 28. For what value(s) of a is a2 2 2a 1 1 a2 29. Find the value of 30. If x y a and y2 2 6y 1 9 y2 2 9 y 4 z 5 1 a undefined? 4 a2 2 1 a when y 70. 10y 2 30 y2 1 3y 4 , what is the value of x z? 14-5 ADDING OR SUBTRACTING ALGEBRAIC FRACTIONS We know that the sum (or difference) of two arithmetic fractions that have the same denominator is another fraction whose numerator is the sum (or difference) of the numerators and whose denominator is the common denominator of the given fractions. We use the same rule to add algebraic fractions that have the same nonzero denominator. Thus: Adding or Subtracting Algebraic Fractions 551 Arithmetic fractions Algebraic fractions Procedure To add (or subtract) fractions that have the same denominator: 1. Write a fraction whose numerator is the sum (or difference) of the numerators and whose denominator is the common denominator of the given fractions. 2. Reduce the resulting fraction to lowest terms. Add and reduce the answer to lowest terms: 4x 1 9 5 4x 4x 1 9 5 4x 5 5 1 9 4x 5 14 4x 5 7 2x EXAMPLE 1 Solution Answer 7 2x (x 0) EXAMPLE 2 Subtract: 4x 1 7 6x 2 2x 2 4 6x Solution 4x 1 7 6x 2 2x 2 4 6x 5 (4x 1 7) 2 (2x 2 4) 6x 5 4x 1 7 2 2x 1 4 6x 5 2x 1 11 6x Answer 2x 1 11 6x (x 0) Note: In Example 2, since the fraction bar is a symbol of grouping, we enclose numerators in parentheses when the difference is written as a single fraction. In this way, we can see all the signs that need to be changed for the subtraction. In arithmetic, in order to add (or subtract) fractions that have different denominators, we change these fractions to equivalent fractions that have the same denominator, called the common denominator. Then we add (or subtract) the equivalent fractions. 3 4 For example, to add and , we use any common denominator that has 4 and 1 6 6 as factors. METHOD 1. Use the product of the denominators as the common denominator. Here, a common denominator is 4 6, or 24. 3 4 1 1 22 24 5 4 24 5 18 24 Answer 6 3 4 3 11 12 552 Algebraic Fractions, and Equations and Inequalities Involving Fractions METHOD 2. To simplify our work, we use the least common denominator (LCD), that is, the least common multiple of the given denominators. The LCD of is 12. 3 4 and 12 1 2 12 5 11 12 Answer To find the least common denominator of two fractions, we factor the denominators of the fractions completely. The LCD is the product of all of the factors of the first denominator times the factors of the second denominator that are not factors of the first. 4 2 2 3 6 2 LCD 2 2 3 Then, to change each fraction to an equivalent form that has the LCD as the , where x is the number by which the original denominator, we multiply by denominator must be multiplied to obtain the LCD 12 5 9 12 1 6 1 6 5 12 5 2 12 x x 2 2 B B A A Note
that the LCD is the smallest possible common denominator. Procedure To add (or subtract) fractions that have different denominators: 1. Choose a common denominator for the fractions. 2. Change each fraction to an equivalent fraction with the chosen common denominator. 3. Write a fraction whose numerator is the sum (or difference) of the numerators of the new fractions and whose denominator is the common denominator. 4. Reduce the resulting fraction to lowest terms. Algebraic fractions are added in the same manner as arithmetic fractions, as shown in the examples that follow. Adding or Subtracting Algebraic Fractions 553 EXAMPLE 3 Add: 5 a2b 1 2 ab2 Solution How to Proceed (1) Find the LCD of the fractions: (2) Change each fraction to an equivalent fraction with the least common denominator, a2b2: (3) Write a fraction whose numerator is the sum of the numerators of the new fractions and whose denominator is the common denominator: a2b a a b ab2 a b b LCD a a b b a2b2 5 a2b ab2 ? 1 2 1 2 b b a a ab2 5 5 5 5b a2b ? a2b2 1 2a a2b2 5 5b 1 2a a2b2 Answer 5b 1 2a a2b2 (a 0, b 0) EXAMPLE 4 Solution Subtract: 2x 1 5 3 2 x 2 2 4 2x 1 5 3 2 x 2 2 LCD 3 4 12 2x 1 5 4 ? 3 5 8x 1 20 12 4 5 4 2 3 3 ? 2 3x 2 6 12 x 2 2 4 (8x 1 20) 2 (3x 2 6) 12 5 5 8x 1 20 2 3x 1 6 12 5 5x 1 26 Answer 12 EXAMPLE 5 Express as a fraction in simplest form: y 1 1 2 1 y 2 1 Solution LCD ? Answer y2 2 2 y 2 1 (y 1) 2 1 y 2 1 5 y2 y2 2 1 2 1 y 2 1 5 y2 554 Algebraic Fractions, and Equations and Inequalities Involving Fractions EXAMPLE 6 Solution Subtract: 6x x2 2 4 2 3 x 2 2 x2 4 (x 2)(x 2) x 2 (x 2) LCD (x 2)(x 2) 6x x2 2 4 2 3 x 2 2 3(x 1 2) (x 2 2)(x 1 2) 6x 5 (x 2 2)(x 1 2) 2 6x 2 3(x 1 2) 5 (x 2 2)(x 1 2) 5 6x 2 3x 2 6 (x 2 2)(x 1 2) 3x 2 6 (x 2 2)(x 1 2) 3(x 2 2) (x 2 2)(x 1 2) 5 5 5 3 x 1 2 Answer 3 x 1 2 (x 2, 2) EXERCISES Writing About Mathematics 1. In Example 2, the answer is 2x 1 11 6x . Can we divide 2x and 6x by 2 to write the answer in lowest terms as ? Explain why or why not. x 1 11 3x . Joey said that if a x. Do you agree with Joey? Explain why or why not. Developing Skills In 3–43, add or subtract the fractions as indicated. Reduce each answer to lowest terms. In each case, list the values of the variables for which the fractions are not defined. 3. 6. 9. 12. 15. 18. 4c 2 6 4c 1 5 11 4c 5x 2 6 6x 2 5 x2 2 1 x2 2 1 5x 6 2 2x 3 8x 4 1 7x 5 2 3x 10 9 4x 1 3 2x 4. 7. 10. 13. 16. 19. t 2 2s 5r t 7 2 2y 6y 2 4 4y 1 3 1 4y 1 3 1 8 2 r2 r2 1 4r r2 3a 5a 4 1 x 1 1 2x 2 1 8x r2 2 r 2 6 5. 8. 11. 14. 17. 10c 1 9 10c 2 3 6 10c 9d 1 6 2d 1 1 2 7d 1 5 2d 1 1 x 3 1 x 2 5 1 ab ab 4 a 7 1 b 14 8b 2 3a 4b 20. 9a Adding or Subtracting Algebraic Fractions 555 21. 24. 27. 30. 33. 36. 39. d 1 7 5d b 2 3 5b 2 b 1 2 10b 4y 1 4 1x 8x 2 8 2 3x 4x 2 4 4 3x 1 18 x x2 2 36 x 2 5 2 x 2 x 1 3 22. 25. 28. 31. 34. 37. 40 3y 2 5 4y2 1 3y 5 2 2x x 1 y 2 3a 2 1 1 7 15a y2 2 9 y 2 3 1 2 1 2x 2 1 x 1 2 1 2x 2 3 23. 26. 29. 32. 35. 38. 41. 2 y 2 2 4 2 3c 2 3 6c2 7 2x 2 6 3y 2 4 5 3c 2 7 2c 5 x 2 3 1 10 3 2x 2 4 3x 2 6 1 6 y2 2 16 2y 3x 1 12y 2 6x 2 y x2 1 3xy 2 4y2 42. 7a (a 2 1)(a 1 3) 1 2a 2 5 (a 1 3)(a 1 2) 43. 2a 1 7 a2 2 2a 2 15 2 3a 2 4 a2 2 7a 1 10 Applying Skills In 44–46, represent the perimeter of each polygon in simplest form. 44. The lengths of the sides of a triangle are represented by x 2 , 3x 5 , and 7x 10 . 45. The length of a rectangle is represented by x 1 3 4 , and its width is represented by x 2 4 3 . 46. Each leg of an isosceles triangle is represented by 6x 2 18 21 . 2x 2 3 7 , and its base is represented by In 47 and 48, find, in each case, the simplest form of the indicated length. 47. The perimeter of a triangle is the length of the third side. 17x 24 , and the lengths of two of the sides are 3x 8 and 2x 2 5 12 . Find 48. The perimeter of a rectangle is 14x 15 , and the measure of each length is x 1 2 3 . Find the mea- sure of each width. 49. The time t needed to travel a distance d at a rate of speed r can be found by using the for- mula t 5 d r . a. For the first part of a trip, a car travels x miles at 45 miles per hour. Represent the time that the car traveled at that speed in terms of x. b. For the remainder of the trip, the car travels 2x 20 miles at 60 miles per hour. Represent the time that the car traveled at that speed in terms of x. c. Express, in terms of x, the total time for the two parts of the trip. 50. Ernesto walked 2 miles at a miles per hour and then walked 3 miles at (a 1) miles per hour. Represent, in terms of a, the total time that he walked. 51. Fran rode her bicycle for x miles at 10 miles per hour and then rode (x 3) miles farther at 8 miles per hour. Represent, in terms of x, the total time that she rode. 556 Algebraic Fractions, and Equations and Inequalities Involving Fractions 14-6 SOLVING EQUATIONS WITH FRACTIONAL COEFFICIENTS The following equations contain fractional coefficients: 1 2x 5 10 x 2 5 10 3x 1 60 5 5 1 6x 3 1 60 5 5x x 6 Each of these equations can be solved by finding an equivalent equation that does not contain fractional coefficients. This can be done by multiplying both sides of the equation by a common denominator for all the fractions present in the equation. We usually multiply by the least common denominator, the LCD. , since a dec- 1 Note that the equation 0.5x 10 can also be written as 2x 5 10 imal fraction can be replaced by a common fraction. Procedure To solve an equation that contains fractional coefficients: 1. Find the LCD of all coefficients. 2. Multiply both sides of the equation by the LCD. 3. Solve the resulting equation using the usual methods. 4. Check in the original equation. EXAMPLE 1 Solve and check: 3 1 x x 5 5 8 Solution How to Proceed (1) Write the equation: (2) Find the LCD: 3 1 x x 5 5 8 LCD 3 5 15 (3) Multiply both sides of the equation by the LCD: 15 (4) Use the distributive property: (5) Simplify: (6) Solve for x: 15 x 3 A B Answer x 15 5 15(8) 5 15(8 15 A B B 5x 3x 120 8x 120 x 15 Check 3 1 x x 15 3 1 15 5 3 5 5 8 5 5? 8 5? 8 8 8 ✔ EXAMPLE 2 Solution Solve: a. 4 5 20 1 x 3x 4 a. 4 5 20 1 x 3x 4 LCD 4 4 4 A A 5 4 20 1 x 4 B 5 4(20) 1 4 3x 4 3x 4 3x 5 80 1 x B B A x 4 A B 2x 5 80 x 5 40 Answer Solving Equations with Fractional Coefficients 557 b. 2x 1 7 6 2 2x 2 9 10 5 3 b. 30 2x 1 7 6 2 2x 2 9 10 5 3 LCD 30 B B 30 A 2x 1 7 6 2x 1 7 6 2 30 2 2x 2 9 10 2x 2 9 10 5 30(3) 5 30(3) B A A 5(2x 7) 3(2x 9) 90 10x 35 6x + 27 90 4x 62 90 4x 28 x 7 Answer In Example 2, the check is left to you. EXAMPLE 3 A woman purchased stock in the PAX Company over 3 months. In the first month, she purchased one-half of her present number of shares. In the second month, she bought two-fifths of her present number of shares. In the third month, she purchased 14 shares. How many shares of PAX stock did the woman purchase? Solution Let x total number of shares of stock purchased. 1 Then number of shares purchased in month 1, 2x 2 number of shares purchased in month 2, 5x 14 number of shares purchased in month 3. The sum of the shares purchased over 3 months is the total number of shares. month 1 1 month 2 1 month 3 5 total 10 A 2 5x 2 5x 14 x 1 2x 1 2x 5x 4x 140 10x 9x 140 10x 14 B 10(x) 140 x 558 Algebraic Fractions, and Equations and Inequalities Involving Fractions 70, month 2 2 5(140) 70 56 14 140 ✔ 56, month 3 14 Check month 1 1 2(140) Answer 140 shares EXAMPLE 4 In a child’s coin bank, there is a collection of nickels, dimes, and quarters that amounts to $3.20. There are 3 times as many quarters as nickels, and 5 more dimes than nickels. How many coins of each kind are there? Solution Let x the number of nickels. Then 3x the number of quarters, and x 5 the number of dimes. Also, 0.05x the value of the nickels, 0.25(3x) the value of the quarters, and 0.10(x 5) the value of the dimes. Write the equation for the value of the coins. To simplify the equation, which contains coefficients that are decimal fractions with denominators of 100, multiply each side of the equation by 100. The total value of the coins is $3.20. 0.05x 0.25(3x) 0.10(x 5) 3.20 100[0.05x 0.25(3x) 0.10(x 5)] 100(3.20) 5x 25(3x) 10(x 5) 320 5x 75x 10x 50 320 90x 50 320 90x 270 x 3 Check There are 3 nickels, 3(3) 9 quarters, and 3 5 8 dimes. The value of 3 nickels is $0.05(3) $0.15 The value of 9 quarters is $0.25(9) $2.25 The value of 8 dimes is $0.10(8) $0.80 $3.20 ✔ Answer There are 3 nickels, 9 quarters, and 8 dimes. Solving Equations with Fractional Coefficients 559 Note: In a problem such as this, a chart such as the one shown below can be used to organize the information: Coins Number of Coins Value of One Coin Total Value Nickels Quarters Dimes x 3x x 5 0.05 0.25 0.10 0.05x 0.25(3x) 0.10(x 5) EXERCISES Writing About Mathematics 1. Abby solved the equation 0.2x 0.84 3x as follows: 3x 0.2x 0.84 0.1x 0.2x 0.84 0.2x 8.4 x Is Abby’s solution correct? Explain why or why not. 2. In order to write the equation 0.2x 0.84 3x as an equivalent equation with integral coefficients, Heidi multiplied both sides of the equation by 10. Will Heidi’s method lead to a correct solution? Explain why or why not. Compare Heidi’s method with multiplying by 100 or multiplying by 1,000. Developing Skills In 3–37, solve each equation and check. 3. 6. 15. 12. 2x 1 1 5y 2 30 4 5 6 7 5 0 3 5 6x 2 9 3 1 x 10 26 6 2 t 2 25 t 2 3 5 5 4 21. 24. 0.4x 0.08 4.24 27. 1.7x 30 0.2x 18. 1 6t 5 18 m 2 2 9 5 3 5x 2 5 15 4 3y 1 1 4 5 44 2 y 5 4. 7. 10. 13. 16. 19. 3 2 r r 7y 12 2 1 3m 1 1 6 5 2 4 5 2y 2 5 3 4 5 2 2 3 2 2m 6 22. 25. 2c 0.5c 50 28. 0.02(x 5) 8 5. 3x 5 5 15 2r 1 6 8. 17. 11. 14. 5 5 24 35 15 4 2 6 5 t 3t 12 20. 2 23. 0.03y 1.2 8.7 26. 0.08y 0.9 0.02y 29. 0.05(x 8) 0.07x 560 Algebraic Fractions, and Equations and Inequalities Involving Fractions 30. 0.4(x 9) 0.3(x 4) 32. 0.04x 0.03(2,000 x) 75 34. 0.05x 10 0.06(x 50) 36. 3 1 0.2a 0.4a 4 5 2 31. 0.06(x 5) 0.04(x 8) 33. 0.02x 0.04(1,500 x) 48 35. 0.08x 0.03(x 200) 4 37. 6 2 0.3a 0.1a 4 5 3 38. The sum of one-half of a number and one-third of that number is 25. Find the number. 39. The difference between
one-fifth of a positive number and one-tenth of that number is 10. Find the number. 40. If one-half of a number is increased by 20, the result is 35. Find the number. 41. If two-thirds of a number is decreased by 30, the result is 10. Find the number. 42. If the sum of two consecutive integers is divided by 3, the quotient is 9. Find the integers. 43. If the sum of two consecutive odd integers is divided by 4, the quotient is 10. Find the integers. 44. In an isosceles triangle, each of the congruent sides is two-thirds of the base. The perimeter of the triangle is 42. Find the length of each side of the triangle. 45. The larger of two numbers is 12 less than 5 times the smaller. If the smaller number is equal to one-third of the larger number, find the numbers. 46. The larger of two numbers exceeds the smaller by 14. If the smaller number is equal to three-fifths of the larger, find the numbers. 47. Separate 90 into two parts such that one part is one-half of the other part. 48. Separate 150 into two parts such that one part is two-thirds of the other part. Applying Skills 49. Four vegetable plots of unequal lengths and of equal widths are arranged as shown. The length of the third plot is one-fourth the length of the second plot. 1 2 3 4 The length of the fourth plot is one-half the length of the second plot. The length of the first plot is 10 feet more than the length of the fourth plot. If the total length of the four plots is 100 feet, find the length of each plot. 50. Sam is now one-sixth as old as his father. In 4 years, Sam will be one-fourth as old as his father will be then. Find the ages of Sam and his father now. 51. Robert is one-half as old as his father. Twelve years ago, he was one-third as old as his father was then. Find their present ages. Solving Equations with Fractional Coefficients 561 52. A coach finds that, of the students who try out for track, 65% qualify for the team and 90% of those who qualify remain on the team throughout the season. What is the smallest number of students who must try out for track in order to have 30 on the team at the end of the season? 53. A bus that runs once daily between the villages of Alpaca and Down makes only two stops in between, at Billow and at Comfort. Today, the bus left Alpaca with some passengers. At Billow, one-half of the passengers got off, and six new ones got on. At Comfort, again onehalf of the passengers got off, and, this time, five new ones got on. At Down, the last 13 passengers on the bus got off. How many passengers were aboard when the bus left Alpaca? 54. Sally spent half of her money on a present for her mother. Then she spent one-quarter of the cost of the present for her mother on a treat for herself. If Sally had $6.00 left after she bought her treat, how much money did she have originally? 55. Bob planted some lettuce seedlings in his garden. After a few days, one-tenth of these seedlings had been eaten by rabbits. A week later, one-fifth of the remaining seedlings had been eaten, leaving 36 seedlings unharmed. How many lettuce seedlings had Bob planted originally? 56. May has 3 times as many dimes as nickels. In all, she has $1.40. How many coins of each type does she have? 57. Mr. Jantzen bought some cans of soup at $0.39 per can, and some packages of frozen vegetables at $0.59 per package. He bought twice as many packages of vegetables as cans of soup. If the total bill was $9.42, how many cans of soup did he buy? 58. Roger has $2.30 in dimes and nickels. There are 5 more dimes than nickels. Find the number of each kind of coin that he has. 59. Bess has $2.80 in quarters and dimes. The number of dimes is 7 less than the number of quarters. Find the number of each kind of coin that she has. 60. A movie theater sold student tickets for $5.00 and full-price tickets for $7.00. On Saturday, the theater sold 16 more full-price tickets than student tickets. If the total sales on Saturday were $1,072, how many of each kind of ticket were sold? 61. Is it possible to have $4.50 in dimes and quarters, and have twice as many quarters as dimes? Explain. 62. Is it possible to have $6.00 in nickels, dimes, and quarters, and have the same number of each kind of coin? Explain. 63. Mr. Symms invested a sum of money in 7% bonds. He invested $400 more than this sum in 8% bonds. If the total annual interest from these two investments is $257, how much did he invest at each rate? 64. Mr. Charles borrowed a sum of money at 10% interest. He borrowed a second sum, which was $1,500 less than the first sum, at 11% interest. If the annual interest on these two loans is $202.50, how much did he borrow at each rate? 562 Algebraic Fractions, and Equations and Inequalities Involving Fractions 14-7 SOLVING INEQUALITIES WITH FRACTIONAL COEFFICIENTS In our modern world, many problems involve inequalities. A potential buyer may offer at most one amount for a house, while the seller will accept no less than another amount. Inequalities that contain fractional coefficients are handled in much the same way as equations that contain fractional coefficients. The chart on the right helps us to translate words into algebraic symbols. Procedure Words Symbols a is greater than b a is less than b a is at least b a is no less than b a is at most b a is no greater than b a b a b a b a b To solve an inequality that contains fractional coefficients: 1. Find the LCD, a positive number. 2. Multiply both sides of the inequality by the LCD. 3. Solve the resulting inequality using the usual methods. EXAMPLE 1 Solve the inequality, and graph the solution set on a number line: a. 3 2 x x 6 . 2 Solution a. 6 . 2 . 6(2 2x 2 x . 12 . 12 A b. b. 8 2 4y 7 # 3 8 2 4y 3y 2 1 7 # 3 3y 2 1 3y 2 1 A 1 14 # 14(3) 14 8 2 4y 7 8 2 4y 7 B 21y 1 16 2 8y # 42 # 42 B A B 14 3y 2 A x . 12 Since no domain was given, use the domain of real numbers. Answer: x 12 0 2 4 6 8 10 12 14 16 13y # 26 y # 2 Since no domain was given, use the domain of real numbers. Answer: y 2 –1 0 1 2 3 Solving Inequalities with Fractional Coefficients 563 EXAMPLE 2 Two boys want to pool their money to buy a comic book. The younger of the boys has one-third as much money as the older. Together they have more than $2.00. Find the smallest possible amount of money each can have. Solution Let x the number of cents that the older boy has. Then the number of cents that the younger boy has. 1 3x The sum of their money in cents is greater than 200. 1 x 200 3x x 1 1 3x . 3(200) 3 A B 3x + x 600 4x 600 x 150 50 1 3x The number of cents that the younger boy has must be an integer greater than 50. The number of cents that the older boy has must be a multiple of 3 that is greater than 150. The younger boy has at least 51 cents. The older boy has at least 153 cents. The sum of 51 and 153 is greater than 200. Answer The younger boy has at least $0.51 and the older boy has at least $1.53. EXERCISES Writing About Mathematics 1. Explain the error in the following solution of an inequality. x 23 . 2 2 x 23 . 23(22) B x . 6 23 A 2. In Example 2, what is the domain for the variable? Developing Skills In 3–23, solve each inequality, and graph the solution set on a number line. 3. 6. 9. 5x . 9 4x 2 1 1 20 8 # 5 x 4 2 x 8 t 10 # 4 1 t 5 4. 7. 10. y 2 2 3y , 5 y y 6 $ 12 1 1 1 1 2x 3 $ x 2 5. 5 6c . 1 3c . 36 11. 2.5x 1.7x 4 564 Algebraic Fractions, and Equations and Inequalities Involving Fractions 12. 2y 3 0.2y 15. 18. 21. 2d 1 1 3x 2 30 4 , 7d 6 , x 2 r 2 3 12 1 5 3 3 2 2 3 # 2 2r 2 3 5 13. 16. 19. 22. 3x 4c 2 . 37 6x 2 3 3 $ 2t 1 4 2 1 7 6 10 1 x 1 2 5 1 5t 2 1 3t 2 4 9 6 14. 17. 20. 23. 1 1 5y 2 30 7 # 0 3 $ 7 2 m 2m 2y , 10 5 2 2a 1 3 6 24. If one-third of an integer is increased by 7, the result is at most 13. Find the largest possible integer. 25. If two-fifths of an integer is decreased by 11, the result is at least 4. Find the smallest possi- ble integer. 26. The sum of one-fifth of an integer and one-tenth of that integer is less than 40. Find the greatest possible integer. 27. The difference between three-fourths of a positive integer and one-half of that integer is greater than 28. Find the smallest possible integer. 28. The smaller of two integers is two-fifths of the larger, and their sum is less than 40. Find the largest possible integers. 29. The smaller of two positive integers is five-sixths of the larger, and their difference is greater than 3. Find the smallest possible integers. Applying Skills 30. Talk and Tell Answering Service offers customers two monthly options. OPTION 1 Measured Service base rate is $15 each call costs $0.10 OPTION 2 Unmeasured Service base rate is $20 no additional charge per call Find the least number of calls for which unmeasured service is cheaper than measured service. 31. Paul earned some money mowing lawns. He spent one-half of this money for a book, and then one-third for a CD. If he had less than $3 left, how much money did he earn? 32. Mary bought some cans of vegetables at $0.89 per can, and some cans of soup at $0.99 per can. If she bought twice as many cans of vegetables as cans of soup, and paid at least $10, what is the least number of cans of vegetables she could have bought? 33. A coin bank contains nickels, dimes, and quarters. The number of dimes is 7 more than the number of nickels, and the number of quarters is twice the number of dimes. If the total value of the coins is no greater than $7.20, what is the greatest possible number of nickels in the bank? 34. Rhoda is two-thirds as old as her sister Alice. Five years from now, the sum of their ages will be less than 60. What is the largest possible integral value for each sister’s present age? Solving Fractional Equations 565 35. Four years ago, Bill was 11 4 times as old as his cousin Mary. The difference between their present ages is at least 3. What is the smallest possible integral value for each cousin’s present age? 36. Mr. Drew invested a sum of money at 71 2% interest. He invested a second sum, which was $200 less than the fir
st, at 7% interest. If the total annual interest from these two investments is at least $160, what is the smallest amount he could have invested at 71 2% ? 37. Mr. Lehtimaki wanted to sell his house. He advertised an asking price, but knew that he would accept, as a minimum, nine-tenths of the asking price. Mrs. Patel offered to buy the house, but her maximum offer was seven-eighths of the asking price. If the difference between the seller’s lowest acceptance price and the buyer’s maximum offer was at least $3,000, find: a. the minimum asking price for the house; b. the minimum amount Mr. Lehtimaki, the seller, would accept; c. the maximum amount offered by Mrs. Patel, the buyer. 38. When packing his books to move, Philip put the same number of books in each of 12 boxes. Once packed, the boxes were too heavy to lift so Philip removed one-fifth of the books from each box. If at least 100 books in total remain in the boxes, what is the minimum number of books that Philip originally packed in each box? 14-8 SOLVING FRACTIONAL EQUATIONS An equation is called an algebraic equation when a variable appears in at least one of its sides. An algebraic equation is a fractional equation when a variable appears in the denominator of one, or more than one, of its terms. For example, a2 11 3d 1 1 2 6d y2 1 2y 2 3 y 2 2 are all fractional equations. To simplify such an equation, clear it of fractions by multiplying both sides by the least common denominator of all fractions in the equation. Then, solve the simpler equation. As is true of all algebraic fractions, a fractional equation has meaning only when values of the variable do not lead to a denominator of 0. KEEP IN MIND When both sides of an equation are multiplied by a variable expression that may represent 0, the resulting equation may not be equivalent to the given equation. Such equations will yield extraneous solutions, which are solutions that satisfy the derived equation but not the given equation. Each solution, therefore, must be checked in the original equation. 566 Algebraic Fractions, and Equations and Inequalities Involving Fractions EXAMPLE 1 Solve and check: 3 1 1 1 x 5 1 2 Solution Multiply both sides of the equation by the least common denominator, 6x. 6x x 5 1 2 5 6x 6x B B 2x 1 6 5 3x 5 6x 6x 1 3 A B Answer x 6 EXAMPLE 2 Solve and check: 5x 1 10 x 1 2 5 7 Solution Multiply both sides of the equation by the least common denominator, x 2. 5x 1 10 x 1 2 5 7 (x 1 2) A 5x 1 10 x 1 2 5x 1 10 5 7x 1 14 5 (x 1 2)(7) B 22x 5 4 x 5 22 Check ? 1 3 1 1 1 2 6 5? 1 2 1 2 5 1 2 ✔ Check 5x 1 10 x 1 2 5 7 5(22) 1 10 22 1 2 210 1 10 0 5? 5? 7 7 0 0 5 7 ✘ The only possible value of x is a value for which the equation has no meaning because it leads to a denominator of 0. Therefore, there is no solution for this equation. Answer The solution set is the empty set, or { }. EXAMPLE 3 Solve and check: 2 x 5 6 2 x 4 Solution METHOD 1 METHOD 2 Solving Fractional Equations 567 Multiply both sides of the equation by the LCD, 4x: 2 x 5 4x 4x (6 2 x) 8 5 6x 2 x2 x2 2 6x 1 8 5 0 (x 2 2)(x 2 4 Use the rule for proportion: the product of the means equals the product of the extremes. x 5 6 2 x 2 4 x(6 2 x) 5 8 6x 2 x2 5 8 2x2 1 6x 2 8 5 0 x2 2 6x 1 8 5 0 (x 2 2)(x 2 4 Check Answer x 2 or x 4 EXAMPLE 4 Solve and check(x 2 1) Solution Multiply both sides of the equation by the LCD, 2(x 1 2)(x 2 1) : 2(x 1 2)(x 2 1) 2(x 1 2)(x 2 1) x 1 2 2(x 1 2)(x 2 1) x 1 2 5 1 2(x 2 1(x 1 2)(x 2 1) x 2 1 2(x 1 2)(x 2 1) x 2 1 2(x 2 1) 1 2(x 1 2) 5 x 1 2 B A 2(x 1 2)(x 2 1) 2(x 2 1) 2(x 1 2)(x 2 1) 2(x 2 1) 1 5 5 2(x 1 2)(x 2 1) 4x 1 2 5 x 1 2 3x 5 0 x 5 0 Answer x 0 Check ? 2 1 1 1 21 5? 21 2 5 21 2 1 2(21) ✔ 1 2(0 2 1) 568 Algebraic Fractions, and Equations and Inequalities Involving Fractions EXERCISES Writing About Mathematics 1. Nathan said that the solution set of r 2 5 5 10 2 5r 2 25 agree with Nathan? Explain why or why not. is the set of all real numbers. Do you 2. Pam multiplied each side of the equation y 1 5 y2 2 25 5 3 y 1 5 by (y + 5)(y 5) to obtain the 5 3 equation y 5 3y 15, which has as its solution y 10. Pru said that the equation y 1 5 is a proportion and can be solved by writing the product of the means equal y2 2 25 to the product of the extremes. She obtained the equation 3(y2 25) (y 5)2, which has as its solution 10 and 5. Both girls used a correct method of solution. Explain the difference in their answers. y 1 5 Developing Skills In 3–6, explain why each fractional equation has no solution. 3. 6x x 5 3 4. 4a 1 4 a 1 1 5 5 5. x 5 4 1 2 2 x 6 In 7–45, solve each equation, and check. 7. 10. 13. 16. 19. 22. 25. 28. 31. 34. 37. 40. 43. 10 x 5 5 15 4x 5 1 8 9 2x 5 7 2x 6x 5 3 2 1 x 5x 1 1 30 6 3x 2 1 5 3 4 5 2 3a 12 y 5 1 3 8. 11. 14. 17. 20. 23. 26. 29. 32. 35. 38. 41. 44. 15 y 5 3 10 18 30 2x y 1 9 2y 1 3 5 15 3x 2 4 5 1 4 4z 7 1 5z 15 12 x2 18 9. 12. 15. 18. 21. 24. 27. 30. 33. 36. 39. 42. 45. 3 2x 5 1 2 15 y 2 3 y 5 4 y 2 2 2y 5 3 8 3a 1 5 12 5 2 1 a 5 1 x 2x 2x 1 2x 1 1 3x 5 1 5x 3m 2 1 12y 8y 2b 1 1 5 b 2 2x 1 4 5 23 1 x 1 2 1 2x 1 2 3 In 46–49, solve each equation for x in terms of the other variables. Chapter Summary 569 46. t x 2 k 5 0 by c , y 5 c2 a x 5 47. t x 2 k 5 5k 48. a 1 b x 5 c 49, and c 0, is it possible to know the numerical value of x 50. If without knowing numerical values of a, b, c, and y? Explain your answer Applying Skills 51. If 24 is divided by a number, the result is 6. Find the number. 52. If 10 is divided by a number, the result is 30. Find the number. 53. The sum of 20 divided by a number, and 7 divided by the same number, is 9. Find the num- ber. 54. When the reciprocal of a number is decreased by 2, the result is 5. Find the number. 55. The numerator of a fraction is 8 less than the denominator of the fraction. The value of the fraction is . Find the fraction. 3 5 56. The numerator and denominator of a fraction are in the ratio 3 : 4. When the numerator is decreased by 4 and the denominator is increased by 2, the value of the new fraction, in simplest form, is . Find the original fraction. 1 2 57. The ratio of boys to girls in the chess club is 4 to 5. After 2 boys leave the club and 2 girls join, the ratio is 1 to 2. How many members are in the club? 58. The length of Emily’s rectangular garden is 4 feet greater than its width. The width of Sarah’s rectangular garden is equal to the length of Emily’s and its length is 18 feet. The two gardens are similar rectangles, that is, the ratio of the length to the width of Emily’s garden equals the ratio of the length to the width of Sarah’s garden. Find the possible dimensions of each garden. (Two answers are possible.) CHAPTER SUMMARY An algebraic fraction is the quotient of two algebraic expressions. If the algebraic expressions are polynomials, the fraction is called a rational expression or a fractional expression. An algebraic fraction is defined only if values of the variables do not result in a denominator of 0. Fractions that are equal in value are called equivalent fractions. A fraction is reduced to lowest terms when an equivalent fraction is found such that its numerator and denominator have no common factor other than 1 or 1. This fraction is considered a lowest terms fraction. 570 Algebraic Fractions, and Equations and Inequalities Involving Fractions Operations with algebraic fractions follow the same rules as operations with arithmetic fractions: Multiplication Division a x ? y 5 ab b xy (x 0, y 0 ay bx (x 0, y 0, b 0) Addition/subtraction with the same denominator c 0), c 0) Addition/subtraction with different denominators (first, obtain the common denominator): bd 1 bc bd 5 ad 1 bc b 5 ad b d 1 c d d ? b ? bd (b 0, d 0) b 1 c a d 5 a A fractional equation is an equation in which a variable appears in the denominator of one or more than one of its terms. To simplify a fractional equation, or any equation or inequality containing fractional coefficients, multiply both sides by the least common denominator (LCD) to eliminate the fractions. Then solve the simpler equation or inequality and check for extraneous solutions. VOCABULARY 14-1 Algebraic fraction • Fractional expression • Rational expression 14-2 Reduced to lowest terms • Lowest terms fraction • Equivalent fractions • Division property of a fraction • Cancellation • Multiplication property of a fraction 14-3 Cancellation method 14-5 Common denominator • Least common denominator 14-8 Algebraic equation • Fractional equation • Extraneous solution REVIEW EXERCISES 1. Explain the difference between an algebraic fraction and a fractional expression. 2. What fractional part of 1 centimeter is x millimeters? 3. For what value of y is the fraction 4. Factor completely: 12x3 27x y 2 1 y 2 4 undefined? Review Exercises 571 In 5–8, reduce each fraction to lowest terms. 5. 8bg 12bg 6. 14d 7d2 7. 5x2 2 60 5 8. 8y2 2 12y 8y In 9–23, in each case, perform the indicated operation and express the answer in lowest terms. 3xy 2x 9. 10. 12. 13. 15. 2x 2 2 3x2 8 4 ? 9x 7b 4 18a 3a 35 3 1 ax ax 4 x2 2 5x x ? x2 2x 2 10 x2 2 25 12 4 x2 2 10x 1 25 3y ? 6 2 m 5m 6 5 xy 2 2 yz a 1 b 1 2b 2a a 1 b c 2 3 12 1 c 1 3 8 b 24. If the sides of a triangle are represented by , 2 perimeter of the triangle in simplest form. 16. 22. 21. 19. 18. 4 11. 14. 17. 20. 23. 6c2 3x 1 2x 1 1 4x 1 5 2x y 1 3 y 1 7 5 2 4 3a 2 9a2 a 4 (1 2 9a2) 5b 6 , and 2b 3 , express the 25. If a 2, b 3, and c 4, what is the sum of b a 1 a c ? In 26–31, solve each equation and check. 26. 29. 20 5 3 k 4 6 m 5 20 m 2 2 27. 30. x 2 3 10 5 4 5 2t 5 2 t 2 2 10 5 2 28. 31 20 In 32–34, solve each equation for r in terms of the other variables. 32. S h 5 2pr 33. c 2r 5 p 34. a r 2 n 5 0 35. Mr. Vroman deposited a sum of money in the bank. After a few years, he found that the interest equaled one-fourth of his original deposit and he had a total sum, deposit plus interest, of $2,400 in the bank. What was the original deposit? 36. One-third of the result obtained by adding 5 to a certain number is equal to o
ne-half of the result obtained when 5 is subtracted from the number. Find the number. 37. Of the total number of points scored by the winning team in a basketball game, one-fifth was scored in the first quarter, one-sixth was scored in the second quarter, one-third was scored in the third quarter, and 27 was scored in the fourth quarter. How many points did the winning team score? 38. Ross drove 300 miles at r miles per hour and 360 miles at r 10 miles per hour. If the time needed to drive 300 miles was equal to the time needed to drive 360 miles, find the rates at which Ross drove. (Express the time t 5 d needed for each part of the trip as r .) 572 Algebraic Fractions, and Equations and Inequalities Involving Fractions 39. The total cost, T, of n items that cost a dollars each is given by the equa- tion T na. a. Solve the equation T na for n in terms of T and a. b. Use your answer to a to express n1, the number of cans of soda that cost $12.00 if each can of soda costs a dollars. c. Use your answer to a to express n2, the number of cans of soda that cost $15.00 if each can of soda costs a dollars. d. If the number of cans of soda purchased for $12.00 is 4 less than the number purchased for $15.00, find the cost of a can of soda and the number of cans of soda purchased. 40. The cost of two cups of coffee and a bagel is $1.75. The cost of four cups of coffee and three bagels is $4.25. What is the cost of a cup of coffee and the cost of a bagel? 41. A piggybank contains nickels, dimes, and quarters. The number of nickels is 4 more than the number of dimes, and the number of quarters is 3 times the number of nickels. If the total value of the coins is no greater than $8.60, what is the greatest possible number of dimes in the bank? Exploration Some rational numbers can be written as terminating decimals and others as infinitely repeating decimals. (1) Write each of the following fractions as a decimal: 1 2, 1 4, 1 5, 1 8, 1 10, 1 16, 1 20, 1 25, 1 50, 1 100 (2) What do you observe about the decimals written in (1)? (3) Write each denominator in factored form. (4) What do you observe about the factors of the denominators? (5) Write each of the following fractions as a decimal: 1 3, 1 6, 1 9, 1 11, 1 12, 1 15, 1 18, 1 22, 1 24, 1 30 What do you observe about the decimals written in (5)? (7) Write each denominator in factored form. (8) What do you observe about the factors of the denominators? (9) Write a statement about terminating and infinitely repeating decimals based on your observations. CUMULATIVE REVIEW Part I Cumulative Review 573 CHAPTERS 1–14 Answer all questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. 1. The product of 3a2 and 5a5 is (1) 15a10 (2) 15a7 (3) 8a10 (4) 8a7 2. In the coordinate plane, the point whose coordinates are (2, 1) is in quadrant (1) I (2) II (3) III (4) IV 3. In decimal notation, 3.75 10–2 is (1) 0.0375 (2) 0.00375 4. The slope of the line whose equation is 3x y 5 is (3) 37.5 (1) 5 (2) 5 (3) 3 5. Which of the following is an irrational number? 2 3 (1) 1.3 (3) (2) 9 " 6. The factors of x2 7x 18 are (1) 9 and 2 (2) 9 and 2 (3) (x 9) and (x 2) (4) (x 9) and (x 2) (4) 375 (4) 3 (4) 5 " 7. The dimensions of a rectangular box are 8 by 5 by 9. The surface area is (1) 360 cubic units (2) 360 square units (3) 157 square units (4) 314 square units 8. The length of one leg of a right triangle is 8 and the length of the hypotenuse is 12. The length of the other leg is (1) 4 9. The solution set of (1) {1, 2} 4 5 (2) " a2 2) {2} is (3) 4 13 " (3) {0, 1} (4) 80 (4) {1, 2} 10. In the last n times a baseball player was up to bat, he got 3 hits and struck out the rest of the times. The ratio of hits to strike-outs is (1) 3 n Part II (2) n 2 3 n (3) 3 n 2 3 (4) n 2 3 3 Answer all questions in this part. Each correct answer will receive 2 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 574 Algebraic Fractions, and Equations and Inequalities Involving Fractions 11. Mrs. Kniger bought some stock on May 1 for $3,500. By June 1, the value of the stock was $3,640. What was the percent of increase of the cost of the stock? 12. A furlong is one-eighth of a mile. A horse ran 10 furlongs in 2.5 minutes. What was the speed of the horse in feet per second? Part III Answer all questions in this part. Each correct answer will receive 3 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 13. Two cans of soda and an order of fries cost $2.60. One can of soda and two orders of fries cost $2.80. What is the cost of a can of soda and of an order of fries? 14. a. Draw the graph of y x2 2x . b. From the graph, determine the solution set of the equation x2 2x 3. Part IV Answer all questions in this part. Each correct answer will receive 4 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 15. If the measure of the smallest angle of a right triangle is 32° and the length of the shortest side is 36.5 centimeters, find the length of the hypotenuse of the triangle to the nearest tenth of a centimeter. 16. The area of a garden is 120 square feet. The length of the garden is 1 foot less than twice the width. What are the dimensions of the garden? PROBABILITY Mathematicians first studied probability by looking at situations involving games of chance.Today, probability is used in a wide variety of fields. In medicine, it helps us to determine the chances of catching an infection or of controlling an epidemic, and the likelihood that a drug will be effective in curing a disease. In industry, probability tells us how long a manufactured product should last or how many defective items may be expected in a production run. In biology, the study of genes inherited from one’s parents and grandparents is a direct application of probability. Probability helps us to predict when more tellers are needed at bank windows, when and where traffic jams are likely to occur, and what kind of weather we may expect for the next few days. While the list of applications is almost endless, all of them demand a strong knowledge of higher mathematics. As you study this chapter, you will learn to solve problems such as the following: A doctor finds that, as winter approaches, 45% of her patients need flu shots, 20% need pneumonia shots, and 5% need both. What is the probability that the next patient that the doctor sees will need either a flu shot or a pneumonia shot? Like the early mathematicians, we will begin a formal study of probability by looking at games and other rather simple applications. CHAPTER 15 CHAPTER TABLE OF CONTENTS 15-1 Empirical Probability 15-2 Theoretical Probability 15-3 Evaluating Simple Probabilities 15-4 The Probability of (A and B) 15-5 The Probability of (A or B) 15-6 The Probability of (Not A) 15-7 The Counting Principle, Sample Spaces, and Probability 15-8 Probabilities With Two or More Activities 15-9 Permutations 15-10 Permutations With Repetition 15-11 Combinations 15-12 Permutations, Combinations, and Probability Chapter Summary Vocabulary Review Exercises Cumulative Review 575 576 Probability 15-1 EMPIRICAL PROBABILITY Probability is a branch of mathematics in which the chance of an event happening is assigned a numerical value that predicts how likely that event is to occur. Although this prediction tells us little about what may happen in an individual case, it can provide valuable information about what to expect in a large number of cases. A decision is sometimes reached by the toss of a coin: “Heads, we’ll go to the movies; tails, we’ll go bowling.” When we toss a coin, we don’t know whether the coin will land with the head side or the tail side facing upward. However, we believe that heads and tails have equal chances of happening whenever we toss a fair coin. We can describe this situation by saying that the probability of heads is and the probability of tails is , symbolized as: 1 2 1 2 P(heads) 5 1 2 or P(H) 5 1 2 P(tails) 5 1 2 or P(T) 5 1 2 Before we define probability, let us consider two more situations. 1. Suppose we toss a coin and it lands heads up. If we were to toss the coin a second time, would the coin land tails up? Is your answer “I don’t know”? Good! We cannot say that the coin must now be tails because we cannot predict the next result with certainty. 2. Suppose we take an index card Card and fold it down the center. If we then toss the card and let it fall, there are only three possible results. The card may land on its side, it may land on its edge, or it may form a tent when it lands. Can we say P(edge) , P(side) 5 1 3 , and P(tent) 5 1 3 ? 5 1 3 Side Edge Tent Again, your answer should be “I don’t know.” We cannot assign a number as a probability until we have some evidence to support our claim. In fact, if we were to gather evidence by tossing this card, we would find that the probabili3, 1 1 ties are not because, unlike the result of tossing the coin, each result is not equally likely to occur. 3, and 1 3 Variables that might affect the experiment include the dimensions of the index card, the weight of the cardboard, and the angle measure of the fold. (An index card with a 10° opening would be much less likely to form a tent than an index card with a 110° opening.) Empirical Probability 577 An Empirical Study Let us go back to the problem of tossing a coin. While we cannot predict the 1 result of one toss of a coin, we can still say that the probability of heads is 2 based on observations made in an empirical stu
dy. In an empirical study, we perform an experiment many times, keep records of the results, and then analyze these results. 20 20 20 20 20 20 20 20 20 20 Number of Heads Number of Tosses For example, ten students decided to take turns tossing a coin. Each student completed 20 tosses and the number of heads was recorded as shown at the right. If we look at the results and think of the probability of heads as a fraction comparing the number of heads to the total number of tosses, only Maria, with 10 heads out of 20 tosses, had results 1 where the probability was , or . This 2 fraction is called the relative frequency. Elizabeth had the lowest rel- 10 20 Albert Peter Thomas Maria Elizabeth Joanna Kathy Jeanne 8 13 12 10 6 12 11 7 6 20 ative frequency of heads, . Peter and Debbie tied for the highest relative . Maria’s relative frefrequency with 1 quency was 2 other students were incorrect; the coins simply fell that way. 13 20 or . This does not mean that Maria had correct results while the James 10 20 9 Debbie 13 The students decided to combine their results, by expanding the chart, to see what happened when all 200 tosses of the coin were considered. As shown in columns 3 and 4 of the table on the next page, each cumulative result is found by adding all the results up to that point. For example, in the second row, by adding the 8 heads that Albert tossed and the 13 heads that Peter tossed, we find the cumulative number of heads up to this point to be 21, the total number of heads tossed by Albert and Peter together. Similarly, when the 20 tosses that Albert made and the 20 tosses that Peter made are added, the cumulative number of tosses up to this point is 40. Since each student completed 20 coin tosses, the cumulative number of tosses should increase by 20 for each row. The cumulative number of heads should increase by varying amounts for each row since each student experienced different results. 578 Probability (COL. 1) (COL. 2) (COL. 3) (COL. 4) (COL. 5) Number Number of Tosses of Heads Albert Peter Thomas Maria Elizabeth Joanna Kathy Jeanne Debbie James 8 13 12 10 6 12 11 7 13 9 20 20 20 20 20 20 20 20 20 20 Cumulative Cumulative Cumulative Number of Tosses Relative Frequency Number of Heads 8 21 33 43 49 61 72 79 92 101 20 40 60 80 100 120 140 160 180 200 8 20 5 21 40 5 33 60 5 43 80 5 49 100 5 61 120 5 72 140 5 79 160 5 92 180 5 101 200 5 .400 .525 .550 .538 .490 .508 .514 .494 .511 .505 In column 5, the cumulative relative frequency is found by dividing the total number of heads at or above a row by the total number of tosses at or above that row. The cumulative relative frequency is shown as a fraction and then, for easy comparison, as a decimal. The decimal is given to the nearest thousandth. 6 20 for Peter and Debbie, the cumulative relative frequency, after While the relative frequency for individual students varied greatly, from for Elizabeth to 1 all 200 tosses were combined, was a number very close to . 2 A graph of the results of columns 4 and 5 will tell us even more. In the graph, the horizontal axis is labeled “Number of tosses” to show the cumulative results of column 4; the vertical axis is labeled “Cumulative relative frequency of heads” to show the results of column 5. 13 20 CUMULATIVE RELATIVE FREQUENCY OF HEADS FOR 0 TO 200 TOSSES OF A COIN .6 0.5 0.4 0.3 0.2 0.1 0 0 20 40 60 80 140 100 Number of tosses 120 160 180 200 Empirical Probability 579 On the graph, we have plotted the points that represent the data in columns 4 and 5 of the preceding table, and we have connected these points to form a line graph. Notice how the line moves up and down around the relative frequency of 0.5, or . The graph shows that the more times the coin is tossed, the 1 closer the relative frequency comes to . In other words, the line seems to level 2 out at a relative frequency of . We say that the cumulative relative frequency converges to the number and the coin will land heads up about one-half of the time. 1 2 1 2 1 2 Even though the cumulative relative frequency of , we sense that the line will approach the number . When we use carefully collected evidence about tossing a fair coin to guess that the probability of heads is , we have arrived at this conclusion empirically, that is, by experimentation and observation. is not exactly 1 2 1 2 101 200 1 2 Empirical probability may be defined as the most accurate scientific estimate, based on a large number of trials, of the cumulative relative frequency of an event happening. Experiments in Probability A single attempt at doing something, such as tossing a coin only once, is called a trial. We perform experiments in probability by repeating the same trial many times. Experiments are aimed at finding the probabilities to be assigned to the occurrences of an event, such as heads coming up on a coin. The objects used in an experiment may be classified into one of two categories: 1. Fair and unbiased objects have not been weighted or made unbalanced. An object is fair when the different possible results have equal chances of happening. Objects such as coins, cards, and spinners will always be treated in this book as fair objects, unless otherwise noted. 2. Biased objects have been tampered with or are weighted to give one result a better chance of happening than another. The folded index card described earlier in this section is a biased object because the probability of each of three results is not . The card is weighted so that it will fall on its side more often than it will fall on its edge. 1 3 You have seen how to determine empirical probability by the tables and graph previously shown in this section. Sometimes, however, it is possible to guess the probability that should be assigned to the result described before you start an experiment. In Examples 1–5, use common sense to guess the probability that might be assigned to each result. (The answers are given without comment here. You will learn how to determine these probabilities in the next section.) 580 Probability EXAMPLE 1 A die is a six-sided solid object (a cube). Each side (or face) is a square. The sides are marked with 1, 2, 3, 4, 5, and 6 pips, respectively. (The plural of die is dice.) In rolling a fair die, getting a 4 means that the side showing four pips is facing up. Find the probability of getting a 4, or P(4). Die Faces of a die Answer P(4) 5 1 6 EXAMPLE 2 A standard deck of cards contains 52 cards. There are four suits: hearts, diamonds, spades, and clubs. Each suit contains 13 cards: 2, 3, 4, 5, 6, 7, 8, 9, 10, jack, queen, king, and ace. The diamonds and hearts are red; the spades and clubs are black. In selecting a card from the deck without looking, find the probability of drawing: a. the 7 of diamonds b. a 7 c. a diamond Answers a. There is only one 7 of diamonds. P(7 of diamonds) 5 1 52 b. There are four 7s. P(7) 5 4 52 or 1 13 c. There are 13 diamonds. P(diamond) 5 13 52 or 1 4 EXAMPLE 3 There are 10 digits in our numeral system: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. In selecting a digit without looking, what is the probability it will be: a. the 8? b. an odd digit? Answers a. P(8) 5 1 10 b. P(odd) 5 10 or 1 2 EXAMPLE 4 A jar contains eight marbles: three are white and the remaining five are blue. In selecting a marble without looking, what is the probability it will be blue? (All marbles are the same size.) Answer P(blue) 5 5 8 Empirical Probability 581 EXAMPLE 5 The English alphabet contains 26 letters. There are 5 vowels (A, E, I, O, U). The other 21 letters are consonants. If a person turns 26 tiles from a word game facedown and each tile represents a different letter of the alphabet, what is the probability of turning over: a. the A? b. a vowel? c. a consonant? Answers a. P(A) 5 1 26 b. P(vowel) 5 5 26 c. P(consonant) 5 21 26 EXERCISES Writing About Mathematics 1. Alicia read that, in a given year, one out of four people will be involved in an automobile accident. There are four people in Alicia’s family. Alicia concluded that this year, one of the people in her family will be involved in an automobile accident. Do you agree with Alicia’s conclusion? Explain why or why not. 2. A library has a collection of 25,000 books. Is the probability that a particular book will be checked out 1 25,000 ? Explain why or why not. Developing Skills In 3–8, in each case, a fair, unbiased object is involved. These questions should be answered without conducting an experiment; take a guess. 3. The six sides of a number cube are labeled 1, 2, 3, 4, 5, and 6. Find P(5), the probability of getting a 5 when the die is rolled. 4. In drawing a card from a standard deck without looking, find P(any heart). 5. Each of 10 pieces of paper contains a different number from the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. The pieces of paper are folded and placed in a bag. In selecting a piece of paper without looking, find P(7). 6. A jar contains five marbles, all the same size. Two marbles are black, and the other three are white. In selecting a marble without looking, find P(black). 7. Using the lettered tiles from a word game, a boy places 26 tiles, one tile for each letter of the alphabet, facedown on a table. After mixing up the tiles, he picks one. What is the probability that the tile contains one of the letters in the word MATH? 582 Probability 8. A tetrahedron is a four-sided object. Each side, or face, is an equilateral triangle. The numerals 1, 2, 3, and 4 are used to number the different faces, as shown in the figure. A trial consists of rolling the tetrahedron and reading the number that is facedown. Find P(4). 1 3 2 4 Tetrahedron Faces of a tetrahedron 9. By yourself, or with some classmates, conduct any of the experiments described in Exercises 3–8 to verify that you have assigned the correct probability to the event or result described. A good experiment should contain at least 100 trials. 10. The figure at the right shows a spinner that has an equal chance of landing on one of four sectors. The regions
are equal in size and are numbered 1, 2, 3, and 4. An experiment was conducted by five people to find the probability that the arrow will land on the 2. Each person spun the arrow 100 times. When the arrow landed on a line, the result did not count and the arrow was spun again. a. Before doing the experiment, what probability would you assign to the arrow landing on the 2? (In symbols, P(2) ?) 1 3 2 4 b. Copy and complete the table below to find the cumulative results of this experiment. In the last column, record the cumulative relative frequencies as fractions and as decimals to the nearest thousandth. Number of Times Arrow Number of Landed on 2 Spins Cumulative Number of Times Arrow Landed on 2 Cumulative Number of Spins Barbara Tom Ann Eddie Cathy 29 31 19 23 24 100 100 100 100 100 29 60 100 200 Cumulative Relative Frequency 29 100 5 .290 c. Did the experiment provide evidence that the probability you assigned in part a was correct? d. Form a group of five people and repeat the experiment, making a table similar to the one shown above. Do your results provide evidence that the probability you assigned in a was correct? Empirical Probability 583 In 11–15, a biased object is described. A probability can be assigned to a result only by conducting an experiment to determine the cumulative relative frequency of the event. While you may wish to guess at the probability of the event before starting the experiment, conduct at least 100 trials to determine the best probability to be assigned. 11. An index card is folded in half and tossed. As described earlier in this section, the card may land in one of three positions: on its side, on its edge, or in the form of a tent. In tossing the folded card, find P(tent), the probability that the card will form a tent when it lands. 12. A paper cup is tossed. It can land in one of three positions: on its top, on its bottom, or on its side, as shown in the figure. In tossing the cup, find P(top), the probability that the cup will land on its top. Top Bottom Side 13. A nickel and a quarter are glued or taped together so that the two faces seen are the head of the quarter and the tail of the nickel. (This is a very crude model of a weighted coin.) In tossing the coin, find P(head). 14. A paper cup in the shape of a cone is tossed. It can land in one of two positions: on its base or on its side, as shown in the figure. In tossing this cup, find P(side), the probability that the cup will land on its side. Base Side 15. A thumbtack is tossed. It may land either with the pin up or with the pin down, as shown in the figure. In tossing the thumbtack, find P(pin up). Pin up Pin down 16. The first word is selected from a page of a book written in English. a. What is the probability that the word contains at least one of the letters a, e, i, o, u, or y? b. In general, what is the largest possible probability? c. What is the probability that the word does not contain at least one of the letters a, e, i, o, u, or y? d. In general, what is the smallest possible probability? Applying Skills 17. An insurance company’s records show that last year, of the 1,000 cars insured by the com- pany, 210 were involved in accidents. What is the probability that an insurance policy, chosen at random from their files, is that of a car that was not involved in an accident? 18. A school’s attendance records for last year show that of the 885 students enrolled, 15 had no absences for the year. What is the probability that a student, chosen at random, had no absences? 19. A chess club consists of 45 members of whom 24 are boys and 21 are girls. If a member of the club is chosen at random to represent the club at a tournament, what is the probability that the person chosen is a boy? 584 Probability 15-2 THEORETICAL PROBABILITY An empirical approach to probability is necessary whenever we deal with biased objects. However, common sense tells us that there is a simple way to define the probability of an event when we deal with fair, unbiased objects. For example, let us suppose that Alma is playing a game in which each player must roll a die. To win, Alma must roll a number greater than 4. What is the probability that Alma will win on her next turn? Common sense tells us that: 1. The die has an equal chance of falling in any one of six ways: 1, 2, 3, 4, 5, and 6. 2. There are two ways for Alma to win: rolling a 5 or a 6. 3. Therefore: P(Alma wins) number of winning results number of possible results 5 2 6 5 1 3 . Terms and Definitions Using the details of the preceding example, let us examine the correct terminology to be used. An outcome is a result of some activity or experiment. In rolling a die, 1 is an outcome, 2 is an outcome, 3 is an outcome, and so on. There are six outcomes when rolling a six-sided die. A sample space is a set of all possible outcomes for the activity. When rolling a die, there are six possible outcomes in the sample space: 1, 2, 3, 4, 5, and 6. We say that the sample space is {1, 2, 3, 4, 5, 6}. An event is a subset of the sample space. We use the term event in two ways. In ordinary conversation, it means a situation or happening. In the technical language of probability, it is the subset of the sample space that lists all of the outcomes for a given situation. When we focus on a particular event, such as heads facing up when we toss a coin, we refer to it as the favorable event. The other event or events in the sample space, in this case tails, are called unfavorable. When we roll a die, we may define many different situations. Each of these is called an event. 1. For Alma, the event of rolling a number greater than 4 contains only two outcomes: 5 and 6. 2. For Lee, a different event might be rolling a number less than 5. This event contains four outcomes: 1, 2, 3, and 4. Theoretical Probability 585 3. For Sandi, the event of rolling a 2 contains only one outcome: 2. When there is only one outcome, we call this a singleton event. We can now define theoretical probability for fair, unbiased objects: The theoretical probability of an event is the number of ways that the event can occur, divided by the total number of possibilities in the sample space. In symbolic form, we write: where P(E) n(E) n(S) P(E) n(E) n(S) represents the probability of event E; represents the number of ways event E can occur or the number of outcomes in event E; represents the total number of possibilities, or the number of outcomes in sample space S. Since theoretical probability relies on calculation as opposed to experimen- tation, it is sometimes referred to as calculated probability. Let us now use this formula to write the probabilities of the three events described above. 1. For Alma, there are two ways to roll a number greater than 4, and there are six possible ways that the die may fall. We say: E the set of numbers on a die that are greater than 4: {5, 6}. n(E) 2 since there are 2 outcomes in this event. S the set of all possible outcomes: {1, 2, 3, 4, 5, 6}. n(S) 6 since there are six outcomes in the sample space. Therefore: 2. For Lee, the probability of rolling a number less than 5 is P(E) n(E) n(S) 5 2 6 5 1 3 n(E) n(S) 5 4 3. For Sandi, the probability of rolling a 2 is P(E) 6 5 2 3 P(E) n(E) n(S) 5 1 6 586 Probability Uniform Probability A sample space is said to have uniform probability, or to contain equally likely outcomes, when each of the possible outcomes has an equal chance of occurring. In rolling a die, there are six possible outcomes in the sample space; each is equally likely to occur. Therefore(6) ; P(5) ; P(4) ; P(3) ; P(2) P(1) We say that the die has uniform probability. If, however, a die is weighted to make it biased, then one or more sides will have a probability greater than while one or more sides will have a probabil- 1 6, ity less than theoretical probability does not apply to weighted objects. . A weighted die does not have uniform probability; the rule for 1 6 Random Selection When we select an object from a collection of objects without knowing any of the special characteristics of the object, we are making a random selection. Random selections are made when drawing a marble from a bag, when taking a card from a deck, or when picking a name out of a hat. In the same way, we may use the word random to describe the outcomes when tossing a coin or rolling a die; the outcomes happen without any special selection on our part. Procedure To find the simple probability of an event: 1. Count the total number of outcomes in the sample space S: n(S). 2. Count all the possible outcomes of the event E: n(E). 3. Substitute these values in the formula for the probability of event E: P(E) n(E) n(S) Note: The probability of an event is usually written as a fraction. A standard calculator display, however, is in decimal form. Therefore, if you use a calculator when working with probability, it is helpful to know fraction-decimal equivalents. Recall that some common fractions have equivalent decimals that are ter- minating decimals: 1 2 5 0.5 1 4 5 0.25 1 5 5 0.2 1 8 5 0.125 Theoretical Probability 587 Others have equivalent decimals that are repeating decimals: 1 3 5 0.3 2 3 5 0.6 1 6 5 0.16 1 9 5 0.1 The graphing calculator has a function that will change a decimal fraction to a common fraction in lowest terms. For example, the probability of a die showing a number greater than 4 can be displayed on a calculator as follows: ENTER: 2 6 ENTER ENTER: MATH 1 ENTER DISPLAY: 2 / 6 DISPLAY EXAMPLE 1 A standard deck of 52 cards is shuffled. Daniella draws a single card from the deck at random. What is the probability that the card is a jack? Solution S sample space of all possible outcomes, or 52 cards. Thus, n(S) 52. J event of selecting a jack. There are four jacks in the deck: jack of hearts, of diamonds, of spades, and of clubs. Thus, n(J) 4. P(J) number of possible jacks number of possible cards 5 n(J) n(S) 5 4 52 5 1 13 Answer Calculator Solution ENTER: 4 52 MATH 1 ENTER DISPLAY EXAMPLE
2 An aquarium at a pet store contains 8 goldfish, 6 angelfish, and 7 zebrafish. David randomly chooses a fish to take home for his fishbowl. a. How many possible outcomes are in the sample space? b. What is the probability that David takes home a zebrafish? Solution a. Each fish represents a distinct outcome. Therefore, if S is the sample space of all possible outcomes, then n(S) 8 6 7 21. b. Since there are 7 zebrafish: P(zebrafish) 5 7 21 5 1 3 . Answers a. 21 b. 1 3 588 Probability EXAMPLE 3 A spinner contains eight regions, numbered 1 through 8, as shown in the figure. The arrow has an equally likely chance of landing on any of the eight regions. If the arrow lands on a line, the result is not counted and the arrow is spun again. a. How many possible outcomes are in the sample space S. What is the probability that the arrow lands on the 4? That is, what is P(4)? c. List the set of possible outcomes for event O, in which the arrow lands on an odd number. d. Find the probability that the arrow lands on an odd number. Answers a. n(S) 8 b. Since there is only one region numbered 4 out of eight regions: P(4) c. Event O {1, 3, 5, 7} d. Since O {1, 3, 5, 7}, n(O) 4. P(O) n(O) n(S EXERCISES Writing About Mathematics 1. A spinner is divided into three sections numbered 1, 2, and 3, as shown in the figure. Explain why the probability of the arrow landing on the region num1 bered 1 is not . 3 2 3 1 2. Mark said that since there are 50 states, the probability that the next baby born in the United States will be born in New Jersey is why not. 1 50 . Do you agree with Mark? Explain why or Developing Skills 3. A fair coin is tossed. a. List the sample space. b. What is P(head), the probability that a head will appear? c. What is P(tail)? Theoretical Probability 589 In 4–9, a fair die is tossed. For each question: a. List the possible outcomes for the event. b. State the probability of the event. 4. The number 3 appears. 5. An even number appears. 6. A number less than 3 appears. 7. An odd number appears. 8. A number greater than 3 appears. 9. A number greater than or equal to 3 appears. In 10–15, a spinner is divided into five equal regions, numbered 1 through 5, as shown below. The arrow is spun and lands in one of the regions. For each question: a. List the outcomes for the event. b. State the probability of the event. 10. The arrow lands on number 3. 11. The arrow lands on an even number. 12. The arrow lands on a number less than 3. 13. The arrow lands on an odd number. 14. The arrow lands on a number greater than 3. 15. The arrow lands on a number greater than or equal to 3. 1 2 5 3 4 16. A standard deck of 52 cards is shuffled, and one card is drawn. What is the probability that the card is: a. the queen of hearts? b. a queen? d. a red card? g. an ace? e. the 7 of clubs? h. a red 7? c. a heart? f. a club? i. a black 10? j. a picture card (king, queen, jack)? 17. A person does not know the answer to a test question and takes a guess. Find the probabil- ity that the answer is correct if the question is: a. a multiple-choice question with four choices b. a true-false question c. a question where the choices given are “sometimes, always, or never” 18. A marble is drawn at random from a bag. Find the probability that the marble is green if the bag contains marbles whose colors are: a. 3 blue, 2 green d. 6 blue, 4 green b. 4 blue, 1 green e. 3 green, 9 blue c. 5 red, 2 green, 3 blue f. 5 red, 2 green, 9 blue 19. The digits of the number 1,776 are written on disks and placed in a jar. What is the probabil- ity that the digit 7 will be chosen on a single random draw? 20. A letter is chosen at random from a given word. Find the probability that the letter is a vowel if the word is: a. APPLE b. BANANA c. GEOMETRY d. MATHEMATICS 590 Probability Applying Skills 21. There are 16 boys and 14 girls in a class. The teacher calls students at random to the chalk- board. What is the probability that the first person called is: a. a boy? b. a girl? 22. There are 840 tickets sold in a raffle. Jay bought five tickets, and Lynn bought four tickets. What is the probability that: a. Jay has the winning ticket? b. Lynn has the winning ticket? 23. The figures that follow are eight polygons: a square; a rectangle; a parallelogram that is not a rectangle; a right triangle; an isosceles triangle that does not contain a right angle; a trapezoid that does not contain a right angle; an equilateral triangle; a regular hexagon. One of the figures is selected at random. What is the probability that this polygon: a. contains a right angle? b. is a quadrilateral? c. is a triangle? d. has at least one acute angle? e. has all sides congruent? f. has at least two sides congruent? g. has fewer than five sides? h. has an odd number of sides? i. has four or more sides? j. has at least two obtuse angles? 15-3 EVALUATING SIMPLE PROBABILITIES We have called an event for which there is only one outcome a singleton. For example, when rolling a die only once, getting a 3 is a singleton. However, when rolling a die only once, some events are not singletons. For example: 1. The event of rolling an even number on a die is {2, 4, 6}. 2. The event of rolling a number less than 6 on a die is {1, 2, 3, 4, 5}. Evaluating Simple Probabilities 591 The Impossible Case On a single roll of a die, what is the probability that the number 7 will appear? We call this case an impossibility because there is no way in which this event can occur. In this example, event E rolling a 7; so, E { } or , and n(E) 0. The sample space S for rolling a die contains six possible outcomes, and n(S) 6. Therefore: P(E) number of ways to roll a 7 number of outcomes for the die 5 n(E) n(S) 5 0 6 5 0 In general, for any sample space S containing k possible outcomes, we say n(S) k. For any impossible event E, which cannot occur in any way, we say n(E) 0. Thus, the probability of an impossible event is: P(E) n(E) n(S) 5 0 k 5 0 and we say: The probability of an impossible event is 0. There are many other impossibilities where the probability must equal zero. For example, the probability of selecting the letter E from the word PROBABILITY is 0. Also, selecting a coin worth 9 cents from a bank containing a nickel, a dime, and a quarter is an impossible event. The Certain Case On a single roll of a die, what is the probability that a whole number less than 7 will appear? We call this case a certainty because every one of the possible outcomes in the sample space is also an outcome for this event. In this example, the event E rolling a whole number less than 7, so n(E) = 6. The sample space S for rolling a die contains six possible outcomes, so n(S) 6. Therefore: P(E) number of ways to roll a number less than 7 number of outcomes for the die 5 n(E) n(S) 5 6 6 5 1 When an event E is certain, the event E is the same as the sample space S, that is, E 5 S and n(E) n(S). In general, for any sample space S containing k possible outcomes, n(S) k. When the event E is certain, every possible outcome for the sample space is also an outcome for event E, or n(E) = k. Thus, the probability of a certainty is given as: P(E) n(E) n(S) 5 k k 5 1 and we say: The probability of an event that is certain to occur is 1. 592 Probability There are many other certainties where the probability must equal 1. Examples include the probability of selecting a consonant from the letters JFK or selecting a red sweater from a drawer containing only red sweaters. The Probability of Any Event The smallest possible probability is 0, for an impossible case; no probability can be less than 0. The largest possible probability is 1, for a certain event; no probability can be greater than 1. Many other events, as seen earlier, however, have probabilities that fall between 0 and 1. Therefore, we conclude: The probability of any event E must be equal to or greater than 0, and less than or equal to 1: 0 P(E) 1 Subscripts in Sample Spaces A sample space may sometimes contain two or more objects that are exactly alike. To distinguish one object from another, we make use of subscripts. A subscript is a number, usually written in smaller size to the lower right of a term. For example, a box contains six jellybeans: two red, three green, and one yellow. Using R, G, and Y to represent the colors red, green, and yellow, respectively, we can list this sample space in full, using subscripts: {R1, R2, G1, G2, G3, Y1} Since there is only one yellow jellybean, we could have listed the last out- come as Y instead of Y1. EXAMPLE 1 An arrow is spun once and lands on one of three equally likely regions, numbered 1, 2, and 3, as shown in the figure. a. List the sample space for this experiment. b. List all eight possible events for one spin of the arrow. 1 2 3 Solution a. The sample space S {1, 2, 3}. b. Since events are subsets of the sample space S, the eight possible events are the eight subsets of S: { }, the empty set for impossible events. The arrow lands on a number other than 1, 2, or 3. {1}, a singleton. The arrow lands on 1 or the arrow lands on a number less than 2. Evaluating Simple Probabilities 593 {2}, a singleton. The arrow lands on 2 or the arrow lands on an even number. {3}, a singleton. The arrow lands on 3 or the arrow lands on a number greater than 2. {1, 2}, an event with two possible outcomes. The arrow does not land on 3 or does land on a number less than 3. {1, 3}, an event with two possible outcomes. The arrow lands on an odd number or does not land on 2. {2, 3}, an event with two possible outcomes. The arrow lands on a number greater than 1 or does not land on 1. {1, 2, 3}, the sample space itself for events that are certain. The arrow lands on a whole number less than 4 or greater than 0. Answers a. S {1, 2, 3} b. { }, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3} EXAMPLE 2 A piggy bank contains a nickel, two dimes, and a quarter. A person selects one of the coins. What is the probability that the coin is worth: a. exactly 10 cents? b. a
t least 10 cents? c. exactly 3 cents? d. more than 3 cents? Solution The sample space for this example is {N, D1, D2, Q}. Therefore, n(S) 4. a. There are two coins worth exactly 10 cents, D1 and D2. Therefore, n(E) 2 and P(coin worth 10 cents) n(E) n(S) 5 2 4 5 1 2 . b. There are three coins worth at least 10 cents, D1, D2, and Q. Therefore, n(E) 3 and P(coin worth at least 10 cents) n(E) n(S) 5 3 4 . c. There is no coin worth exactly 3 cents. This is an impossible event. Therefore, P(coin worth 3 cents) 0. d. Each of the four coins is worth more than 3 cents. This is a certain event. Therefore, P(coin worth more than 3 cents) 1. Answers a. 1 2 b. 3 4 c. 0 d. 1 594 Probability EXAMPLE 3 In the Sullivan family, there are two more girls than boys. At random, Mrs. Sullivan asks one of her children to go to the store. If she is equally likely to ask any one of her children, and the probability that she asks a girl is , how many boys and how many girls are there in the Sullivan family? 2 3 Check number of girls number of children P(girl) 5 2 3 5? 4 6 2 3 5 2 3 ✔ Solution Let x the number of boys x 2 the number of girls 2x 2 the number of children. Then: P(girl) 5 number of girls number of children 3 5 x 1 2 2 2x 1 2 2(2x 1 2) 5 3(x 1 2) 4x 1 4 5 3x 1 6 x 5 2 Then x 2 4 and 2x 2 6 Answer There are two boys and four girls. EXERCISES Writing About Mathematics 1. Describe three events for which the probability is 0. 2. Describe three events for which the probability is 1. Developing Skills 3. A fair coin is tossed, and its sample space is S {H, T}. a. List all four possible events for the toss of a fair coin. b. Find the probability of each event named in part a. In 4–11, a spinner is divided into seven equal sectors, numbered 1 through 7. An arrow is spun to fall into one of the regions. For each question, find the probability that the arrow lands on the number described. 4. the number 5 6. a number less than 5 5. an even number 7. an odd number Evaluating Simple Probabilities 595 8. a number greater than 5 9. a number greater than 1 10. a number greater than 7 11. a number less than 8 12. A marble is drawn at random from a bag. Find the probability that the marble is black if the bag contains marbles whose colors are: a. 5 black, 2 green b. 2 black, 1 green c. 3 black, 4 green, 1 red d. 9 black e. 3 green, 4 red f. 3 green 13. Ted has two quarters, three dimes, and one nickel in his pocket. He pulls out a coin at ran- dom. Find the probability that the coin is worth: a. exactly 5 cents b. exactly 10 cents c. exactly 25 cents d. exactly 50 cents e. less than 25 cents f. less than 50 cents g. more than 25 cents h. more than l cent i. less than l cent 14. A single fair die is rolled. Find the probability for each event. a. The number 8 appears. b. A whole number appears. c. The number is less than 5. d. The number is less than 1. e. The number is less than 10. f. The number is negative. 15. A standard deck of 52 cards is shuffled, and a card is picked at random. Find the probability that the card is: a. a jack d. a red club b. a club c. a star e. a card from the deck f. a black club g. the jack of stars h. a 17 i. a red 17 In 16–20, a letter is chosen at random from a given word. For each question: a. Write the sample space, using subscripts to designate events if needed. b. Find the probability of the event. 16. Selecting the letter E from the word EVENT 17. Selecting the letter S from the word MISSISSIPPI 18. Selecting a vowel from the word TRIANGLE 19. Selecting a vowel from the word RECEIVE 20. Selecting a consonant from the word SPRY Applying Skills 21. There are 15 girls and 10 boys in a class. The teacher calls on a student in the class at random to answer a question. Express, in decimal form, the probability that the student called upon is: a. a girl b. a boy c. a pupil in the class d. a person who is not a student in the class 596 Probability 22. The last digit of a telephone number can be any of the following: 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9. Express, as a percent, the probability that the last digit is: a. 7 b. odd c. greater than 5 d. a whole number e. the letter R 23. A girl is holding five cards in her hand: the 3 of hearts, the 3 of diamonds, the 3 of clubs, the 4 of diamonds, the 7 of clubs. A player to her left takes one of these cards at random. Find the probability that the card selected from the five cards in the girl’s hand is: a. a 3 d. a black 4 g. a 5 b. a diamond c. a 4 e. a club f. the 4 of hearts h. the 7 of clubs i. a red card j. a number card k. a spade l. a number greater than 1 and less than 8. 24. A sack contains 20 marbles. The probability of drawing a green marble is . How many 2 5 green marbles are in the sack? 25. There are three more boys than girls in the chess club. A member of the club is to be chosen at random to play in a tournament. Each member is equally likely to be chosen. If the probability that a girl is chosen is , how many boys and how many girls are members of the club? 3 7 26. A box of candy contains caramels and nut clusters. There are six more caramels than nut clusters. If a piece of candy is to be chosen at random, the probability that it will be a caramel is . How many caramels and how many nut clusters are in the box? 3 5 27. At a fair, each ride costs one, two, or four tickets. The number of rides that cost two tickets is three times the number of rides that cost one ticket. Also, seven more rides cost four tickets than cost two tickets. Tycella, who has a book of tickets, goes on a ride at random. If the probability that the ride cost her four tickets is , how many rides are there at the fair? 4 7 15-4 THE PROBABILITY OF (A AND B) If a fair die is rolled, we can find simple probabilities, since we know that S {1, 2, 3, 4, 5, 6}. For example, let event A be rolling an even number. Then A {2, 4, 6} and P(A) Let event B be rolling a number less than 3. Then B {1, 2} and n(S) 5 3 6 n(A) P(B) n(B) n(S) 5 2 6 Now, what is the probability of obtaining a number on the die that is even and less than 3? We may think of this as event (A and B), which consists of those elements of the sample space that are in A and also in B. In set notation we say, The Probability of (A and B) 597 (A and B) A B {2} Only 2 is both even and less than 3. Notice that we use the symbol for inter- section () to denote and. Since n(A and B) 1, P(A and B) n(A and B) n(S) 5 1 6 Let us consider another example in which a fair die is rolled. What is the probability of rolling a number that is both odd and a 4? Event C {1, 3, 5}. Three numbers on a die are odd. P(C) n(C) n(S) 5 3 6 Event D {4}. One number on the die is 4. n(D) n(S) 5 1 6 (C and D) {numbers on a die that are odd and 4}= C D , the empty P(D) set. Since there is no outcome common to both C and D, n(C and D) 0. Therefore, P(C D) Conclusions n(C d D) n(S) 5 0 6 5 0 There is no simple rule or formula whereby the n(A) and n(B) can be used to find n(A and B). We must simply count the number of outcomes that are common in both events or write the intersection of the two sets and count the number of elements in that intersection. KEEP IN MIND Event (A and B) consists of the outcomes that are in event A and in event B. Event (A and B) may be regarded as the intersection of sets, namely, A B. EXAMPLE 1 A fair die is rolled once. Find the probability of obtaining a number that is greater than 3 and less than 6. Solution Event A {numbers greater than 3} {4, 5, 6}. Event B {numbers less than 6} {1, 2, 3, 4, 5}. Event (A and B) {numbers greater than 3 and less than 6} {4, 5}. n(A d B) n(S) 5 2 6 Therefore: P(A and B) or P(A B) n(A and B) n(S) 5 2 6 . Answer P(number greater than 3 and less than 6) or 1 3 2 6 598 Probability EXERCISES Writing About Mathematics 1. Give an example of two events A and B, such that P(A and B) P(A). 2. If P(A and B) P(A), what must be the relationship between set A and set B? Developing Skills 3. A fair die is rolled once. The sides are numbered 1, 2, 3, 4, 5, and 6. Find the probability that the number rolled is: a. greater than 2 and odd b. less than 4 and even c. greater than 2 and less than 4 d. less than 2 and even e. less than 6 and odd f. less than 4 and greater than 3 4. From a standard deck of cards, one card is drawn. Find the probability that the card is: a. the king of hearts b. a red king c. a king of clubs d. a black jack g. a 2 of spades e. a 10 of diamonds f. a red club h. a black 2 i. a red picture card Applying Skills 5. A set of polygons consists of an equilateral triangle, a square, a rhombus that is not a square, and a rectangle. One of the polygons is selected at random. Find the probability that the polygon contains: a. all sides congruent and all angles congruent b. all sides congruent and all right angles c. all sides congruent and two angles not congruent d. at least two congruent sides and at least two congruent angles e. at least three congruent sides and at least two congruent angles 6. In a class of 30 students, 23 take science, 28 take math, and all take either science or math. a. How many students take both science and math? b. A student from the class is selected at random. Find: (1) P(takes science) (2) P(takes math) (3) P(takes science and math) The Probability of (A or B) 599 7. At a karaoke party, some of the boys and girls take turns singing songs. Of the five boys, Patrick and Terence are teenagers while Brendan, Drew, and Kevin are younger. Of the seven girls, Heather and Claudia are teenagers while Maureen, Elizabeth, Gwen, Caitlin, and Kelly are younger. Find the probability that the first song is sung by: a. a girl c. a teenager e. a boy under 13 g. a teenage girl b. a boy d. someone under 13 years old f. a girl whose initial is C h. a girl under 13 i. a boy whose initial is C j. a teenage boy 15-5 THE PROBABILITY OF (A OR B) If a fair die is rolled, we can find simple probabilities. For example, let event A be rolling an even number. Then: P(A) n(A) n(S) 5 3 6 Let event C be ro
lling a number less than 2. Then: P(C) n(C) n(S) 5 1 6 Now, what is the probability of obtaining a number on the die that is even or less than 2? We may think of this as event (A or C). For the example above, there are four outcomes in the event (A or C): 1, 2, 4, and 6. Each of these numbers is either even or less than 2 or both. Since n(A or C) 4 and there are six elements in the sample space: Observe that P(A) 5 3 6 , P(C) 5 1 6 that P(A or C) n(A or C) n(S) 5 4 6 and P(A or C) 5 4 6 . In this case, it appears P(A)1P(C) 5 P(A or C) . Will this simple addition rule hold true for all problems? Before you say “yes,” consider the next example, in which a fair die is rolled. Event A {even numbers on a die} {2, 4, 6}. P(A) n(A) n(S) 5 3 6 Event B {numbers less than 3 on a die} {1, 2}. P(B) Then, event (A or B) {numbers that are even or less than 3}. n(B) n(S) 5 2 6 P(A or B) n(A or B) n(S) 5 4 6 600 Probability 5 3 6 5 2 6 Here P(A) . In this case, the simple rule of addition does not work: P(A) P(B) P(A or B). What makes this example different from P(A or C), shown previously? , and P(A or B) , P(B) 5 4 6 A Rule for the Probability of (A or B) Probability is based on the number of outcomes in a given event. For the event (A or B) in our example, we observe that the outcome 2 is found in event A and in event B. Therefore, we may describe rolling a 2 as the event (A and B). The simple addition rule does not work for the event (A or B) because we have counted the event (A and B) twice: first in event A, then again in event B. In order to count the event (A and B) only once, we must subtract the number of shared elements, , from the overall number of elements in (A or B). n(A and B) Thus, the rule becomes: n(A or B) n(A) n(B) n(A and B) For this example: n(A or B) 3 2 1 4 Dividing each term by n(S), we get an equivalent equation: For this example: n(A or B) n(S) n(A or B) n(S) n(A) n(S) 3 6 n(B) n(S) 2 6 n(A and B) n(S) 1 6 4 6 Since P(A or B) n(A or B) n(S) , we can write a general rule: P(A or B) P(A) P(B) P(A and B) In set terminology, the rule for probability becomes: P(A or B) P(A B) P(A) P(B) P(A B) " Note the use of the union symbol () to indicate or. Mutually Exclusive Events We have been examining three different events that occur when a die is tossed. Event A {an even number} {2, 4, 6} Event B {a number less than 3} {1, 2} Event C {a number less than 2} {1} The Probability of (A or B) 601 We found that P(A or C) P(A) P(C) P(A or B) P(A) P(B) P(A and B) Why are these results different? Of these sets, A and C are disjoint, that is they have no element in common. Events A and C are said to be mutually exclusive events because only one of the events can occur at any one throw of the dice. Events that are disjoint sets are mutually exclusive. If two events A and C are mutually exclusive: P(A or C) P(A) P(C) Sets A and B are not disjoint sets. They have element 2 in common. Events A and B are not mutually exclusive events because both can occur at any one throw of the dice. Events that are not disjoint sets are not mutually exclusive. If two events A and B are not mutually exclusive: P(A or B) P(A) P(B) P(A and B) Some examples of mutually exclusive events include the following. • Drawing a spade or a red card when one card is drawn from a standard deck. • Choosing a ninth-grade boy or a tenth-grade girl from a student body that consists of boys and girls in each grade 9 through 12. • Drawing a quarter or a dime when one coin is drawn from a purse that contains three quarters, two dimes, and a nickel. • Drawing a consonant or a vowel when a letter is drawn from the word PROBABILITY. Some examples of events that are not mutually exclusive include the fol- lowing. • Drawing an ace or a red card when one card is drawn from a standard deck. • Choosing a girl or a ninth-grade student from a student body that consists of boys and girls in each grade 9 through 12. • Drawing a dime or a coin worth more than five cents when one coin is drawn from a purse that contains a quarter, two dimes, and a nickel. • Drawing a Y or a letter that follows L in the alphabet when a letter is drawn from the word PROBABILITY. 602 Probability EXAMPLE 1 A standard deck of 52 cards is shuffled, and one card is drawn at random. Find the probability that the card is: a. a king or an ace b. red or an ace Solution a. There are four kings in the deck, so P(king) 5 4 52 . There are four aces in the deck, so P(ace) 5 4 52 . These are mutually exclusive events. The set of kings and the set of aces are disjoint sets, having no elements in common. P(king or ace) P(king) P(ace) 4 52 1 4 52 5 8 52 or 2 13 Answer b. There are 26 red cards in the deck, so P(red) 5 26 52 . There are four aces in the deck, so P(ace) 5 4 52 . There are two red aces in the deck, so P(red and ace) 5 2 52 . These are not mutually exclusive events. Two cards are both red and ace. P(red or ace) P(red) P(ace) P(red and ace) 52 2 2 52 5 28 52 or 7 13 52 1 4 Answer 5 26 Alternative Solution There are 26 red cards and two more aces not already counted (the ace of spades, the ace of clubs). Therefore, there are 26 2, or 28, cards in this event. Then: P(red or ace) 52 or 7 28 13 Answer EXAMPLE 2 There are two events, A and B. Given that P(A) .3, P(B) .5, and P(A d B) 5 .1 , find P(A B). Solution Since the probability of A B is not 0, A and B are not mutually exclusive. P(A B) P(A) P(B) P(A B) .3 .5 .1 .7 Answer P(A B) .7 EXAMPLE 3 A town has two newspapers, the Times and the Chronicle. One out of every two persons in the town subscribes to the Times, three out of every five persons in the town subscribe to the Chronicle, and three out of every ten persons in the The Probability of (A or B) 603 town subscribe to both papers. What is the probability that a person in the town chosen at random subscribes to the Times or the Chronicle? Solution Subscribing to the Times and subscribing to the Chronicle are not mutually exclusive events. 5 1 2 P(Times) 5 3 5 P(Times or Chronicle) P(Times) P(Chronicle) P(Times and Chronicle) P(Times and Chronicle) P(Chronicle) 5 3 10 5 1 5 5 2 1 3 10 1 6 5 2 3 10 10 2 3 10 5 8 10 or 4 5 Answer EXERCISES Writing About Mathematics 1. Let A and B be two events. Is it possible for P(A or B) to be less that P(A)? Explain why or why not. 2. If B is a subset of A, which of the following is true: P(A or B) P(A), P(A or B) P(A)? Explain your answer. P(A or B) 5 P(A) , Developing Skills 3. A spinner consists of five equal sectors of a circle. The sectors are numbered 1 through 5, and when an arrow is spun, it is equally likely to stop on any sector. For a single spin of the arrow, determine whether or not the events are mutually exclusive and find the probability that the number on the sector is: a. 3 or 4 b. odd or 2 c. 4 or less d. 2 or 3 or 4 e. odd or 3 4. A fair die is rolled once. Determine whether or not the events are mutually exclusive and find the probability that the number rolled is: a. 3 or 4 e. odd or 3 b. odd or 2 f. less than 2 or more than 5 c. 4 or less than 4 g. less than 5 or more than 2 d. 2, 3, or 4 h. even or more than 3 5. From a standard deck of cards, one card is drawn. Determine whether or not the events are mutually exclusive and find the probability that the card will be: a. a queen or an ace b. a queen or a 7 c. a heart or a spade d. a queen or a spade e. a queen or a red card f. a jack or a queen or a king g. a 7 or a diamond h. a club or a red card i. an ace or a picture card 604 Probability 6. A bank contains two quarters, six dimes, three nickels, and five pennies. A coin is drawn at random. Determine whether or not the events are mutually exclusive and find the probability that the coin is: a. a quarter or a dime b. a dime or a nickel c. worth 10 cents or more than 10 cents d. worth 10 cents or less e. worth 1 cent or more f. a quarter, a nickel, or a penny In 7–12, in each case choose the numeral preceding the expression that best completes the statement or answers the question. 7. If a single card is drawn from a standard deck, what is the probability that it is a 4 or a 9? (1) 2 52 (2) 8 52 (3) 13 52 (4) 26 52 8. If a single card is drawn from a standard deck, what is the probability that it is a 4 or a diamond? 8 52 (1) (2) 16 52 (3) 17 52 9. If P(A) .2, P(B) .5, and P(A B) .1, then P(A B) = (1) .6 (2) .7 (3) .8 10. If P(A) 5 1 3 , P(B) 5 1 2 , and P(A and B) 5 1 6 , then P(A or B) = (1) 2 5 (2) 2 3 (3) 5 6 11. If P(A) 5 1 4 , P(B) 5 1 2 , and P(A B) 5 1 8 , then P(A B) = (1) 1 8 (2) 5 8 (3) 3 4 12. If P(A) .30, P(B) .35, and (A B) , then P(A B) = (1) .05 (2) .38 (3) .65 (4) 26 52 (4) .9 (4) 1 (4) 7 8 (4) 0 Applying Skills 13. Linda and Aaron recorded a CD together. Each sang some solos, and both of them sang some duets. Aaron recorded twice as many duets as solos, and Linda recorded six more solos than duets. If a CD player selects one of these songs at random, the probability that it 1 will select a duet is . Find the number of: 4 a. solos by Aaron b. solos by Linda c. duets 14. In a sophomore class of 340 students, some students study Spanish, some study French, some study both languages, and some study neither language. If P(Spanish) .70, P(French) .40, and P(Spanish and French) .25, find: a. the probability that a sophomore studies Spanish or French b. the number of sophomores who study one or more of these languages The Probability of (Not A) 605 15. The Greenspace Company offers lawn care services and snow plowing in the appropriate seasons. Of the 600 property owners in town, 120 have contracts for lawn care, 90 for snow plowing, and 60 for both with the Greenspace Company. A new landscape company, the Earthpro Company offers the same services and begins a telephone campaign to attract customers, choosing telephone numbers of property owners at random. What is the probability that the Earthpro Company reaches someone who has a contract for lawn care or snow plowing with the Greenspace Company? 15-6 THE PROBABILITY OF (NOT
A) In rolling a fair die, we know that P(4) since there is only one outcome for the event (rolling a 4). We can also say that P(not 4) since there are five outcomes for the event (not rolling a 4): 1, 2, 3, 5, and 6. 5 6 5 1 6 We can think of these probabilities in another way. The event (4) and the event (not 4) are mutually exclusive events. Also, the event (4 or not 4) is a certainty whose probability is 1. P(4 or not 4) P(4) P(not 4) 1 P(4) P(not 4) 1 P(4) P(not 4) 1 1 6 5 5 6 5 P(not 4) P(not 4) The event (not A) is the complement of event A, when the universal set is the sample space, S. In general, if P(A) is the probability that some given result will occur, and P(not A) is the probability that the given result will not occur, then: 1. P(A) P(not A) 1 2. P(A) 1 P(not A) 3. P(not A) 1 P(A) Probability as a Sum When sets are disjoint, we have seen that the probability of the union can be found by the rule P(A B) = P(A) P(B). Since the possible outcomes that are singletons represent disjoint sets, we can say: The probability of any event is equal to the sum of the probabilities of the singleton outcomes in the event. 606 Probability For example, when we draw a card from a standard deck, there are 52 sin. Since all singleton events are dis- gleton outcomes, each with a probability of joint, we can say: 1 52 P(king) P(king of hearts) P(king of diamonds) P(king of spades) P(king of clubs) 1 1 52 52 52 or 1 4 13 1 52 1 52 P(king) We also say: The sum of the probabilities of all possible singleton outcomes for any sam- ple space must always equal 1. For example, in tossing a coin, P(S) P(heads) P (tails) 2 1 1 1 2 1. Also, in rolling a die, P(S) P(1) P(2) P(3) P(4) P(5) P(6 EXAMPLE 1 Dr. Van Brunt estimates that 4 out every 10 patients that he will see this week will need a flu shot. What is the probability that the next patient he sees will not need a flu shot? Solution The probability that a patient will need a flu shot is 4 10 or 2 5 The probability that a patient will not need a flu shot is 1 . 5 5 3 2 5 . Answer EXAMPLE 2 A letter is drawn at random from the word ERROR. a. Find the probability of drawing each of the different letters in the word. b. Demonstrate that the sum of these probabilities is 1. Solution a. P(E) 5 1 5 ; P(R) 5 3 5 ; P(O Answer 5 5 5 5 b. P(E) P(R) P(O) 1 Answer The Probability of (Not A) 607 EXERCISES Writing About Mathematics 1. If event A is a certainty, P(A) 1. What must be true about P(not A)? Explain your answer. 2. If A and B are disjoint sets, what is P(not A or not B)? Explain your answer. Developing Skills 3. A fair die is rolled once. Find each probability: a. P(3) b. P(not 3) c. P(even) d. P(not even) e. P(less than 3) f. P(not less than 3) g. P(odd or even) h. P[not (odd or even)] i. P[not (2 or 3)] 4. From a standard deck of cards, one card is drawn. Find the probability that the card is: a. a heart d. not a picture card b. not a heart e. not an 8 c. a picture card f. not a red 6 g. not the queen of spades h. not an 8 or a 6 5. One letter is selected at random from the word PICNICKING. a. Find the probability of drawing each of the different letters in the word. b. Demonstrate that the sum of these probabilities is 1. 6. If the probability of an event happening is , what is the probability of that event not 1 7 happening? 7. If the probability of an event happening is .093, what is the probability of that event not happening? 8. A jar contains seven marbles, all the same size. Three are red and four are green. If a marble is chosen at random, find each probability: a. P(red) b. P(green) c. P(not red) d. P(red or green) e. P(red and green) f. P[not (red or green)] 9. A box contains 3 times as many black marbles as green marbles, all the same size. If a mar- ble is drawn at random, find the probability that it is: a. black b. green c. not black d. black or green e. not green 10. A letter is chosen at random from the word PROBABILITY. Find each probability: a. P(A) d. P(A or B) b. P(B) e. P(A or I) c. P(C) f. P(a vowel) g. P(not a vowel) h. P(A or B or L) i. P(A or not A) 608 Probability 11. A single card is drawn at random from a well-shuffled deck of 52 cards. Find the probability that the card is: a. a 6 d. a 6 or a club g. a 6 or a 7 j. a black 6 Applying Skills b. a club e. not a club h. not the 6 of clubs c. the 6 of clubs f. not a 6 i. a 6 and a 7 k. a 6 or a black card l. a black card or not a 6 12. A bank contains three quarters, four dimes, and five nickels. A coin is drawn at random. a. Find the probability of drawing: (1) a quarter (2) a dime (3) a nickel b. Demonstrate that the sum of the three probabilities given as answers in part a is 1. c. Find the probability of not drawing: (1) a quarter (2) a dime (3) a nickel 13. The weather bureau predicted a 30% chance of rain. Express in fractional form: a. the probability that it will rain b. the probability that it will not rain 14. Mr. Jacobsen’s mail contains two letters, three bills, and five ads. He picks up the first piece of mail without looking at it. Express, in decimal form, the probability that this piece of mail is: a. a letter b. a bill d. a letter or an ad e. a bill or an ad g. not an ad h. a bill and an ad c. an ad f. not a bill 15. The square dartboard shown at the right, whose side measures 30 inches, has at its center a shaded square region whose side measures 10 inches. If darts directed at the board are equally likely to land anywhere on the board, what is the probability that a dart does not land in the shaded region? 30 in. 10 in. 10 in. 30 in. 16. A telephone keypad contains the ten digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Mabel is calling a friend. Find the probability that the last digit in her friend’s telephone number is: a. 6 b. 6 or a larger number c. a number smaller than 6 d. 6 or an odd number e. 6 or a number smaller than 6 f. not 6 The Counting Principle, Sample Spaces, and Probability 609 g. 6 and an odd number h. a number not larger than 6 i. a number smaller than 2 and larger than 6 j. a number smaller than 2 or larger than 6 k. a number smaller than 6 and larger than 2 l. a number smaller than 6 or larger than 2 15-7 THE COUNTING PRINCIPLE, SAMPLE SPACES, AND PROBABILITY So far, we have looked at simple problems involving a single activity, such as rolling one die or choosing one card. More realistic problems occur when there are two or more activities, such as rolling two dice or dealing a hand of five cards. An event consisting of two or more activities is called a compound event. Before studying the probability of events based on two or more activities, let us study ways to count the number of elements or outcomes in a sample space for two or more activities. For example: A store offers five flavors of ice cream: vanilla, chocolate, strawberry, peach, and raspberry. A sundae can be made with either a hot fudge topping or a marshmallow topping. If a sundae consists of one flavor of ice cream and one topping, how many different sundaes are possible? We will use letters to represent the five flavors of ice cream (V, C, S, P, R) and the two toppings (F, M). We can show the number of elements in the sample space in three ways: 1. Tree Diagram. The tree diagram at the right first branches out to show five flavors of ice cream. For each of these flavors the tree again branches out to show the two toppings. In all, there are 10 paths or branches to follow, each with one flavor of ice cream and one topping. These 10 branches show that the sample space consists of 10 possible outcomes, in this case, sundaes. V C S P R Tree diagram . List of Ordered Pairs. It is usual to order a sundae by telling the clerk the flavor of ice cream first and then the type of topping. This suggests a listing of ordered pairs. The first component of the ordered pair is the icecream flavor, and the second component is the type of topping. The set of pairs (ice cream, topping) is shown below. {(V, F), (C, F), (S, F), (P, F), (R, F), (V, M), (C, M), (S, M), (P, M), (R, M)} These 10 ordered pairs show that the sample space consists of 10 possible sundaes. 610 Probability 3. Graph of Ordered Pairs. Instead of listing pairs, we may construct a graph of the ordered pairs. At the right, the five flavors of ice cream appear on the horizontal scale or line, and the two toppings are on the vertical line. Each point in the graph represents an ordered pair. For example, the point circled shows the ordered pair (P, F), that is, (peach ice cream, fudge topping). M R V C F P S This graph of 10 points, or 10 ordered pairs, shows that the sample space consists of 10 possible sundaes. Graph of ordered pairs Whether we use a tree diagram, a list of ordered pairs, or a graph of ordered pairs, we recognize that the sample space consists of 10 sundaes. The number of elements in the sample space can be found by multiplication: number of flavors number of toppings number of of ice cream possible sundaes e 5 f f 2 10 Suppose the store offered 30 flavors of ice cream and seven possible top- pings. To find the number of elements in the sample space, we multiply: 30 7 210 possible sundaes This simple multiplication procedure is known as the counting principle, because it enables us to count the number of elements in a sample space. The Counting Principle: If one activity can occur in any of m ways and, following this, a second activity can occur in any of n ways, then both activities can occur in the order given in m n ways. We can extend this rule to include three or more activities by extending the multiplication process. We can also display three or more activities by extending the branches on a tree diagram, or by listing ordered elements such as ordered triples and ordered quadruples. For example, a coin is tossed three times in succession. • On the first toss, the coin may fall in either of two ways: a head or a tail. • On the second toss, the coin may also fall in either of two ways. • On the third toss, the coin may still fall
in either of two ways. By the counting principle, the sample space contains 2 2 2 or 8 possible outcomes. By letting H represent a head and T represent a tail, we can illustrate the sample space by a tree diagram or by a list of ordered triples: We did not attempt to draw a graph of this sample space because we would need a horizontal scale, a vertical scale, and a third scale, making the graph The Counting Principle, Sample Spaces, and Probability 611 Three Tosses of a Coin H, H, H) (H, H, T) (H, T, H) (H, T, T) (T, H, H) (T, H, T) (T, T, H) (T, T, T) Tree diagram List of ordered triples three-dimensional. Although such a graph can be drawn, it is too difficult at this time. We can conclude that: 1. Tree diagrams, or lists of ordered elements, are effective ways to indicate any compound event of two or more activities. 2. Graphs should be limited to ordered pairs, or to events consisting of exactly two activities. EXAMPLE 1 The school cafeteria offers four types of salads, three types of beverages, and five types of desserts. If a lunch consists of one salad, one beverage, and one dessert, how many possible lunches can be chosen? Solution By the counting principle, we multiply the number of possibilities for each choice: 4 3 5 12 5 60 possible lunches Answer Independent Events The probability of rolling 5 on one toss of a die is . What is the probability of rolling a pair of 5’s when two dice are tossed? 1 6 When we roll two dice, the number obtained on the second die is in no way determined by of the result obtained on the first die. When we toss two coins, the face that shows on the second coin is in no way determined by the face that shows on the first coin. When the result of one activity in no way influences the result of a second activity, the results of these activities are called independent events. In cases where two events are independent, we may extend the counting principle to find the probability that both independent events occur at the same time. 612 Probability For instance, what is the probability that, when two dice are thrown, a 5 will appear on each of the dice? Let S represent the sample space and F represent the event (5 on both dice). (1) Use the counting principle to find the number of elements in the sample space. There are 6 ways in which the first die can land and 6 ways in which the second die can land. Therefore, there are 6 6 or 36 pairs of numbers in the sample space, that is, n(S) 36. (2) There is only one face on each die that has a 5. Therefore there is 1 1 or 1 pair in the event F, that is n(F) 1. (3) P(5 on both dice) n(F) n(S) 5 1 36 The probability of 5 on both dice can also be determined by using the prob- ability of each of the independent events. P(5 on first die) P(5 on second die) 1 6 1 6 P(5 on both dice) P(5 on first) P(5 on second) 6 3 1 1 6 5 1 36 We can extend the counting principle to help us find the probability of any two or more independent events. The Counting Principle for Probability: E and F are independent events. The probability of event E is m (0 m 1) and the probability of event F is n (0 n 1). The probability of the event in which E and F occur jointly is the product m n. Note 1: The product m n is within the range of values for a probability, namely, 0 # m 3 n # 1 . Note 2: Not all events are independent, and this simple product rule cannot be used to find the probability when events are not independent. EXAMPLE 2 Mr. Gillen may take any of three buses, A or B or C, to get to the train station. He may then take the 6th Avenue train or the 8th Avenue train to get to work. The buses and trains arrive at random and are equally likely to arrive. What is the probability that Mr. Gillen takes the B bus and the 6th Avenue train to get to work? Solution P(B bus) 1 3 1 and P(6th Ave. train) 2 Since the train taken is independent of the bus taken: P(B bus and 6th Ave. train) P(B bus) P(6th Ave. train) Answer 231 1 3 1 6 The Counting Principle, Sample Spaces, and Probability 613 EXERCISES Writing About Mathematics 1. Judy said that if a quarter and a nickel are tossed, there are three equally likely outcomes; two heads, two tails, or one head and one tail. Do you agree with Judy? Explain why or why not. 2. a. When a green die and a red die are rolled, is the probability of getting a 2 on the green die and a 3 on the red die the same as the probability of getting 3 on both dice? Explain why or why not. b. When rolling two fair dice, is the probability of getting a 2 and a 3 the same as the probability of getting two 3’s? Explain why or why not. Developing Skills 3. A quarter and a penny are tossed simultaneously. Each coin may fall heads or tails. The tree diagram at the right shows the sample space involved. a. List the sample space as a set of ordered pairs. b. Use the counting principle to demonstrate that there are four outcomes in the sample space. c. In how many outcomes do the coins both fall heads up? H T H T H T Quarter Penny d. In how many outcomes do the coins land showing one head and one tail? 4. Two dice are rolled simultaneously. Each die may land with any one of the six numbers faceup. a. Use the counting principle to determine the number of outcomes in this sample space. b. Display the sample space by drawing a graph of the set of ordered pairs. 5. A sack contains four marbles: one red, one blue, one white, one green. One marble is drawn and placed on the table. Then a second marble is drawn and placed to the right of the first. a. How many possible marbles can be selected on the first draw? b. How many possible marbles can be selected on the second draw? c. How many possible ordered pairs of marbles can be drawn? d. Draw a tree diagram to show all of the possible outcomes for this experiment. Represent the marbles by R, B, W, and G. 6. A sack contains four marbles: one red, one blue, one white, one green. One marble is drawn, its color noted and then it is replaced in the sack. A second marble is drawn and its color noted to the right of the first. a. How many possible colors can be noted on the first draw? b. How many possible colors can be noted on the second draw? c. How many possible ordered pairs of colors can be noted? d. Draw a tree diagram to show all of the possible outcomes for this experiment. Represent the marbles by R, B, W, and G. 614 Probability 7. A fair coin and a six-sided die are tossed simultaneously. What is the probability of obtain- ing: a. a head on the coin? b. a 4 on the die? c. a head on the coin and a 4 on the die on a single throw? 8. A fair coin and a six-sided die are tossed simultaneously. What is the probability of obtain- ing on a single throw: a. a head and a 3? b. a head and an even number? c. a tail and a number less than 5? d. a tail and a number greater than 4? 9. Two fair coins are tossed. What is the probability that both land heads up? 10. Three fair coins are tossed. a. Find P(H, H, H). b. Find P(T, T, T). 11. A fair spinner contains congruent sectors, numbered 1 through 8. If the arrow is spun twice, find the probability that it lands: a. (7, 7) e. (2, 8) b. not (7, 7) f. not (2, 8) c. (not 7, not 7) g. (not 2, not 8) d. (7, not 7) or (not 7, 7) h. (2, not 8) or (not 2, 8) 12. A fair coin is tossed 50 times and lands heads up each time. What is the probability that it will land heads up on the next toss? Explain your answer. Applying Skills 13. Tell how many possible outfits consisting of one shirt and one pair of pants Terry can choose if he owns: a. 5 shirts and 2 pairs of pants b. 10 shirts and 4 pairs of pants c. 6 shirts and 6 pairs of pants 14. There are 10 doors into the school and eight staircases from the first floor to the second. How many possible ways are there for a student to go from outside the school to a classroom on the second floor? 15. A tennis club has 15 members: eight women and seven men. How many different teams may be formed consisting of one woman and one man on the team? 16. A dinner menu lists two soups, seven main courses, and three desserts. How many different meals consisting of one soup, one main course, and one dessert are possible? The Counting Principle, Sample Spaces, and Probability 615 17. The school cafeteria offers the menu shown. Main Course Dessert Beverage Pizza Yogurt Frankfurter Fruit salad Milk Juice Ham sandwich Jello Tuna sandwich Apple pie Veggie burger a. How many meals consisting of one main course, one dessert, and one beverage can be selected from this menu? b. Joe does not like ham and tuna. How many meals (again, one main course, one dessert, and one beverage) can Joe select, not having ham and not having tuna? c. If the pizza, frankfurters, yogurt, and fruit salad have been sold out, how many differ- ent meals can JoAnn select from the remaining menu? 18. A teacher gives a quiz consisting of three questions. Each question has as its answer either true (T) or false (F). a. Using T and F, draw a tree diagram to show all possible ways the questions can be answered. b. List this sample space as a set of ordered triples. 19. A test consists of multiple-choice questions. Each question has four choices. Tell how many possible ways there are to answer the questions on the test if the test consists of the following number of questions: a. 1 question b. 3 questions c. 5 questions d. n questions 20. Options on a bicycle include two types of handlebars, two types of seats, and a choice of 15 colors. The bike may also be ordered in ten-speed, in three-speed, or standard. How many possible versions of a bicycle can a customer choose from, if he selects a specific type of handlebars, type of seat, color, and speed? 21. A state issues license plates consisting of letters and numbers. There are 26 letters, and the letters may be repeated on a plate; there are 10 digits, and the digits may be repeated. Tell how many possible license plates the state may issue when a license consists of each of the following: a. 2 letters, followed by 3 numbers b. 2 numbers, fol
lowed by 3 letters c. 4 numbers, followed by 2 letters 22. In a school cafeteria, the menu rotates so that P(hamburger) , P(apple pie) , and 4 5 P(soup) . The selection of menu items is random so that the appearance of hamburgers, apple pie, and soup are independent events. On any given day, what is the probability that the cafeteria offers hamburger, apple pie, and soup on the same menu? 2 3 1 4 616 Probability 23. A quiz consists of true-false questions only. Harry has not studied, and he guesses every answer. Find the probability that he will guess correctly to get a perfect score if the test conc. n questions sists of: a. 1 question b. 4 questions 24. The probability of the Tigers beating the Cougars is . The probability of the Tigers beating the Mustangs is . If the Tigers play one game with the Cougars and one game with the Mustangs, find the probability that the Tigers: a. win both games b. lose both games 1 4 2 3 25. On Main Street in Pittsford, there are two intersections that have traffic lights. The lights are not timed to accommodate traffic. They are independent of one another. At each of the intersections, P(red light) .3 and P(green light) .7 for cars traveling along Main Street. Find the probability that a car traveling on Main Street will be faced with each set of given conditions at the two traffic lights shown. a. Both lights are red. b. Both lights are green. c. The first light is red, and the second is green. d. The first light is green, and the second is red. e. At least one light is red, that is, not both lights are green. f. Both lights are the same color. 26. A manufacturer of radios knows that the probability of a defect in any of his products is If 10,000 radios are manufactured in January, how many are likely to be defective? 1 400 . 27. Past records from the weather bureau indicate that the probability of rain in August on Cape Cod is . If Joan goes to Cape Cod for 2 weeks in August, how many days will it probably rain if the records hold true? 2 7 28. A nationwide fast-food chain has a promotion, distributing to customers 2,000,000 coupons for the prizes shown below. Each coupon awards the customer one of the following prizes. 1 Grand Prize: $25,000 cash 2 Second-Place Prizes: New car 100 Third-Place Prizes: New TV a. Find the probability of winning: Fourth-Place Prizes: Free meal Consolation Prizes: 25 cents off any purchase (1) the grand prize (2) a new car (3) a new TV b. If a customer has one coupon, what is the probability of winning one of the first three prizes (cash, a car, or a TV)? c. If the probability of winning a free meal is 1 400 , how many coupons are marked as fourth-place prizes? d. How many coupons are marked “25 cents off any purchase”? e. If a customer has one coupon, what is the probability of not winning one of the first three prizes? 15-8 PROBABILITIES WITH TWO OR MORE ACTIVITIES Probabilities with Two or More Activities 617 Without Replacement Two cards are drawn at random from an ordinary pack of 52 cards. In this situation, a single card is drawn from a deck of 52 cards, and then a second card is drawn from the remaining 51 cards in the deck. What is the probability that both cards drawn are kings? On the first draw, there are four kings in the deck of 52 cards, so P(first king) 4 52 If a second card is drawn without replacing the first king selected, there are now only three kings in the 51 cards remaining. Therefore we are considering the probability of drawing a king, given that a king has already been drawn. P(second king) 3 51 By the counting principle for probabilities: P(both kings) P(first king) P(second king) 52 3 3 4 51 5 1 13 3 1 17 5 1 221 This result could also have been obtained using the counting principle. The number of elements in the sample space is 52 51. The number of elements in the event (two kings) is . Then, 4 3 3 P(both kings) 5 4 3 3 52 3 51 5 12 2,652 5 1 221 This is called a problem without replacement because the first king drawn was not placed back into the deck. These are dependent events because the probability of a king on the second draw depends on whether or not a king appeared on the first draw. In general, if A and B are two dependent events: P(A and B) P(A) P(B given that A has occurred) Earlier in the chapter we discussed P(A and B) where (A and B) is a single event that satisfies both conditions. Here (A and B) denotes two dependent events with A the outcome of one event and B the outcome of the other. The conditions of the problem will indicate which of these situations exists. When A and B are dependent events, P(A and B) can also be written as P(A, B). With Replacement A card is drawn at random from an ordinary deck, the card is placed back into the deck, and a second card is then drawn and replaced. In this situation, it is clear the deck contains 52 cards each time that a card is drawn and that the same card could be drawn twice. What is the probability that the card drawn each time is a king? 618 Probability On the first draw, there are four kings in the deck of 52 cards. P(first king) 4 52 If the first king drawn is now placed back into the deck, then, on the second draw, there are again four kings in the deck of 52 cards. P(second king) 4 52 By the counting principle for probabilities: P(both kings) P(first king) P(second king) 4 52 3 4 52 5 1 13 3 1 13 5 1 169 This is called a problem with replacement because the first card drawn was placed back into the deck. In this case, the events are independent because the probability of a king on the second draw does not depend on whether or not a king appeared on the first draw. Since the card drawn is replaced, the number of cards in the deck and the number of kings in the deck remain constant. In this case, P(B given that A has occurred) P(B). In general, if A and B are two independent events, P(A and B) P(A) P(B) Rolling two dice is similar to drawing two cards with replacement because the number of faces on each die remains constant, as did the number of cards in the deck. Typical problems with replacement include rolling dice, tossing coins (each coin always has two sides), and spinning arrows. KEEP IN MIND 1. If the problem does not specifically mention with replacement or without replacement, ask yourself: “Is this problem with or without replacement?” or “Are the events dependent or independent?” 2. For many compound events, the probability can be determined most easily by using the counting principle. 3. Every probability problem can always be solved by: • counting the number of elements in the sample space, n(S); • counting the number of outcomes in the event, n(E); • substituting these numbers in the probability formula, P(E) n(E) n(S) Probabilities with Two or More Activities 619 Conditional Probability The previous discussion involved the concept of conditional probability. For both dependent and independent events, in order to find the probability of A followed by B, it is necessary to calculate the probability that B occurs given that A has occurred. Notation for conditional probability is P(B given that A has occurred) P(B A). Then the following statement is true for both dependent and independent events: P(A and B) P(A) P(B A) If A and B independent events, P(B A) P(B). Therefore, for independent events: P(A and B) P(A) P(B A) P(A) P(B) The general formula P(A and B) P(A) P(B A) can be solved for P(B A): P(B A) P(A and B) P(A) For example, suppose a box contains one red marble, one blue marble, one green marble, and one yellow marble. Two marbles are drawn without replacement. Let R be the event {red marble} and Y be the event {yellow marble}. The probability of R is and the probability of Y is . If we want to find the probability of drawing a red marble followed by a yellow marble, R and Y are depen1 dent events. We need the probability of Y given that R has occurred, which is 3 since once the red marble has been drawn, only 3 marbles remain, one of which is yellow. 1 4 1 4 P(R followed by Y) P(R and Y) P(R) P(Y \| R) 1 3 1 4 1 12 This result can be shown by displaying the sample space. (R, B) (B, R) (G, R) (Y, R) (R, G) (B, G) (G, B) (Y, B) (R, Y) (B, Y) (G, Y) (Y, G) There are 12 possible outcomes in the sample space and one of them is (R, Y). Therefore, P(R, Y) .1 12 620 Probability EXAMPLE 1 Three fair dice are thrown. What is the probability that all three dice show a 5? Solution These are independent events. There are six possible faces that can come up on each die. 1 On the first die, there is one way to obtain a 5 so P(5) . 6 1 On the second die, there is one way to obtain a 5 so P(5) . 6 1 On the third die, there is one way to obtain a 5 so P(5) . 6 By the counting principle: P(5 on each die) P(5 on first) P(5 on second) P(5 on third) 1 6 3 1 1 216 6 3 1 6 Answer EXAMPLE 2 If two cards are drawn from an ordinary deck without replacement, what is the probability that the cards form a pair (two cards of the same face value but different suits)? Solution These are dependent events. On the first draw, any card at all may be chosen, so: P(any card) 52 52 There are now 51 cards left in the deck. Of these 51, there are three that match the first card taken, to form a pair, so: P(second card forms a pair) 3 51 Then: P(pair) P(any card) P(second card forms a pair) EXAMPLE 3 52 52 3 3 51 1 3 1 17 Answer 1 17 A jar contains four white marbles and two blue marbles, all the same size. A marble is drawn at random and not replaced. A second marble is then drawn from the jar. Find the probability that: a. both marbles are white b. both marbles are blue c. both marbles are the same color Probabilities with Two or More Activities 621 Solution These are dependent events. a. On the first draw: P(white) 4 6 Since the white marble drawn is not replaced, five marbles, of which three are white, are left in the jar. On the second draw: Then: P(white given that the first was white) 3 5 P(both white) 6 3 3 4 5 5 12 30 or Answer 2 5 b. Start with a full jar of six m
arbles of which two are blue. On the first draw: P(blue) 2 6 Since the blue marble drawn is not replaced, five marbles, of which only one is blue, are left in the jar. On the second draw: P(blue given that the first was blue) 1 5 Then: P(both blue) 2 6 3 1 5 5 2 30 or 1 15 Answer c. If both marbles are the same color, both are white or both are blue. These are disjoint or mutually exclusive events: P(A or B) P(A) P(B) Therefore: P(both white or both blue) P(both white) P(both blue) 2 5 6 15 7 15 Answer 1 15 1 15 Note: By considering the complement of the set named in part c, we can easily determine the probability of drawing two marbles of different colors. P[not (both white or both blue)] 1 P(both white or both blue) 5 1 2 7 15 5 8 15 622 Probability EXAMPLE 4 A fair die is rolled. a. Find the probability that the die shows a 4 given that the die shows an even number. b. Find the probability that the die shows a 1 given that the die shows a num- ber less than 5. Solution a. Use the formula for conditional probability. EXAMPLE 5 P(4 given an even number) P(4 even) 5 P(4 and even) P(even) The event “4 and even” occurs whenever the outcome is 4. Therefore, P(4 and even) P(4). P(4 even) P(4 and even) P(even) 5 P(4) P(even) 5 1 6 3 6 5 1 3 Answer b. The event “1 and less than 5” occurs whenever the outcome is 1. Therefore, P(1 and less than 5) P(1). P(1 less than 5) P(1 and less than 5) P(less than 5) 5 P(1) P(less than 5) 5 1 6 5 6 5 1 5 Answer Fred has two quarters and one nickel in his pocket. The pocket has a hole in it, and a coin drops out. Fred picks up the coin and puts it back into his pocket. A few minutes later, a coin drops out of his pocket again. a. Draw a tree diagram or list the sample space for all possible pairs that are outcomes to describe the coins that fell. b. What is the probability that the same coin fell out of Fred’s pocket both times? a. What is the probability that the two coins that fell have a total value of 30 cents? b. What is the probability that a quarter fell out at least once? Solution a. Because there are two quarters, use subscripts. The three coins are {Q1, Q2, N}, where Q represents a quarter and N represents a nickel. This is a problem with replacement. The events are independent. {(Q1, Q1), (Q1, Q2), (Q1, N), (Q2, Q1), (Q2, Q2), (Q2, N), (N, Q1), (N, Q2), (N, N)} Q1 Q2 N Q1 Q2 N Q1 Q2 N Q1 Q2 N Probabilities with Two or More Activities 623 b. Of the nine outcomes, three name the same coin both times: (Q1, Q1), (Q2, Q2) and (N, N). Therefore: P(same coin) or Answer 3 9 1 3 c. Of the nine outcomes, the four that consist of a quarter and a nickel total 30 cents: (Q1, N), (Q2, N), (N, Q1), (N, Q2). P(total value of 30 cents) Answer 4 9 d. Of the nine outcomes, only (N, N) does not include a quarter. Eight contain at least one quarter, that is, one or more quarters. P(at least one quarter) Answer 8 9 EXAMPLE 6 Of Roosevelt High School’s 1,000 students, 300 are athletes, 200 are in the Honor Roll, and 120 play sports and are in the Honor Roll. What is the probability that a randomly chosen student who plays a sport is also in the Honor Roll? Solution Let A the event that the student is an athlete, and B the event that the student is in the Honor Roll. Then, A and B the event that a student is an athlete and is in the Honor Roll. Therefore, the conditional probability that the student is in the Honor Roll given that he or she is an athlete is: P(A and B) P(A) P(B A) 120 1,000 300 1,000 120 300 5 .4 Answer EXERCISES Writing About Mathematics 1. The name of each person who attends a charity luncheon is placed in a box and names are drawn for first, second, and third prize. No one can win more than one prize. Is the probability of winning first prize greater than, equal to, or less than the probability of winning second prize? Explain your answer. 2. Six tiles numbered 1 through 6 are placed in a sack. Two tiles are drawn. Is the probability of drawing a pair of tiles whose numbers have a sum of 8 greater than, equal to, or less than the probability of obtaining a sum of 8 when rolling a pair of dice? Explain your answer. 624 Probability Developing Skills 3. A jar contains two red and five yellow marbles. If one marble is drawn at random, what is the probability that the marble drawn is: a. red? b. yellow? 4. A jar contains two red and five yellow marbles. A marble is drawn at random and then replaced. A second draw is made at random. Find the probability that: a. both marbles are red b. both marbles are yellow c. both marbles are the same color d. the marbles are different in color 5. A jar contains two red and five yellow marbles. A marble is drawn at random. Then without replacement, a second marble is drawn at random. Find the probability that: a. both marbles are red b. both marbles are yellow c. both marbles are the same color d. the marbles are different in color e. the second marble is red given that the first is yellow 6. In an experiment, an arrow is spun twice on a circular board containing four congruent sectors numbered 1 through 4. The arrow is equally likely to land on any one of the sectors. a. Indicate the sample space by drawing a tree diagram or writing a set of ordered pairs. b. Find the probability of spinning the digits 2 and 3 in that order. c. Find the probability that the same digit is spun both times. d. What is the probability that the first digit spun is larger than the second? 7. A jar contains nine orange disks and three blue disks. A girl chooses one at random and then, without replacing it, chooses another. Let O represent orange and B represent blue. a. Find the probability of each of the following outcomes: (1) (O, O) (2) (O, B) (3) (B, O) (4) (B, B) b. Now, find for the disks chosen, the probability that: (1) neither was orange (3) at least one was orange (5) at most one was orange (7) the second is orange given that the first was blue (2) only one was blue (4) they were the same color (6) they were the same disk 8. A card is drawn at random from a deck of 52 cards. Given that it is a red card, what is the probability that it is: a. a heart? b. a king? 9. A fair coin is tossed three times. a. What is the probability of getting all heads, given that the first toss is heads? b. What is the probability of getting all heads, given that the first two tosses are heads? Applying Skills 10. Sal has a bag of hard candies: three are lemon (L) and two are grape (G). He eats two of the candies while waiting for a bus, selecting them at random one after another. Probabilities with Two or More Activities 625 a. Using subscripts, draw a tree diagram or list the sample space of all possible outcomes showing which candies are eaten. b. Find the probability of each of the following outcomes: (1) both candies are lemon (3) the candies are the same flavor (2) neither candy is lemon (4) at least one candy is lemon c. What is the probability that the second candy that Sal ate was lemon given that the first was grape? 11. Carol has five children: three girls and two boys. One of her children was late for lunch. Later that day, one of her children was late for dinner. a. Indicate the sample space by a tree diagram or list of ordered pairs showing which children were late. b. If each child was equally likely to be late, find the probability of each outcome below. (1) Both children who were late were girls. (2) Both children who were late were boys. (3) The same child was late both times. (4) At least one of the children who was late was a boy. c. What is the probability that the child who was late for dinner was a boy given that the child who was late for lunch was a girl? 12. Several players start playing a game, each with a full deck of 52 cards. Each player draws two cards at random, one at a time without replacement. Find the probability that: a. Flo draws two jacks c. Jerry draws two red cards e. Ann does not draw a pair b. Frances draws two hearts d. Mary draws two picture cards f. Stephen draws two black kings g. Carrie draws a 10 on her second draw given that the first was a 5 h. Bill draws a heart and a club in that order i. Ann draws a king on her second draw given that the first was not a king 13. Saverio has four coins: a half dollar, a quarter, a dime, and a nickel. He chooses one of the coins at random and puts it in a bank. Later he chooses another coin and also puts that in the bank. a. Indicate the sample space of coins saved. b. If each coin is equally likely to be saved, find the probability that: (1) the coins saved will be worth a total of 35 cents (2) the coins saved will add to an even amount (3) the coins saved will include the half-dollar (4) the coins saved will be worth a total of less than 30 cents (5) the second coin saved will be worth more than 20 cents given that the first coin saved was worth more than 20 cents. (6) the second coin saved will be worth more than 20 cents given that the first coin saved was worth less than 20 cents. 626 Probability 14. Farmer Brown must wake up before sunrise to start his chores. Dressing in the dark, he reaches into a drawer and pulls out two loose socks. There are eight white socks and six red socks in the drawer. a. Find the probability that both socks are: (1) white (2) red (3) the same color (4) not the same color b. Find the minimum number of socks Farmer Brown must pull out of the drawer to guarantee that he will get a matching pair. 15. Tillie is approaching the toll booth on an expressway. She has three quarters and four dimes in her purse. She takes out two coins at random from her purse. Find the probability of each outcome. a. Both coins are quarters. b. Both coins are dimes. c. The coins are a dime and a quarter, in any order. d. The value of the coins is enough to pay the 35-cent toll. e. The value of the coins is not enough to pay the 35-cent toll given that one of the coins is a dime. f. The value of the coins was 35 cents given that one coin was a quarter. g. At least one of the coins picke
d was a quarter. 16. One hundred boys and one hundred girls were asked to name the current Secretary of State. Thirty boys and sixty girls knew the correct name. One of these boys and girls is selected at random. a. What is the probability that the person selected knew the correct name? b. What is the probability that the person selected is a girl, given that that person knew the correct name? c. What is the probability that the person selected knew the correct name, given that the person is a boy? 17. In a graduating class of 400 seniors, 200 were male and 200 were female. The students were asked if they had ever downloaded music from an online music store. 75 of the male students and 70 of the female students said that they had downloaded music. a. What is the probability that a randomly chosen senior has downloaded music? b. What is the probability that a randomly chosen senior has downloaded music given that the senior is male? 18. Gracie baked one dozen sugar cookies and two dozen brownies. She then topped two-thirds of the cookies and half the brownies with chocolate frosting. a. What is the probability that a randomly chosen treat is an unfrosted brownie? b. What is the probability that a baked good chosen at random is a cookie given that it is frosted? Permutations 627 19. In a game of Tic-Tac-Toe, the first player can put an X in any of the four corner squares, four edge squares, or the center square of the grid. The second player can then put an O in any of the eight remaining open squares. a. What is the probability that the second player will put an O in the center square given the first player has put an X in an edge square? b. If the first player puts an X in a corner square, what is the probability that the second player will put an O in a corner square? c. In a game of Tic-Tac-Toe, the first player puts an X in the center square and the second player puts an O in a corner square. What is the probability that the first player will put her next X in an edge square? 20. Of the 150 members of the high school marching band, 30 play the trumpet, 40 are in the jazz band, and 18 play the trumpet and are also in the jazz band. a. What is the probability that a randomly chosen member of the marching band plays the trumpet but is not in the jazz band? b. What is the probability that a randomly chosen member of the marching band is also in the jazz band but does not play the trumpet? 15-9 PERMUTATIONS A teacher has announced that Al, Betty, and Chris, three students in her class, will each give an oral report today. How many possible ways are there for the teacher to choose the order in which these students will give their reports? Al Betty Chris Betty Chris Al Chris Al Betty Chris Betty Chris Al Betty Al A tree diagram shows that there are six possible orders or arrangements. For example, Al, Betty, Chris is one possible arrangement; Al, Chris, Betty is another. Each of these arrangements is called a permutation. A permutation is an arrangement of objects or things in some specific order. (In discussing permutations, the words “objects” or “things” are used in a mathematical sense to include all elements in question, whether they are people, numbers, or inanimate objects.) The six possible permutations in this case may also be shown as a set of ordered triples. Here, we let A represent Al, B represent Betty, and C represent Chris: {(A, B, C), (A, C, B), (B, A, C), (B, C, A), (C, A, B), (C, B, A)} 628 Probability Let us see, from another point of view, why there are six possible orders. We know that any one of three students can be called on to give the first report. Once the first report is given, the teacher may call on any one of the two remaining students. After the second report is given, the teacher must call on the one remaining student. Using the counting principle, we see that there are 3 2 1 or 6 possible orders. Consider another situation. A chef is preparing a recipe with 10 ingredients. He puts all of one ingredient in a bowl, followed by all of another ingredient, and so on. How many possible orders are there for placing the 10 ingredients in a bowl? Using the counting principle, we have: 10 ,628,800 possible ways If there are more than 3 million possible ways of placing 10 ingredients in a bowl, can you imagine in how many ways 300 people who want to buy tickets for a football game can be arranged in a line? Using the counting principle, we find the number of possible orders to be 300 299 298 3 2 1. To symbolize such a product, we make use of the factorial symbol, !. We represent the product of these 300 numbers by the symbol 300!, read as “three hundred factorial” or “factorial 300.” Factorials In general, for any natural number n, we define n factorial or factorial n as follows: DEFINITION n! n(n 1)(n 2)(n 3) 3 2 1 Note that 1! is the natural number 1. Calculators can be used to evaluate factorials. On a graphing calculator, the factorial function is found by first pressing and then using the left arrow key to highlight the PRB menu. For example, to evaluate 5!, use the following sequence of keys: MATH ENTER: 5 MATH 4 ENTER DISPLAY: 5 ! 1 2 0 Of course, whether a calculator does or does not have a factorial function, we can always use repeated multiplication to evaluate a factorial: 5! 5 4 3 2 1 Permutations 629 When using a calculator, we must keep in mind that factorial numbers are usually very large. If the number of digits in a factorial exceeds the number of places in the display of the calculator, the calculator will shift from standard decimal notation to scientific notation. For example, evaluate 15! on a calculator: ENTER: 15 MATH 4 ENTER DISPLAY The number in the display can be written in scientific notation or in decimal notation. 1.307674368 E 12 1.307674368 1012 1,307,674,368,000 Representing Permutations We have said that permutations are arrangements of objects in different orders. For example, the number of different orders in which four people can board a bus is 4! or 4 3 2 1 or 24. There are 24 permutations, that is, 24 different orders or arrangements, of these four people, in which all four of them get on the bus. We may also represent this number of permutations by the symbol 4P4. The symbol 4P4 is read as: “the number of permutations of four objects taken four at a time.” Here, the letter P represents the word permutation. The small 4 written to the lower left of P tells us that four objects are avail- able to be used in an arrangement, (four people are waiting for a bus). The small 4 written to the lower right of P tells us how many of these objects are to be used in each arrangement, (four people getting on the bus). 4! 4 3 2 1 24. Thus, 4P4 Similarly, 5P5 In the next section, we will study examples where not all the objects are used in the arrangement. We will also examine a calculator key used with permutations. For now, we make the following observation: 5! 5 4 3 2 1 120. For any natural number n, the number of permutations of n objects taken n at a time can be represented as: nPn n! n(n 1)(n 2) 3 2 1 630 Probability EXAMPLE 1 Compute the value of each expression. a. 6! b. 2P2 c. 7! 3! Solution a. 6! 6 5 4 3 2 1 720 2! 2 1 2 b. 2P2 3! 3 3 2 3 1 c 840 Calculator Solution a. ENTER: 6 MATH 4 ENTER c. ENTER: 7 MATH 4 4 ENTER ENTER 3 MATH DISPLAY DISPLAY: 6 ! Answers a. 120 b. 2 c. 840 EXAMPLE 2 Paul wishes to call Virginia, but he has forgotten her unlisted telephone number. He knows that the exchange (the first three digits) is 555, and that the last four digits are 1, 4, 7, and 9, but he cannot remember their order. What is the maximum number of telephone calls that Paul may have to make in order to dial the correct number? Solution The telephone number is 555- . Since the last four digits will be an arrangement of 1, 4, 7, and 9, this is a permutation of four numbers, taken four at a time. 4P4 4! 4 3 2 1 24 possible orders Answer The maximum number of calls that Paul may have to make is 24. Permutations That Use Some of the Elements At times, we deal with situations involving permutations in which we are given n objects, but we use fewer than n objects in each arrangement. For example, a teacher has announced that he will call students from the first row to explain homework problems at the board. The students in the first row are George, Helene, Jay, Karla, Lou, and Marta. If there are only two homework problems, and each problem is to be explained by a different student, in how many orders may the teacher select students to go to the board? We know that the first problem can be assigned to any of six students. Once this problem is explained, the second problem can be assigned to any of the five Permutations 631 remaining students. We use the counting principle to find the number of possible orders in which the selection can be made. 6 5 30 possible orders If there are three problems, then after the first two students have been selected, there are four students who could be selected to explain the third problem. Extend the counting principle to find the number of possible orders in which the selection can be made. 6 5 4 120 possible orders Note that the starting number is the number of persons in the group from which the selection is made. Each of the factors is one less than the preceding factor. The number of factors is the number of choices to be made. Using the language of permutations, we say that the number of permuta- tions of six objects taken three at a time is 120. The Symbols for Permutations In general, if we have a set of n different objects, and we make arrangements of r objects from this set, we represent the number of arrangements by the symbol nPr. The subscript, r, representing the number of factors being used, must be less than or equal to n, the total number of objects in the set. Thus: For numbers n and r, where r n, the permutation of n objects, taken r at a time, is found by the formula: nPr 5 n(n 2 1)(n 2 2) ??? g r factors This formula ca
n also be written as: nPr n(n 1)(n 2) (n r 1) Note that when there are r factors, the last factor is (n r 1). In the example given above, in which three students were selected from a group of six, n 6, r 3, and the last factor is n r 1 6 3 1 4. Permutations and the Calculator There are many ways to use a calculator to evaluate a permutation. In the three solutions presented here, we will evaluate the permutation 8P3, the order in which, from a set of 8 elements, 3 can be selected. 632 Probability METHOD 1. Use the Multiplication Key, . The number permutation is simply the product of factors. In 8P3, the first factor, 8, is multiplied by (8 1), or 7, and then by (8 2), or 6. Here, exactly three factors have been multiplied. ENTER: 8 7 6 ENTER DISPLAY: 8 * 7 * 6 3 3 6 This approach can be used for any permutation. In the general permutation nPr, where r n, the first factor n is multiplied by (n 1), and then by (n – 2), and so on, until exactly r factors have been multiplied. METHOD 2. Use the Factorial Function. When evaluating 8P3, it appears as if we start to evaluate 8! but then we stop after multiplying only three factors. There is a way to write the product 8 7 6 using factorials. As shown below, when we divide 8! by 5!, all but the first three factors will cancel, leaving or 336. Since the factorial in the denominator uses all but three factors, 8 3 5 can be used to find that factorial: 8 3 7 3 6 8P3 ! 5! 5 8! (8 2 3)! We now evaluate this fraction on a calculator. ENTER: 8 MATH 4 5 MATH 4 ENTER DISPLAY: 8 ! / 5 ! 3 3 6 This approach shows us that there is another formula that can be used for the general permutation nPr, where r n: nPr n! (n 2 r)! Permutations 633 METHOD 3. Use the Permutation Function. Calculators have a special function to evaluate permutations. On a graphing and then using the left arrow key calculator, nPr is found by first pressing MATH to highlight the PRB menu. The value of n is entered first, then the nPr symbol is entered followed by the value of r. ENTER: 8 MATH 2 3 ENTER DISPLAY: 8 n P r 3 3 3 6 EXAMPLE 3 Evaluate 6P2. Solution This is a permutation of six objects, taken two at a time. There are two possible formulas to use. 6P2 6 5 30 6P2 4 30 Calculator Solution ENTER: 6 MATH 2 2 ENTER DISPLAY: 6 n P r 2 3 0 Answer 6P2 = 30 EXAMPLE 4 How many three-letter “words” (arrangements of letters) can be formed from the letters L, O, G, I, C if each letter is used only once in a word? Solution Forming three-letter arrangements from a set of five letters is a permutation of five, taken three at a time. Thus: 5P3 5 4 3 60 5P3 Answer 60 words or 2! (5 2 3)! 5 5! 5 4 3 60 634 Probability EXAMPLE 5 A lottery ticket contains a four-digit number. How many possible four-digit numbers are there when: a. a digit may appear only once in the number? b. a digit may appear more than once in the number? Solution a. If a digit appears only once in a four-digit number, this is a permutation of 10 digits, taken four at a time. Thus: 10P4 10 9 8 7 5,040 b. If a digit may appear more than once, we can choose any of 10 digits for the first position, then any of 10 digits for the second position, and so forth. By the counting principle: 10 10 10 10 10,000 Answers a. 5,040 b. 10,000 EXERCISES Writing About Mathematics 1. Show that n! n(n 1)!. 2. Which of these two values, if either, is larger: 9P9 or 9P8? Explain your answer. Developing Skills In 3–17, compute the value of each expression. 3. 4! 8. 3P3 13. 20P2 4. 6! 9. 8P8 14. 11P4 5. 7! 10. 8! 5! 15. 7P6 6. 3! 2! 11. 6P3 16. 255P2 7. (3 2)! 12. 10P2 17. 999! (999 2 5)! In 18–21, in each case, how many three-letter arrangements can be formed if a letter is used only once? 18. LION 19. TIGER 20. MONKEY 21. LEOPARD 22. Write the following expressions in the order of their values, beginning with the smallest: 60P5, 45P6, 24P7, 19P7. Permutations 635 Applying Skills 23. Using the letters E, M, I, T: a. How many arrangements of four letters can be found if each letter is used only once in the “word”? b. List these “words.” 24. In how many different ways can five students be arranged in a row? 25. In a game of cards, Gary held exactly one club, one diamond, one heart, and one spade. In how many different ways can Gary arrange these four cards in his hand? 26. There are nine players on a baseball team. The manager must establish a batting order for the players at each game. The pitcher will bat last. How many different batting orders are possible for the eight remaining players on the team? 27. There are 30 students in a class. Every day the teacher calls on different students to write homework problems on the board, with each problem done by only one student. In how many ways can the teacher call students to the board if the homework consists of: a. only 1 problem? b. 2 problems? c. 3 problems? 28. At the Olympics, three medals are given for each competition: gold, silver, and bronze. Tell how many possible winning orders there are for the gymnastic competition if the number of competitors is: a. 7 b. 9 c. 11 d. n 29. How many different ways are there to label the three vertices of a scalene triangle, using no letter more than once, when: a. we use the letters R, S, T? b. we use all the letters of the English alphabet? 30. A class of 31 students elects four people to office, namely, a president, vice president, secretary, and treasurer. In how many possible ways can four people be elected from this class? 31. How many possible ways are there to write two initials, using the letters of the English alphabet, if: a. an initial may appear only once in each pair? b. the same initial may be used twice? In 32–35: a. Write each answer in factorial form. b. Write each answer, after using a calculator, in scientific notation. 32. In how many different orders can 60 people line up to buy tickets at a theater? 33. We learn the alphabet in a certain order, starting with A, B, C, and ending with Z. How many possible orders are there for listing the letters of the English alphabet? 34. Twenty-five people are waiting for a bus. When the bus arrives, there is room for 18 people to board. In how many ways could 18 of the people who are waiting board the bus? 35. Forty people attend a party at which eight door prizes are to be awarded. In how many orders can the names of the winners be announced? 636 Probability 15-10 PERMUTATIONS WITH REPETITION How many different “words” or arrangements of four letters can be formed using each letter of the word PEAK? This is the number of permutations of four things, taken four at a time. Since 4! 4 3 2 1 24, there are 24 possible words or arrangements. 4P4 PEAK PAKE PKAE KPAE PAEK PEKA PKEA KPEA EAPK APKE AKPE KAEP AEPK EPKA EKPA KEAP EPAK AEKP AKEP KAPE APEK EAKP EKAP KEPA Now consider a related question. How many different words or arrange- ments of four letters can be formed using each letter of the word PEEK? This is an example of a permutation with repetition because the letter E is repeated in the word. We can try to list the different arrangements by simply replacing the A in each of the arrangements given for the word PEAK. Let the E from PEAK be E1 and the E that replaces A be E2. The arrangements can be written as follows: E1E2PK E2PKE1 E2KPE1 KE2E1P E2E1PK E1PKE2 E1KPE2 KE1E2P PE2E1K PE1KE2 PKE1E2 KPE1E2 E2PE1K E1E2KP E1KE2P KE1PE2 E1PE2K E2E1KP E2KE1P KE2PE1 PE1E2K PE2KE1 PKE2E1 KPE2E1 We have 24 different arrangements if we consider E1 to be different from E2. But they are not really different. Notice that if we consider the E’s to be the same, every word in the first column is the same as a word in the second column, every word in the third column is the same as a word in the fourth column, and every word in the fifth column is the same as a word in the sixth column. 24 Therefore, only the first, third, and fifth columns are different and there are 2 or 12 arrangements of four letters when two of them are the same. This is the number of arrangements of four letters divided by the number of arrangements of two letters. Now consider a third word. In how many ways can the letters of EEEK be arranged? We will use the 24 arrangements of the letters of PE1E2K and write E3 in place of P. E3E1E2K E3E2KE1 E3KE2E1 KE3E2E1 E1E2E3K E3E2E1K E2E3KE1 E3E1KE2 E3KE1E2 E2KE3E1 K E3E1E2 KE2E1E3 E2E1E3K E1E3KE2 E1K E3E2 KE1E2E3 E1E3E2K E2E1KE3 E2KE1E3 KE2E3E1 E2E3E1K E1E2KE3 E1KE2E3 KE1E3E2 Notice that each row is the same arrangement if we consider all the E’s to be the same letter. The 4! arrangements of four letters are in groups of 3! or 6, the number of different orders in which the 3 E’s can be arranged among themselves. Therefore the number of possible arrangements is: 3! 3 3 2 3 1 5 24 6 5 4 Permutations with Repetition 637 If we consider all E’s to be the same, the four arrangements are: EEEK EEKE EKEE KEEE In general, the number of permutations of n things, taken n at a time, with r of these things identical, is: n! r! EXAMPLE 1 How many six-digit numerals can be written using all of the following digits: 2, 2, 2, 2, 3, and 5? Solution This is the number of permutations of six things taken six at a time, with 4 of the digits 2, 2, 2, 2 identical. Therefore: 4! 4 3 3 3 2 3 1 4 MATH 4 5 6 3 5 5 30 4 ENTER Calculator Solution ENTER: 6 MATH DISPLAY: 6 ! / 4 ! 3 0 Answer 30 six-digit numerals EXAMPLE 2 Three children, Rita, Ann, and Marie, take turns doing the dishes each night of the week. At the beginning of each week they make a schedule. If Rita does the dishes three times, and Ann and Marie each do them twice, how many different schedules are possible? Solution We can think of this as an arrangement of the letters RRRAAMM, that is, an arrangement of seven letters (for the seven days of the week) with R appearing three times and A and M each appearing twice. Therefore we will divide 7! by 3!, the number of arrangements of Rita’s days; then by 2!, the number of arrangements of Ann’s days; and finally by 2! again, the number of arrangements o
f Marie’s days. Number of arrangements Answer 210 possible schedules 7! 3! 3 2! 3 2 210 638 Probability EXERCISES Writing About Mathematics 1. a. List the six different arrangements or permutations of the letters in the word TAR. b. Explain why exactly six arrangements are possible. 2. a. List the three different arrangements or permutations of the letters in the word TOT. b. Explain why exactly three arrangements are possible. Developing Skills In 3–6, how many five-letter permutations are there of the letters of each given word? 3. APPLE 4. ADDED 5. VIVID 6. TESTS In 7–14: a. How many different six-letter arrangements can be written using the letters in each given word? b. How many different arrangements begin with E? c. If an arrangement is chosen at random, what is the probability that it begins with E? 7. SIMPLE 8. FREEZE 9. SYSTEM 10. BETTER 11. SEEDED 12. DEEDED 13. TATTOO 14. ELEVEN In 15–22, find the number of distinct arrangements of the letters in each word. 15. STREETS 16. INSISTS 17. ESTEEMED 18. DESERVED 19. TENNESSEE 20. BOOKKEEPER 21. MISSISSIPPI 22. UNUSUALLY In 23–26, in each case find: a. How many different five-digit numerals can be written using all five digits listed? b. How many of the numerals formed from the given digits are greater than 12,000 and less than 13,000? c. If a numeral formed from the given digits is chosen at random, what is the probability that it is greater than 12,000 and less than 13,000? 23. 1, 2, 3, 4, 5 24. 1, 2, 2, 2, 2 25. 1, 1, 2, 2, 2 26. 2, 2, 2, 2, 3 In 27–31, when written without using exponents, a2x can be written as aax, axa, or xaa. How many different arrangements of letters are possible for each given expression when written without exponents? 27. b3y 28. a2b5 29. abx6 30. a2by7 31. a4b8 Applying Skills 32. A bookseller has 7 copies of a novel and 3 copies of a biography. In how many ways can these 10 books be arranged on a shelf? 33. In how many ways can 6 white flags and 3 blue flags be arranged one above another on a single rope on a flagstaff? Combinations 639 34. Florence has 6 blue beads, 8 white beads, and 4 green beads, all the same size. In how many ways can she string these beads on a chain to make a necklace? 35. Frances has 8 tulip bulbs, 10 daffodil bulbs, and 7 crocus bulbs. In how many ways can Frances plant these bulbs in a border along the edge of her garden? 36. Anna has 2 dozen Rollo bars and 1 dozen apples as treats for Halloween. In how many ways can Anna hand out 1 treat to each of 36 children who come to her door? 37. A dish of mixed nuts contains 7 almonds, 5 cashews, 3 filberts, and 4 peanuts. In how many different orders can Jerry eat the contents of the dish, one nut at a time? 38. Print your first and last names using capital letters. How many different arrangements of the letters in your full name are possible? 15-11 COMBINATIONS Comparing Permutations and Combinations A combination is a collection of things in which order is not important. Before we discuss combinations, let us start with a problem we know how to solve. Ann, Beth, Carlos, and Dava are the only members of a school club. In how many ways can they elect a president and a treasurer for the club? Any one of the 4 students can be elected as president. After this happens, any one of the 3 remaining students can be elected treasurer. Thus there are 4 3 or 12 possible outcomes. Using the initials to represent the students involved, we can write these 12 arrangements: (A, B) (B, A) (B, C) (C, B) (A, C) (C, A) (B, D) (D, B) (A, D) (D, A) (C, D) (D, C) These are the permutations. We could have found that there are 12 permutations by using the formula: Answer: 12 permutations 4P2 4 3 12 Now let us consider two problems of a different type involving the members of the same club. Ann, Beth, Carlos, and Dava are the only members of a school club. In how many ways can they choose two members to represent the club at a student council meeting? If we look carefully at the list of 12 possible selections given in the answer to problem 1, we can see that while (A, B) and (B, A) are two different choices for president and treasurer, sending Ann and Beth to the student council meet- 640 Probability ing is exactly the same as sending Beth and Ann. For this problem let us match up answers that consist of the same two persons. (A, B) ↔ (B, A) (A, C) ↔ (C, A) (A, D) ↔ (D, A) (B, C) ↔ (C, B) (B, D) ↔ (D, B) (C, D) ↔ (D, C) Although order is important in listing slates of officers in problem 1, there is no reason to consider the order of elements in this problem. In fact, if we think of two representatives to the student council as a set of two club members, we can list the sets of representatives as follows: {A, B} {A, C} {A, D} {B, C} {B, D} {C, D} From this list we can find the number of combinations of 4 things, taken 2 at a time, written in symbols as 4C2.. The answer to this problem is found by dividing the number of permutations of 4 things taken 2 at a time by 2!.Thus: Answer: 6 combinations 4C2 4 P 2 12 2 5 6 Ann, Beth, Carlos, and Dava are the only members of a school club. In how many ways can they choose a three-member committee to work on the club’s next project? Is order important to this answer? If 3 officers were to be elected, such as a president, a treasurer, and a secretary, then order would be important and the number of permutations would be needed. However, a committee is a set of people. In listing the elements of a set, order is not important. Compare the permutations and combinations of 4 persons taken 3 at a time: Permutations Combinations (A, B, C) (A, C, B) (B, A, C) (B, C, A) (C, A, B) (C, B, A) (A, B, D) (A, D, B) (B, A, D) (B, D, A) (D, A, B) (D, B, A) (A, C, D) (A, D, C) (C, A, D) (C, D, A) (D, A, C) (D, C, A) (B, C, D) (B, D, C) (C, B, D) (C, D, B) (D, B, C) (D, C, B) {A, B, C} {A, B, D} {A, C, D} {B, C, D} While there are 24 permutations, written as ordered triples, there are only 4 combinations, written as sets. For example, in the first row of permutations, there are 3! or or 6 ordered triples (slates of officers) including Ann, Beth, and Carlos. 3 3 2 3 1 However, there is only one set (committee) that includes these three persons. Therefore, the number of ways to select a three-person committee from a group of four persons is the number of combinations of 4 things taken 3 at a time. This number is found by dividing the number of permutations of 4 things taken 3 at a time by 3!, the number of arrangements of the three things. Combinations 641 Answer: 4 combinations 4C3 4 P 3 24 6 5 4 In general, for counting numbers n and r, where r n, the number of com- binations of n things taken r at a time is found by using the formula: P r nCr 5 n r! On a graphing calculator, the sequence of keys needed to find nCr is similar to that for nPr. The combination symbol is entry 3 in the PRB menu. ENTER: 4 MATH 3 3 ENTER DISPLAY: 4 n C r 3 4 Note: The notation also represents the number of combinations of n things taken r at a time. Thus: n r A B n r B A nCr or n r A B P n r r! Some Relationships Involving Combinations Given a group of 5 people, how many different 5-person committees can be formed? Common sense tells us that there is only 1 such committee, namely, the committee consisting of all 5 people. Using combinations, we see that 5C5 5 5 P 5 Also, P 3! 5 3 3 2 3 1 3 3C3 5 3 For any counting number n, nCn 3 3 2 3 1 5 1 and 1. 4C4 5 4 P 4 Given a group of 5 people, in how many different ways can we select a committee consisting of no people, or 0 people? Common sense tells us that there is 1. Let us agree to only 1 way to select no one. Thus, using combinations, 5C0 the following generalization: For any counting number n, nC0 1. 642 Probability In how many ways can we select a committee of 2 people from a group of 7 people? Since a committee is a combination, P 2! 5 7 3 6 2 Now, given a group of 7 people, in how many ways can 5 people not be appointed to the committee? Since each set of people not appointed is a combination: 2 3 1 5 21 7C2 . 7 P 5 7C5 Notice that 7C2 = 7C5. In other words, starting with a group of 7 people, the number of sets of 2 people that can be selected is equal to the number of sets of 5 people that can be not selected. In the same way it can be shown that 7C3 In general, starting with n objects, the number of ways to choose r objects for a combination is equal to the number of ways to not choose (n – r) objects for the combination. 7C1, and that 7C7 2 3 1 3 1 5 21 7C4, that 7C6 7C0. For whole numbers n and r, where r n, nCr nCnr KEEP IN MIND PERMUTATIONS COMBINATIONS 1. Order is important. 1. Order is not important. Think of ordered elements such as ordered pairs and ordered triples. 2. An arrangement or a slate of officers indicates a permutation. Think of sets. 2. A committee, or a selection of a group, indicates a combination. EXAMPLE 1 Evaluate: 10C3 Solution This is the number of combinations of 10 things, taken 3 at a time. Thus 10C3 10 P 3! 5 10 720 6 5 120 Calculator Solution ENTER: 10 MATH 3 3 ENTER DISPLAY Answer 120 Combinations 643 EXAMPLE 2 Evaluate: 25 23b a Solution (1) This is an alternative form for the number of combinations of 25 things, taken 23 at a time: 25C23. 25 23b a nCn r, 25C23 (2) Since nCr (3) Using 25C2, perform the shorter computation. Thus: 25C2. 25 23b a 5 25C23 5 25C2 5 25 3 24 2 3 1 5 300 Answer 300 EXAMPLE 3 There are 10 teachers in the science department. How many 4-person committees can be formed in the department if Mrs. Martens and Dr. Blumenthal, 2 of the teachers, must be on each committee? Solution Since Mrs. Martens and Dr. Blumenthal must be on each committee, the prob- lem becomes one of filling 2 positions on a committee from the remaining 8 teachers. 8C2 Answer 28 committees EXAMPLE 4 8 P 2 28 There are six points in a plane, no three of which are collinear. How many straight lines can be drawn using pairs of these three points? Solution Whether joining points A and B, or points B and A, onl
y 1 line exists, namely, . Since order is not important here, this is a combination of 6 points, taken g AB 2 at a time. Answer 15 lines 6C2 6 P 2 15 644 Probability EXAMPLE 5 Lisa Dwyer is a teacher at a local high school. In her class, there are 10 boys and 20 girls. Find the number of ways in which Ms. Dwyer can select a team of 3 students from the class to work on a group project if the team consists of: a. any 3 students b. 1 boy and 2 girls c. 3 girls d. at least 2 girls Solution a. The class contains 10 boys and 20 girls, for a total of 30 students. Since order is not important on a team, this is a combination of 30 students, taken 3 at a time. 30 30C3 P 3! 5 30 3 29 3 28 3 b. This is a compound event. To find the number of ways to select 1 boy out of 10 boys for a team, use 10C1. To find the number of ways to select 2 girls out of 20 for the team, use 20C2. Then, by the counting principle, multiply the results. 3 3 2 3 1 5 4,060 1 3 20 3 19 10 2 3 1 c. This is another compound event, in which 0 boys out of 10 boys and 3 girls 10 190 1,900 10C1 20C2 out of 20 girls are selected. Recall that 10C0 = 1. Thus: 10C0 20C3 1 3 20 3 19 3 18 3 3 2 3 1 1,140 Note that this could also have been thought of as the simple event of selecting 3 girls out of 20 girls: 20 3 19 3 18 3 3 2 3 1 d. A team of at least 2 girls can consist of exactly 2 girls (see part b) or exactly 3 girls (see part c). Since these events are disjoint, add the solutions to parts b and c: 1,140 20C3 Answers a. 4,060 teams b. 1,900 teams c. 1,140 teams d. 3,040 teams 1,900 1,140 3,040 EXERCISES Writing About Mathematics 1. Explain the difference between a permutation and a combination. 2. A set of r letters is to be selected from the 26 letters of the English alphabet. For what value of r is the number of possible sets of numbers greatest? Combinations 645 Developing Skills In 3–14, evaluate each expression. 3. 15C2 7. 13C0 11. 7 3b a 4. 12C3 8. 14C14 12. 9 4b a 5. 10C4 9. 9C8 13. 17 17b a 6. 25C1 10. 200C198 14. 499 2 b a 15. Find the number of combinations of 6 things, taken 3 at a time. 16. How many different committees of 3 people can be chosen from a group of 9 people? 17. How many different subsets of exactly 7 elements can be formed from the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}? 18. For each given number of non-collinear points in a plane, how many straight lines can be drawn? a. 3 b. 4 c. 5 d. 7 e. 8 f. n 19. Consider the following formulas: n! (n 2 r)! r! P n r r! (2) nCr (3) nCr (1) nCr a. Evaluate 8C3 using each of the three formulas. b. Evaluate 11C7 using each of the three formulas. c. Can all three formulas be used to find the combination of n things, taken r at a time? n(n 2 1)(n 2 2) c(n 2 r 1 1) r! Applying Skills 20. A coach selects players for a team. If, while making this first selection, the coach pays no attention to the positions that individuals will play, how many teams are possible? a. Of 14 candidates, Coach Richko needs 5 for a basketball team. b. Of 16 candidates, Coach Jones needs 11 for a football team. c. Of 13 candidates, Coach Greves needs 9 for a baseball team. 21. A disc jockey has 25 recordings at hand, but has time to play only 22 on the air. How many sets of 22 recordings can be selected? 22. There are 14 teachers in a mathematics department. a. How many 4-person committees can be formed in the department? b. How many 4-person committees can be formed if Mr. McDonough, 1 of the 14, must be on the committee? c. How many 4-person committees can be formed if Mr. Goldstein and Mrs. Friedel, 2 of the 14, must be on the committee? 646 Probability 23. There are 12 Republicans and 10 Democrats on a senate committee. From this group, a 3-person subcommittee is to be formed. Find the number of 3-person subcommittees that consist of: a. any members of the senate committee c. 1 Republican and 2 Democrats b. Democrats only d. at least 2 Democrats e. John Clark, who is a Democrat, and any 2 Republicans 24. Sue Bartling loves to read mystery books and car-repair manuals. On a visit to the library, Sue finds 9 new mystery books and 3 car-repair manuals. She borrows 4 of these books. Find the number of different sets of 4 books Sue can borrow if: a. all are mystery books c. only 1 is a mystery book b. exactly 2 are mystery books d. all are car-repair manuals 25. Cards are drawn at random from a 52-card deck. Find the number of different 5-card poker hands possible consisting of: a. any 5 cards from the deck c. 4 queens and any other card e. 2 aces and 3 picture cards b. 3 aces and 2 kings d. 5 spades f. 5 jacks 26. How many committees consisting of 7 people or more can be formed from a group of 10 people? 27. There are 12 roses growing in Heather’s garden. How many different ways can Heather choose roses for a bouquet consisting of more than 8 roses? 15-12 PERMUTATIONS, COMBINATIONS, AND PROBABILITY In this section, a variety of probability questions are presented. In some cases, permutations should be used. In other cases, combinations should be used. When answering these questions, the following should be kept in mind: 1. If the question asks “How many?” or “In how many ways?” the answer will be a whole number. 2. If the question asks “What is the probability?” the answer will be a value between 0 and 1 inclusive. 3. If order is important, use permutations. 4. If order is not important, use combinations. In Examples 1–4, Ms. Fenstermacher must select 4 students to represent the class in a spelling bee. Her best students include 3 girls (Callie, Daretta, and Jessica) and 4 boys (Bandu, Carlos, Sanjit, and Uri). Permutations, Combinations, and Probability 647 EXAMPLE 1 Ms. Fenstermacher decides to select the 4 students from the 7 best by drawing names from a hat. How many different groups of 4 are possible? Solution In choosing a group of 4 students out of 7, order is not important. Therefore, use combinations. 7C4 Answer 35 groups EXAMPLE 2 7 P 4 35 What is the probability that the 4 students selected for the spelling bee will consist of 2 girls and 2 boys? Solution (1) The answer to Example 1 shows 35 possible groups, or n(S) 35. (2) Since order is not important in choosing groups, use combinations. The number of ways to choose 2 girls out of 3 is 3C2 = The number of ways to choose 2 boys out of 4 is 4C2 Event E is the compound event of choosing 2 girls and 2 boys, or 6. 3 6 18. n(E. 3C2 4C2 (3) Thus P(E) C 2 4 C 2 3 3 7C4 5 18 35 . Answer P(2 girls, 2 boys) 18 35 EXAMPLE 3 The students chosen are Callie, Jessica, Carlos, and Sanjit. In how many orders can these 4 students be called upon in the spelling bee? Solution The number of ways in which 4 students can be called upon in a spelling bee means that someone is first, someone else is second, and so on. Since this is a problem about order, use permutations. 4P4 4! 4 3 2 1 24 Answer 24 orders 648 Probability EXAMPLE 4 For the group consisting of Callie, Jessica, Carlos, and Sanjit, what is the probability that the first 2 students called upon will be girls? Solution This question may be answered using various methods, two of which are shown below. METHOD 1. Counting Principle There are 2 girls out of 4 students who may be called first. Once a girl is called upon, only 1 girl remains in the 3 students not yet called upon. Apply the counting principle of probability. P(first 2 are girls) P(girl first) P(girl second) 2 4 3 1 3 5 2 12 5 1 6 METHOD 2. Permutations Let n(S) equal the number of ways to call 2 of the 4 students in order, and let n(E) equal the number of ways to call 2 of the 2 girls in order. P(E) n(E) n(S) 5 2 P 2 4P2 Answer P(first 2 are girls) 1 6 EXERCISES Writing About Mathematics 5 2 3 1 4 3 3 5 2 12 5 1 6 1. A committee of three persons is to be chosen from a group of eight persons. Olivia is one of the persons in that group. Olivia said that since 8C3 8C5, the probability that she will be chosen for that committee is equal to the probability that she will not be chosen. Do you agree with Olivia? Explain why or why not. 2. Four letters are to be selected at random from the alphabet. Jenna found the probability that the four letters followed S in alphabetical order by using permutations. Colin found the probability that the four letters followed S in alphabetical order by using combinations. Who was correct? Explain your answer. Applying Skills 3. The Art Club consists of 4 girls (Jennifer, Anna, Gloria, Teresa) and 2 boys (Mark and Dan). a. In how many ways can the club elect a president and a treasurer? b. Find the probability that the 2 officers elected are both girls. c. How many 2-person teams can be selected to work on a project? d. Find the probability that a 2-person team consists of: (1) 2 girls (2) 2 boys (3) 1 girl and 1 boy (4) Anna and Mark Permutations, Combinations, and Probability 649 4. A committee of 4 is to be chosen at random from 4 men and 3 women. a. How many different 4-member committees are possible? b. How many 4-member committees contain 3 men and 1 woman? c. What is the probability that a 4-member committee will contain exactly 1 woman? d. What is the probability that a man will be on the committee? e. What is the probability that the fourth person chosen is a woman given that 3 men have already been chosen? 5. A committee of 6 people is to be chosen from 9 available people. a. How many 6-person committees can be chosen? b. The committee, when chosen, has 4 students and 2 teachers. Find the probability that a 3-person subcommittee from this group includes: (1) students only (2) exactly 1 teacher (3) at least 2 students (4) a teacher given that 2 students have been chosen 6. A box of chocolate-covered candies contains 7 caramels and 3 creams all exactly the same in appearance. Jim selects 4 pieces of candy. a. Find the number of selections possible of 4 pieces of candy that include: (1) 4 caramels (2) 1 caramel and 3 creams (3) 2 caramels and 2 creams (4) any 4 pieces b. Find the probability that Jim’s selection included: (1) 4 caram
els (2) 1 caramel and 3 creams (3) 2 caramels and 2 creams (4) no caramels (5) a second cream given that 2 caramels and a cream have been selected 7. Two cards are drawn at random from a 52-card deck without replacement. Find the proba- bility of drawing: a. the ace of clubs and jack of clubs in b. a red ace and a black jack in either order c. 2 jacks either order d. 2 clubs e. an ace and a jack in either order f. an ace and a jack in that order g. a heart given that the king of hearts h. a king given that a queen was was drawn drawn 8. Mrs. Carberry has 4 quarters and 3 nickels in her purse. If she takes 3 coins out of her purse without looking at them, find the probability that the 3 coins are worth: a. exactly 75 cents d. exactly 55 cents b. exactly 15 cents c. exactly 35 cents e. more than 10 cents f. less than 40 cents 650 Probability 9. A 3-digit numeral is formed by selecting digits at random from {2, 4, 6, 7} without repetition. Find the probability that the number formed: a. is less than 700 c. contains only even digits b. d. is greater than 600 is an even number 10. a. There are 10 runners on the track team. If 4 runners are needed for a relay race, how many different relay teams are possible? b. Once the relay team is chosen, in how many different orders can the 4 runners run the race? c. If Nicolette is on the relay team, what is the probability that she will lead off the race? 11. Lou Grant is an editor at a newspaper employing 10 reporters and 3 photographers. a. If Lou selects 2 reporters and 1 photographer to cover a story, from how many possible 3-person teams can he choose? b. If Lou hands out 1 assignment per reporter, in how many ways can he assign the first 3 stories to his 10 reporters? c. If Lou plans to give the first story to Rossi, a reporter, in how many ways can he now assign the first 3 stories? d. If 3 out of 10 reporters are chosen at random to cover a story, what is the probability that Rossi is on this team? 12. Chris, Willie, Tim, Matt, Juan, Bob, and Steven audition for roles in the school play. a. If 2 male roles in the play are those of the hero and the clown, in how many ways can the director select 2 of the 7 boys for these roles? b. Chris and Willie got the 2 leading male roles. (1) In how many ways can the director select a group of 3 of the remaining 5 boys to work in a crowd scene? (2) How many of these groups of 3 will include Tim? (3) Find the probability that Tim is in the crowd scene. 13. There are 8 candidates for 3 seats in the student government. The candidates include 3 boys (Alberto, Peter, Thomas) and 5 girls (Elizabeth, Maria, Joanna, Rosa, Danielle). If all candidates have an equal chance of winning, find the probability that the winners include: a. 3 boys c. 1 boy and 2 girls b. 3 girls d. at least 2 girls e. Maria, Peter, and anyone else f. Danielle and any other two candidates g. Alberto, Elizabeth, and Rosa h. Rosa, Peter, and Thomas Chapter Summary 651 14. A gumball machine contains 6 lemon-, 4 lime-, 3 cherry-, and 2 orange-flavored gumballs. Five coins are put into the machine, and 5 gumballs are obtained. a. How many different sets of 5 gumballs are possible? b. How many of these will contain 2 lemon and 3 lime gumballs? c. Find the probability that the 5 gumballs dispensed by the machine include: (1) 2 lemon and 3 lime (2) 3 cherry and 2 orange (3) 2 lemon, 2 lime, and 1 orange (4) lemon only (5) lime only (6) no lemon 15. The letters in the word HOLIDAY are rearranged at random. a. How many 7-letter words can be formed? b. Find the probability that the first 2 letters are vowels. c. Find the probability of no vowels in the first 3 letters. 16. At a bus stop, 5 people enter a bus that has only 3 empty seats. a. In how many different ways can 3 of the 5 people occupy these empty seats? b. If Mrs. Costa is 1 of the 5 people, what is the probability that she will not get a seat? c. If Ann and Bill are 2 of the 5 people, what is the probability that they both will get seats? CHAPTER SUMMARY Probability is a branch of mathematics in which the chance of an event happening is assigned a numerical value that predicts how likely that event is to occur. Empirical probability may be defined as the most accurate scientific estimate, based on a large number of trials, of the cumulative relative frequency of an event happening. An outcome is a result of some activity or experiment. A sample space is a set of all possible outcomes for the activity. An event is a subset of the sample space. The theoretical probability of an event is the number of ways that the event can occur, divided by the total number of possibilities in the sample space. If P(E) represents the probability of event E, n(E) represents the number of outcomes in event E and n(S) represents the number of outcomes in the sample space S. The formula, which applies to fair and unbiased objects and situations is: P(E) n(E) n(S) The probability of an impossible event is 0. The probability of an event that is certain to occur is 1. The probability of any event E must be equal to or greater than 0, and less than or equal to 1: 0 P(E) 1 652 Probability For the shared outcomes of events A and B: n(A d B) n(S) If two events A and C are mutually exclusive: P(A and B) P(A or C) P(A) P(C) If two events A and B are not mutually exclusive: P(A or B) P(A) P(B) P(A and B) For any event A: P(A) P(not A) 1 The sum of the probabilities of all possible singleton outcomes for any sam- ple space must always equal 1. The Counting Principle: If one activity can occur in any of m ways and, following this, a second activity can occur in any of n ways, then both activities can occur in the order given in m n ways. The Counting Principle for Probability: E and F are independent events. The probability of event E is m (0 m 1) and the probability of event F is n (0 n 1), the probability of the event in which E and F occur jointly is the product m n. When the result of one activity in no way influences the result of a second activity, the results of these activities are called independent events. Two events are called dependent events when the result of one activity influences the result of a second activity. If A and B are two events, conditional probability is the probability of B given that A has occurred. The notation for conditional probability is P(B A). P(B A) P(A and B) P(A) The general result P(A and B) P(A) P(B A) is true for both dependent and independent events since for independent events, P(B A) P(B). A permutation is an arrangement of objects in some specific order. The num- ber of permutations of n objects taken n at a time, nPn, is equal to n factorial: nPn = n! n (n 1) (n 2) 3 2 1 The number of permutations of n objects taken r at a time, where r n, is equal to: nPr 5 n(n 2 1)(n 2 2) h r factors or nPr n! (n 2 r)! A permutation of n objects taken n at a time in which r are identical is equal .n! r! to A combination is a set of objects in which order is not important, as in a committee. The formula for the number of combinations of n objects taken r at a time (r n) is Review Exercises 653 nCr P n r r! From this formula, it can be shown that nCn nCnr . Permutations and combinations are used to evaluate n(S) and n(E) in prob- 1, and nCr 1, nC0 ability problems. VOCABULARY 15-1 Probability • Empirical study • Relative frequency • Cumulative rela- tive frequency • Converge • Empirical probability • Trial • Experiment • Fair and unbiased objects • Biased objects • Die • Standard deck of cards 15-2 Outcome • Sample space • Event • Favorable event • Unfavorable event • Singleton event • Theoretical probability • Calculated probability • Uniform probability • Equally likely outcomes • Random selection 15-3 Impossibility • Certainty • Subscript 15-5 Mutually exclusive events 15-7 Compound event • Tree diagram • List of ordered pairs • Graph of ordered pairs • Counting Principle • Independent events 15-8 Without replacement • Dependent events • With replacement • Conditional probability 15-9 Permutation • Factorial symbol (!) • n factorial 15-11 Combination REVIEW EXERCISES 1. In a dish of jellybeans, some are black. Aaron takes a jellybean from the dish at random without looking at what color he is taking. Jake chooses the same color jellybean that Aaron takes. Is the probability that Aaron takes a black jellybean the same as the probability that Jake takes a black jellybean? Explain your answer. 2. The probability that Aaron takes a black jellybean from the dish is 8 25 we conclude that there are 25 jellybeans in the dish and 8 of them are black? Explain why or why not. . Can 654 Probability 3. The numerical value of nPr is the product of r factors. Express the smallest of those factors in terms of n and r. 4. If P(A) .4, P(B) .3, and P(A B) .2, find P(A B). 5. If P(A) .5, P(B) .2, and P(A B) .5, find P(A B). In 6–11, evaluate each expression: 6. 8! 7. 5P5 8. 12P2 9. 5C5 10. 12C3 11. 40C38 12. How many 7-letter words can be formed from the letters in UNUSUAL if each letter is used only once in a word? 13. A SYZYGY is a nearly straight-line configuration of three celestial bodies such as the Earth, Moon, and Sun during an eclipse. a. How many different 6-letter arrangements can be made using the let- ters in the word SYZYGY? b. Find the probability that the first letter in an arrangement of SYZYGY is (1) G (2) Y (3) a vowel (4) not a vowel 14. From a list of 10 books, Gwen selects 4 to read over the summer. In how many ways can Gwen make a selection of 4 books? 15. Mrs. Moskowitz, the librarian, checks out the 4 books Gwen has chosen. In how many different orders can the librarian stamp these 4 books? 16. A coach lists all possible teams of 5 that could be chosen from 8 candi- dates. How many different teams can he list? 17. The probability that Greg will get a hit the next time at bat is 35%. What is the probability that Greg will not get a hit? 18. a. How many 3-digit numerals can be formed using the digits 2, 3, 4, 5
, 6, 7, and 8 if repetition is not allowed? b. What is the probability that such a 3-digit numeral is greater than 400? 19. a. If a 4-member committee is formed from 3 girls and 6 boys in a club, how many committees can be formed? b. If the members of the committee are chosen at random, find the proba- bility that the committee consists of: (1) 2 girls and 2 boys (2) 1 girl and 3 boys (3) 4 boys c. What is the probability that the fourth member chosen is a girl if the first three are 2 boys and a girl? d. What is the probability that a boy is on the committee? Review Exercises 655 20. From a 52-card deck, 2 cards are drawn at random without replacement. Find the probability of selecting: a. a ten and a king, in that order c. a ten and a king, in either order b. 2 tens d. 2 spades e. a king as the second card if the king of hearts is the first card selected. 21. If a card is drawn at random from a 52-card deck, find the probability that the card is: a. an eight or a queen b. an eight or a club c. red or an eight e. red and a club d. not a club f. not an eight or not a club 22. From an urn, 3 marbles are drawn at random with no replacement. Find the probability that the 3 marbles are the same color if the urn contains, in each case, the given marbles: a. 3 red and 2 white c. 3 red and 4 blue e. 10 blue b. 2 red and 2 blue d. 4 white and 5 blue f. 2 red, 1 white, 3 blue 23. From a class of girls and boys, the probability that one student chosen at random to answer a question will be a girl is . If four boys leave the class, the probability that a student chosen at random to answer a question will be a girl is . How many boys and girls are there in the class before the four boys leave? 2 5 1 3 24. A committee of 5 is to be chosen from 4 men and 3 women. a. How many different 5-person committees are possible? b. Find the probability that the committee includes: (1) 2 men and 3 women (3) at least 2 women (2) 3 men and 2 women (4) all women c. In how many ways can this 5-person committee select a chairperson and a secretary? d. If Hilary and Helene are on the committee, what is the probability that one is selected as chairperson and the other as secretary? 25. Assume that P(male) P(female). In a family of three children, what is the probability that all three children are of the same gender? 26. a. Let n(S) the number of five-letter words that can be formed using the letters in the word RADIO. Find n(S). b. Let n(E) the number of five-letter words that can be formed using the letters in the word RADIO, in which the first letter is a vowel Find n(E). 656 Probability c. If the letters in RADIO are rearranged at random, find the probability that the first letter is a vowel by using the answers to parts a and b, d. Find the answer to part c by a more direct method. In 27–32, if the letters in each given word are rearranged at random, use two different methods to find the probability that the first letter in the word is A. 27. RADAR 28. CANVAS 29. AZALEA 30. AA 31. DEFINE 32. CANOE 33. Twenty-four women and eighteen men are standing in a ticket line. What is the probability that the first five persons in line are women? (The answer need not be simplified.) 34. A four-digit code consists of numbers selected from the set of even digits: 0, 2, 4, 6, 8. No digit is used more than once in any code and the code can begin with 0. What is the probability that the code is a number less than 4,000? 35. In a class there are 4 more boys than girls. A student from the class is chosen at random. The probability that the student is a boy is . How many girls and how many boys are there in the class? 3 5 36. The number of seniors in the chess club is 2 less than twice the number of juniors, and the number of sophomores is 7 more than the number of juniors. If one person is selected at random to represent the chess club at a 2 tournament, the probability that a senior is chosen is . Find the number of 5 students from each class who are members of the club. 37. In a dish, Annie has 16 plain chocolates and 34 candy-coated chocolates, of which 4 are blue, 12 are purple, 15 are green, and 3 are gray. a. What is the probability that Annie will randomly choose a plain choco- late followed by a blue chocolate? b. Annie eats 2 green chocolates and then passes the dish to Bob. What is the probability that he will randomly choose a purple chocolate? 38. Find the probability that the next patient of the doctor described in the chapter opener on page 575 will need either a flu shot or a pneumonia shot. Exploration Rachael had a box of disks, all the same size and shape. She removed 20 disks from the box, marked them, and then returned them to the box. After mixing the marked disks with the others in the box, she removed a handful of disks and recorded the total number of disks and the number of marked disks. Then she Cumulative Review 657 returned the disks to the box, mixed them, and removed another handful. She repeated this last step eight times. The chart below shows her results. Disks Marked Disks 12` 3 10 4 15 5 11 4 8 3 10 2 7 2 12 5 9 3 13 5 a. Use the data to estimate the number of disks in the box. b. Repeat Rachael’s experiment using a box containing an unknown number of disks. Compare your estimate with the actual number of disks in the box. c. Explain how this procedure could be used to estimate the number of fish in a pond. CUMULATIVE REVIEW CHAPTERS 1–15 Part I Answer all questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. 1. Which of the following is a rational number? (1) 169 " (2) 12 " (3) 9 1 4 " (4) p 2. The area of a trapezoid is 21 square inches and the measure of its height is 6.0 inches. The sum of the lengths of the bases is (1) 3.5 inches (3) 7.0 inches (2) 5.3 inches 3. A point on the line whose equation is y 3x 1 is (2) (1, –2) (1) (4, 1) (3) (2, 1) (4) 14 inches (4) (1, 2) 4. Which of the following is the graph of an exponential function? (1) (2) y y x x (3) y (4) y x x 658 Probability 5. The height of a cone is 12.0 centimeters and the radius of its base is 2.30 centimeters. What is the volume of the cone to three significant digits? (1) 52.2 square centimeters (2) 52.3 square centimeters (3) 66.5 square centimeters (4) 66.6 square centimeters 6. If 12 3(x 1) 7x, the value of x is (1) 1.5 (2) 4.5 (3) 0.9 (4) 3.75 7. The product of 4x–2 and 5x3 is (1) 20x–6 (2) 20x (3) 9x–6 (4) 9x 8. There are 12 members of the basketball team. At each game, the coach selects a group of five team members to start the game. For how many games could the coach make different selections? (1) 12! (2) 5! (3) 12! 5! (4) 12! 7! ? 5! 9. Which of the following is an example of direct variation? (1) the area of any square compared to the length of its side (2) the perimeter of any square compared to the length of its side (3) the time it takes to drive 500 miles compared to the speed (4) the temperature compared to the time of day 10. Which of the following intervals represents the solution set of the inequal- ity (1) (3, 7) 23 # 2x11 , 7 ? (2) (2, 3) (3) [–3, 7) (4) [–2, 3) Part II Answer all questions in this part. Each correct answer will receive 2 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 11. The vertices of parallelogram ABCD are A(0, 0), B(7, 0), (12, 8), and D(5, 8). Find, to the nearest degree the measure of DAB. 12. Marty bought 5 pounds of apples for 98 cents a pound. A week later, she bought 7 pounds of apples for 74 cents a pound. What was the average price per pound that Marty paid for apples? Part III Answer all questions in this part. Each correct answer will receive 3 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 13. Each student in an international study group speaks English, Japanese, or Spanish. Of these students, 100 speak English, 50 speak Japanese, 100 speak Spanish, 45 speak English and Spanish, 10 speak English and Cumulative Review 659 Japanese, 13 speak Spanish and Japanese, and 5 speak all three languages. How many students do not speak English? 14. Abe, Brian, and Carmela share the responsibility of caring for the family pets. During a seven-day week, Abe and Brian each take three days and Carmela the other one. In how many different orders can the days of a week be assigned? Part IV Answer all questions in this part. Each correct answer will receive 4 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 15. Mrs. Martinez is buying a sweater that is on sale for 25% off of the origi- nal price. She has a coupon that gives her an additional 20% off of the sale price. Her final purchase price is what percent of the original price? 16. The area of a rectangular parcel of land is 720 square meters. The length of the land is 4 meters less than twice the width. a. Write an equation that can be used to find the dimensions of the land. b. Solve the equation written in part a to find the dimensions of the land. STATISTICS Every four years, each major political party in the United States holds a convention to select the party’s nominee for President of the United States. Before these conventions are held, each candidate assembles a staff whose job is to plan a successful campaign.This plan relies heavily on statistics: on the collection and organization of data, on the results of opinion polls, and on information about the factors that influence the way people vote. At the same time, newspaper reporters and television commentators assemble other d
ata to keep the public informed on the progress of the candidates. Election campaigns are just one example of the use of statistics to organize data in a way that enables us to use available information to evaluate the current situation and to plan for the future. CHAPTER 16 CHAPTER TABLE OF CONTENTS 16-1 Collecting Data 16-2 Organizing Data 16-3 The Histogram 16-4 The Mean, the Median, and the Mode 16-5 Measures of Central Tendency and Grouped Data 16-6 Quartiles, Percentiles, and Cumulative Frequency 16-7 Bivariate Statistics Chapter Summary Vocabulary Review Exercises Cumulative Review 660 16-1 COLLECTING DATA Collecting Data 661 In our daily lives, we often deal with problems that involve many related items of numerical information called data. For example, in the daily newspaper we can find data dealing with sports, with business, with politics, or with the weather. Statistics is the study of numerical data. There are three typical steps in a statistical study: STEP 1. The collection of data. STEP 2. The organization of these data into tables, charts, and graphs. STEP 3. The drawing of conclusions from an analysis of these data. When these three steps, which describe and summarize the formation and use of a set of data, are included in a statistical study, the study is often called descriptive statistics. You will study these steps in this first course. In some cases, a fourth step, in which the analyzed data are used to predict trends and future events, is added. Data can be either qualitative or quantitative. For example, a restaurant may ask customers to rate the meal that was served as excellent, very good, good, fair, or poor. This is a qualitative evaluation. Or the restaurant may wish to make a record of each customer tip at different times of the day. This is a quantitative evaluation, which lends itself more readily to further statistical analysis. Data can be collected in a number of ways, including the following: 1. A written questionnaire or list of questions that a person can answer by checking one of several categories or supplying written responses. Categories to be checked may be either qualitative or quantitative. Written responses are usually qualitative. 2. An interview, either in person or by telephone, in which answers are given verbally and responses are recorded by the person asking the questions. Verbal answers are usually qualitative. 3. A log or a diary, such as a hospital chart or an hourly recording of the out- door temperature, in which a person records information on a regular basis. This type of information is usually quantitative. Note: Not all numerical data are quantitative data. For instance, a researcher wishes to investigate the eye color of the population of a certain island. The researcher assigns “blue” to 0, “black” to 1, “brown” to 2, and so on. The resulting data, although numerical, are qualitative since it represents eye color and the assignment was arbitrary. 662 Statistics Sampling A statistical study may be useful in situations such as the following: 1. A doctor wants to know how effective a new medicine will be in curing a disease. 2. A quality-control team wants to know the expected life span of flashlight batteries made by its company. 3. A company advertising on television wants to know the most frequently watched TV shows so that its ads will be seen by the greatest number of people. When a statistical study is conducted, it is not always possible to obtain information about every person, object, or situation to which the study applies. Unlike a census, in which every person is counted, some statistical studies use only a sample, or portion, of the items being investigated. To find effective medicines, pharmaceutical companies usually conduct tests in which a sample, or small group, of the patients having the disease under study receive the medicine. If the manufacturer of flashlight batteries tested the life span of every battery made, the warehouse would soon be filled with dead batteries. The manufacturer tests only a sample of the batteries to determine their average life span. An advertiser cannot contact every person owning a TV set to determine which shows are being watched. Instead, the advertiser studies TV ratings released by a firm that conducts polls based on a small sample of TV viewers. For any statistical study, whether based on a census or a sample, to be useful, data must be collected carefully and correctly. Poorly designed sampling techniques result in bias, that is, the tendency to favor a selection of certain members of the population which, in turn, produces unreliable conclusions. Techniques of Sampling We must be careful when choosing samples: 1. The sample must be fair or unbiased, to reflect the entire population being studied. To know what an apple pie tastes like, it is not necessary to eat the entire pie. Eating a sample, such as a piece of the apple pie, would be a fair way of knowing how the pie tastes. However, eating only the crust or only the apples would be an unfair sample that would not tell us what the entire pie tastes like. 2. The sample must contain a reasonable number of the items being tested or counted. If a medicine is generally effective, it must work for many people. The sample tested cannot include only one or two patients. Similarly, the manufacturer of flashlight batteries cannot make claims based on testing five or 10 batteries. A better sample might include 100 batteries. Collecting Data 663 3. Patterns of sampling or random selection should be employed in a study. The manufacturer of flashlight batteries might test every 1,000th battery to come off the assembly line. Or, the batteries to be tested might be selected at random. These techniques will help to make the sample, or the small group, representative of the entire population of items being studied. From the study of the small group, reasonable conclusions can be drawn about the entire group. EXAMPLE 1 To determine which television programs are the most popular in a large city, a poll is conducted by selecting people at random at a street corner and interviewing them. Outside of which location would the interviewer be most likely to find an unbiased sample? (1) a ball park (2) a concert hall (3) a supermarket Solution People outside a ball park may be going to a game or purchasing tickets for a game in the future; this sample may be biased in favor of sports programs. Similarly, those outside a concert hall may favor musical or cultural programs. The best (that is, the fairest) sample or cross section of people for the three choices given would probably be found outside a supermarket. Answer (3) Experimental Design So far we have focused on data collection. In an experiment, a researcher imposes a treatment on one or more groups. The treatment group receives the treatment, while the control group does not. For instance, consider an experiment of a new medicine for weight loss. Only the treatment group is given the medicine, and conditions are kept as similar as possible for both groups. In particular, both groups are given the same diet and exercise. Also, both groups are of large enough size and are chosen such that they are comprised of representative samples of the general population. However, it is often not enough to have just a control group and a treatment group. The researcher must keep in mind that people often tend to respond to any treatment. This is called the placebo effect. In such cases, subjects would report that the treatment worked even when it is ineffective. To account for the placebo effect, researchers use a group that is given a placebo or a dummy treatment. Of course, subjects in the experimental and placebo groups should not know which group they are in (otherwise, psychology will again confound the results). The practice of not letting people know whether or not they have been given the real treatment is called blinding, and experiments using blinding are said to be single-blind experiments. When the variable of interest is hard to measure or 664 Statistics define, double-blind experiments are needed. For example, consider an experiment measuring the effectiveness of a drug for attention deficit disorder. The problem is that “attention deficiency” is difficult to define, and so a researcher with a bias towards a particular conclusion may interpret the results of the placebo and treatment groups differently. To avoid such problems, the researchers working directly with the test subjects are not told which group a subject belongs to. Interpreting Graphs of Data Oftentimes embellishments to graphs distort the perception of the data, and so you must exercise care when interpreting graphs of data. 1. Two- and three-dimensional figures. As the graph on the right shows, graphs using two- or three-dimensional figures can distort small changes in the data. The lengths show the decrease in crime, but since our eyes tend to focus on the areas, the total decrease appears greater than it really is. The reason is because linear changes are increased in higher dimensions. For instance, if a length doubles in value, say from x to 2x, the area of a square with sides of length x will increase by x2 → (2x)2 4x2, CRIME RATE IN THE U.S. 1990 = 14,475,613 POLICE POLICE 1995 = 13,862,727 POLICE 2000 = 11,876,669 a four-fold increase. Similarly, the volume of a cube with edges of length x will increase by x3 → (2x)3 8x3, an eight-fold increase! 2. Horizontal and vertical scales. The scales used on the vertical and horizontal axes can exaggerate, diminish, and/or distort the nature of the change in the data. For instance, in the graph on the left of the following page, the total change in weight is less than a pound, which is negligible for an adult human. However, the scale used apparently amplifies this amount. While on the right, the unequal horizontal scale makes the population growth appear linear. AVERAGE WEIGHT OF SUBJECTS OVER 6-MONTH P
ERIOD 200.0 199.9 199.8 199.7 199.6 199.5 199.4 199.3 199.2 199.1 199. Collecting Data 665 POPULATION OF ANYTOWN, U.S ( 500 400 300 200 100 Month Year EXERCISES Writing About Mathematics 1. A census attempts to count every person. Explain why a census may be unreliable. 2. A sample of a new soap powder was left at each home in a small town. The occupants were asked to try the powder and return a questionnaire evaluating the product. To encourage the return of the questionnaire, the company promised to send a coupon for a free box of the soap powder to each person who responded. Do you think that the questionnaires that were returned represent a fair sample of all of the persons who tried the soap? Explain why or why not. Developing Skills In 3–10, determine if each variable is quantitative or qualitative. 3. Political affiliation 4. Opinions of students on a new music album 5. SAT scores 6. Nationality 7. Cholesterol level 8. Class membership (freshman, sophomore, etc.) 9. Height 10. Number of times the word “alligator” is used in an essay. In 11–18, in each case a sample of students is to be selected and the height of each student is to be measured to determine the average height of a student in high school. For each sample: a. Tell whether the sample is biased or unbiased. b. If the sample is biased, explain how this might affect the outcome of the survey. 11. The basketball team 13. All 14-year-old students 12. The senior class 14. All girls 666 Statistics 15. Every tenth person selected from an alphabetical list of all students 16. Every fifth person selected from an alphabetical list of all boys 17. The first three students who report to the nurse on Monday 18. The first three students who enter each homeroom on Tuesday In 19–24, in each case the Student Organization wishes to interview a sample of students to determine the general interests of the student body. Two questions will be asked: “Do you want more pep rallies for sports events? Do you want more dances?” For each location, tell whether the Student Organization would find an unbiased sample at that place. If the sample is biased, explain how this might influence the result of the survey. 19. The gym, after a game 20. The library 21. The lunchroom 22. The cheerleaders’ meeting 23. The next meeting of the Junior Prom committee 24. A homeroom section chosen at random 25. A statistical study is useful when reliable data are collected. At times, however, people may exaggerate or lie when answering a question. Of the six questions that follow, find the three questions that will most probably produce the largest number of unreliable answers. (1) What is your height? (3) What is your age? (5) What is your income? (2) What is your weight? (4) In which state do you live? (6) How many people are in your family? 26. List the three steps necessary to conduct a statistical study. 27. Explain why the graph below is misleading. SUMMER OLYMPIC GAMES CHAMPIONS 100-METER RACE 1988 Carl Lewis (USA) 9.92 sec 1992 Linford Christie (GBR) 9.96 sec 1996 Donovan Bailey (CAN) 9.84 sec 2000 Maurice Green (USA) 9.87 sec 2004 Justin Gatlin (USA) 9.85 sec Organizing Data 667 28. Investigators at the University of Kalamazoo were interested in determining whether or not women can determine a man’s preference for children based on the way that he looks. Researchers asked a group of 20 male volunteers whether or not they liked children. The researchers then showed photographs of the faces of the men to a group of 10 female volunteers and asked them to pick out which men they thought liked children. The women correctly identified over 90% of the men who said they liked children. The researchers concluded that women could identify a man’s preference for children based on the way that he looks. Identify potential problems with this experiment. Hands-On Activity Collect quantitative data for a statistical study. 1. Decide the topic of the study. What data will you collect? 2. Decide how the data will be collected. What will be the source(s) of that data? a. Questionnaires b. Personal interviews c. Telephone interviews d. Published materials from sources such as almanacs or newspapers. 3. Collect the data. How many values are necessary to obtain reliable information? Keep the data that you collect to use as you learn more about statistical studies. 16-2 ORGANIZING DATA Data are often collected in an unorganized and random manner. For example, a teacher recorded the number of days each of 25 students in her class was absent last month. These absences were as follows: 0, 3, 1, 0, 4, 2, 1, 3, 5, 0, 2, 0, 0, 0, 4, 0, 1, 1, 2, 1, 0, 7, 3, 1, 0 How many students were absent fewer than 2 days? What was the number of days for which the most students were absent? How many students were absent more than 5 days? To answer questions such as these, we find it helpful to organize the data. One method of organizing data is to write it as an ordered list. In order from least to greatest, the absences become: 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 5, 7 We can immediately observe certain facts from this ordered list: more students were absent 0 days than any other number of days, the same number of students were absent 5 and 7 days. However, for more a quantitative analysis, it is useful to make a table. 668 Statistics Preparing a Table In the left column of the accompanying table, we list the data values (in this case the number of absences) in order. We start with the largest number, 7, at the top and go down to the smallest number, 0. For each occurrence of a data value, we place a tally mark, \|, in the row for that number. For example, the first data value in the teacher’s list is 0, so we place a tally in the 0 row; the second value is 3, so we place a tally in the 3 row. We follow this procedure until a tally for each data value is recorded in the proper row. To simplify counting, we write every fifth tally as a diagonal mark passing through the first four tallies: . Once the data have been organized, we can count the number of tally marks in each row and add a column for the frequency, that is, the number of times that a value occurs in the set of data. When there are no tally marks in a row, as for the row showing 6 absences, the frequency is 0.The sum of all of the frequencies is called the total frequency. In this case, the total frequency is 25. (It is always wise to check the total frequency to be sure that no data value was overlooked or duplicated in tallying.) From the table, called a frequency distribution table, it is now easy to see that 15 students were absent fewer than 2 days, that more students were absent 0 days (9) than any other number of days, and that 1 student was absent more than 5 days. Grouped Data Absences Tally 7 6 5 4 3 2 1 0 Absences Tally Frequency Total frequency 25 A teacher marked a set of 32 test papers. The grades or scores earned by the students were as follows: 90, 85, 74, 86, 65, 62, 100, 95, 77, 82, 50, 83, 77, 93, 73, 72, 98, 66, 45, 100, 50, 89, 78, 70, 75, 95, 80, 78, 83, 81, 72, 75 Organizing Data 669 Because of the large number of different scores, it is convenient to organize these data into groups or intervals, which must be equal in size. Here we will use six intervals: 41–50, 51–60, 61–70, 71–80, 81–90, 91–100. Each interval has a length of 10, found by subtracting the starting point of an interval from the starting point of the next higher interval. Interval Tally Frequency For each test score, we now place a tally mark in the row for the interval that includes that score. For example, the first two scores in the list above are 90 and 85, so we place two tally marks in the interval 81–90. The next score is 74, so we place a tally mark in the interval 71–80. When all of the scores have been tallied, we write the frequency for each interval. This table, containing a set of intervals and the corresponding frequency for each interval, is an example of grouped data. 91–100 81–90 71–80 61–70 51–60 41–50 6 8 11 4 0 3 When unorganized data are grouped into intervals, we must follow certain rules in setting up the intervals: 1. The intervals must cover the complete range of values. The range is the difference between the highest and lowest values. 2. The intervals must be equal in size. 3. The number of intervals should be between 5 and 15. The use of too many or too few intervals does not make for effective grouping of data. We usually use a large number of intervals, for example, 15, only when we have a very large set of data, such as hundreds of test scores. 4. Every data value to be tallied must fall into one and only one interval. Thus, the intervals should not overlap. When an interval ends with a counting number, the following interval begins with the next counting number. 5. The intervals must be listed in order, either highest to lowest or lowest to highest. 670 Statistics These rules tell us that there are many ways to set up tables, all of them correct, for the same set of data. For example, here is another correct way to group the 32 unorganized test scores given at the beginning of this section. Note that the length of the interval here is 8. Interval Tally Frequency 93–100 85–92 77–84 69–76 61–68 53–60 45–52 6 4 9 7 3 0 3 Constructing a Stem-and-Leaf Diagram Another method of displaying data is called a stem-and-leaf diagram. The stemand-leaf diagram groups the data without losing the individual data values. A group of 30 students were asked to record the length of time, in minutes, spent on math homework yesterday. They reported the following data: 38, 15, 22, 20, 25, 44, 5, 40, 38, 22, 20, 35, 20, 0, 36, 27, 37, 26, 33, 25, 17, 45, 22, 30, 18, 48, 12, 10, 24, 27 To construct a stem-and-leaf diagram for the lengths of time given, we begin by choosing part of the data values to be the stem. Since every score is a one- or two-digit number, we will choose the tens digit as a convenient stem. For the one-digit number
s, 0 and 5, the stem is 0; for the other data values, the stem is 1, 2, 3, or 4. Then the units digit will be the leaf. We construct the diagram as follows: STEP 1. List the stems, starting with 4, under one another Stem Leaf to the left of a vertical line beneath a crossbar. STEP 2. Enter each score by writing its leaf (the units digit) to the right of the vertical line, following the appropriate stem (its tens value). For example, enter 38 by writing 8 to the right of the vertical line, after stem 3. 4 3 2 1 0 Stem Leaf 8 4 3 2 1 0 STEP 3. Add the other scores to the diagram until all are entered. Organizing Data 671 Stem Leaf STEP 4. Arrange the leaves in order after Stem Leaf each stem. STEP 5. Add a key to demonstrate the meaning of each value in the diagram Key: 3 0 30 EXAMPLE 1 The following data consist of the weights, in kilograms, of a group of 30 students: 70, 43, 48, 72, 53, 81, 76, 54, 58, 64, 51, 53, 75, 62, 84, 67, 72, 80, 88, 65, 60, 43, 53, 42, 57, 61, 55, 75, 82, 71 a. Organize the data in a table. Use five intervals starting with 40–49. b. Based on the grouped data, which interval contains the greatest number of students? c. How many students weigh less than 70 kilograms? Solution a. Interval Tally Frequency (number) 80–89 70–79 60–69 50–59 40–49 5 7 6 8 4 b. The interval 50–59 contains the greatest number of students, 8. Answer c. The three lowest intervals, namely 40–49, 50–59, and 60–69, show weights less than 70 kilograms. Add the frequencies in these three intervals: 4 8 6 18 Answer 672 Statistics EXAMPLE 2 Draw a stem-and-leaf diagram for the data in Example 1. Solution Let the tens digit be the stem and the units digit the leaf. (1) Enter the data values in the (2) Arrange the leaves in numerical given order: Stem Leaf 3) Add a key indicating unit of measure: order after each stem: Stem Leaf Key: 5 1 51 kg EXERCISES Writing About Mathematics 1. Of the examples given above, which gives more information about the data: the table or the stem-and-leaf diagram? Explain your answer. 2. A set of data ranges from 2 to 654. What stem can be used for this set of data when drawing a stem-and-leaf diagram? What leaves would be used with this stem? Explain your choices. Developing Skills 3. a. Copy and complete the table to group the data, which represent the heights, in centime- ters, of 36 students: 162, 173, 178, 181, 155, 162, 168, 147, 180, 171, 168, 183, 157, 158, 180, 164, 160, 171, 183, 174, 166, 175, 169, 180, 149, 170, 150, 158, 162, 175, 171, 163, 158, 163, 164, 177 b. Use the grouped data to answer the following questions: (1) How many students are less than 160 centimeters in height? (2) How many students are 160 centimeters or more in height? Interval Tally Frequency 180–189 170–179 160–169 150–159 140–149 (3) Which interval contains the greatest number of students? (4) Which interval contains the least number of students? Organizing Data 673 c. Display the data in a stem-and-leaf diagram. Use the first two digits of the numbers as the stems. d. What is the range of the data? e. How many students are taller than 175 centimeters? 4. a. Copy and complete the table to group the data, which gives the lifespan, in hours, of 50 flashlight batteries: 73, 81, 92, 80, 108, 76, 84, 102, 58, 72, 82, 100, 70, 72, 95, 105, 75, 84, 101, 62, 63, 104, 97, 85, 106, 72, 57, 85, 82, 90, 54, 75, 80, 52, 87, 91, 85, 103, 78, 79, 91, 70, 88, 73, 67, 101, 96, 84, 53, 86 b. Use the grouped data to answer the following questions: (1) How many flashlight batteries lasted for 80 or more hours? Interval Tally Frequency 50–59 60–69 70–79 80–89 90–99 (2) How many flashlight batteries lasted fewer than 80 100–109 hours? (3) Which interval contains the greatest number of batteries? (4) Which interval contains the least number of batteries? c. Display the data in a stem-and-leaf diagram. Use the digits from 5 through 10 as the stems. d. What is the range of the data? e. What is the probability that a battery selected at random lasted more than 100 hours? 5. The following data consist of the hours spent each week watching television, as reported by a group of 38 teenagers: 13, 20, 17, 36, 25, 21, 9, 32, 20, 17, 12, 19, 5, 8, 11, 28, 25, 18, 19, 22, 4, 6, 0, 10, 16, 3, 27, 31, 15, 18, 20, 17, 3, 6, 19, 25, 4, 7 a. Construct a table to group these data, using intervals of 0–4, 5–9, 10–14, 15–19, 20–24, 25–29, 30–34, and 35–39. b. Construct a table to group these data, using intervals of 0–7, 8–15, 16–23, 24–31, and 32–39. c. Display the data in a stem-and-leaf diagram. d. What is the range of the data? e. What is the probability that a teenager, selected at random from this group, spends less than 4 hours watching television each week? 6. The following data show test scores for 30 students: 90, 83, 87, 71, 62, 46, 67, 72, 75, 100, 93, 81, 74, 75, 82, 83, 83, 84, 92, 58, 95, 98, 81, 88, 72, 59, 95, 50, 73, 93 674 Statistics a. Construct a table, using intervals of length 10 starting with 91–100. b. Construct a table, using intervals of length 12 starting with 89–100. c. For the grouped data in part a, which interval contains the greatest number of stu- dents? d. For the grouped data in part b, which interval contains the greatest number of stu- dents? e. Do the answers for parts c and d indicate the same general region of test scores, such as “scores in the eighties”? Explain your answer. 7. For the ungrouped data from Exercise 5, tell why each of the following sets of intervals is not correct for grouping the data. a. Interval b. Interval c. Interval d. Interval 25–38 13–24 0–12 30–39 20–29 10–19 5–9 0–4 32–40 24–32 16–24 8–16 0–8 33–40 25–32 17–24 9–16 1–8 Hands-On Activity Organize the data that you collected in the Hands-On Activity for Section 16-1. 1. Use a stem-and-leaf diagram. a. Decide what will be used as stems. b. Decide what will be used as leaves. c. Construct the diagram. d. Check that the number of leaves in the diagram equals the number of values in the data collected. 2. Use a frequency table. a. How many intervals will be used? b. What will be the length of each interval? c. What will be the starting and ending points of each interval? Check that the intervals do not overlap, are equal in size, and that every value falls into only one interval. d. Tally the data. e. List the frequency for each interval. f. Check that the total frequency equals the number of values in the data collected. 3. Decide which method of organization is better for your data. Explain your choice. Keep your organized data to work with as you learn more about statistics. The Histogram 675 16-3 THE HISTOGRAM In Section 16-2 we organized data by grouping them into intervals of equal length. After the data have been organized, a graph can be used to visualize the intervals and their frequencies. The table below shows the distribution of test scores for 32 students in a class. The data have been organized into six intervals of length 10. Test Scores (Intervals) Frequency (Number of Scores) 91–100 81–90 71–80 61–70 51–60 41–50 6 8 11 4 0 3 We can use a histogram to display the data graphically. A histogram is a vertical bar graph in which each interval is represented by the width of the bar and the frequency of the interval is represented by the height of the bar. The bars are placed next to each other to show that, as one interval ends, the next interval begins 12 11 10 TEST SCORES OF 32 STUDENTS 41–50 51–60 61–70 71–80 81–90 91–100 Test scores (intervals) In the above histogram, the intervals are listed on the horizontal axis in the order of increasing scores, and the frequency scale is shown on the vertical axis. The first bar shows that 3 students had test scores in the interval 41–50. Since no student scored in the interval 51–60, there is no bar for this interval. Then, 4 students scored between 61 and 70; 11 between 71 and 80; 8 between 81 and 90; and 6 between 91 and 100. 676 Statistics Except for an interval having a frequency of 0, the interval 51–60 in this example, there are no gaps between the bars drawn in a histogram. Since the histogram displays the frequency, or number of data values, in each interval, we sometimes call this graph a frequency histogram. A graphing calculator can display a frequency histogram from the data on a frequency distribution table. (1) Clear L1 and L2 with the ClrList function by pressing STAT 4 2nd L1 , 2nd L2 ENTER . (2) Press STAT 1 to edit the lists. L1 will L1 L2 contain the minimum value of each interval. Move the cursor to the first entry position in L1. Type the value and then . Type the next value and press ENTER ENTER then press until all the minimum values of the intervals have been entered. . Repeat this process 91 81 71 61 51 41 ------ ------ 6 8 11 4 0 3 L2(7) = 2 L3 ------ (3) Repeat the process to enter the frequencies that correspond to each inter- val in L2. (4) Clear any functions in the Y= menu. (5) Turn on Plot1 from the STAT PLOT menu, and configure it to graph a histogram. Make sure to also set Xlist to L1 and Freq to L2. ENTER: STAT PLOT ENTER 2nd 1 L1 ENTER 2nd 2nd L2 (6) In the WINDOW menu, accessed by , enter Xmin as 31, WINDOW pressing the length of one interval less than the smallest interval value and Xmax as 110, the length of one interval more than the largest interval value. Enter Xscl as 10, the length of the interval. The Ymin is 0 and Ymax is 12 to be greater than the largest frequency = Ty The Histogram 677 (7) Press GRAPH to draw the graph. We can P 1 : L 1 , L 2 view the frequency (n) associated with each interval by pressing TRACE . Use the left and right arrow keys to move between intervals EXAMPLE 1 The table on the right represents the number of miles per gallon of gasoline obtained by 40 drivers of compact cars. Construct a frequency histogram based on the data. Solution (1) Draw and label a vertical scale to show fre- quencies. The scale starts at 0 and increases to include the highest frequency in any one interval (here, it is 11). (2) Draw and la
bel intervals of equal length on a horizontal scale. Label the horizontal scale, telling what the numbers represent. Interval Frequency 16–19 20–23 24–27 28–31 32–35 36–39 40–43 5 11 8 5 7 3 1 (3) Draw the bars vertically, leaving no gaps between the intervals 12 11 10 16–19 20–23 24–27 28–31 32–35 36–39 40–43 Mileage (miles per gallon) for compact cars 678 Statistics Calculator Solution (1) Press 1 STAT to edit the lists and enter the minimum value of each interval into L1: 16, 20, 24, 28, 32, 36, 40. Use the arrow key to move into L2, and enter the corresponding frequencies: 5, 11, 8, 5, 7, 3, 1. (2) Go to the STAT PLOT menu and choose Plot1 by pressing 2nd STAT PLOT 1 . Move the cursor with the arrow keys, then press ENTER to select On and the histogram. Type 2nd L1 into Xlist and 2nd L2 into Freq. (3) Set the Window. Each interval has length 4, so set Xmin to 12 (4 less than the smallest interval value), Xmax to 44 (4 more than the largest interval value), and Xscl to 4. Make Ymin 0 and Ymax 12 to be greater than the largest frequency4) Draw the graph by pressing GRAPH . Press TRACE and use the right and left arrow keys to show the frequencies, the heights of the vertical bars EXAMPLE 2 Use the histogram constructed in Example 1 to answer the following questions: a. In what interval is the greatest frequency found? b. What is the number (or frequency) of cars reporting mileages between 28 and 31 miles per gallon? c. For what interval are the fewest cars reported? d. How many of the cars reported mileage greater than 31 miles per gallon? e. What percent of the cars reported mileage from 24 to 27 miles per gallon? Solution a. 20–23 b. 5 c. 40–43 d. Add the frequencies for the three highest intervals. The interval 32–35 has a frequency of 7; 36–39 a frequency of 3; 40–43 a frequency of 1: 7 3 1 11. e. The interval 24–27 has a frequency of 8. The total frequency for this survey is 40. 8 40 5 1 5 20%. Answers a. 20–23 b. 5 c. 40–43 d. 11 e. 20% The Histogram 679 EXERCISES Writing About Mathematics 1. Compare a stem-and-leaf diagram with a frequency histogram. In what ways are they alike and in what ways are they different? 2. If the data in Example 1 had been grouped into intervals with a lowest interval of 16–20, what would be the endpoints for the other intervals? Would you be able to determine the frequency for each new interval? Explain why or why not. Developing Skills In 3–5, in each case, construct a frequency histogram for the grouped data. Use graph paper or a graphing calculator. 3. Interval Frequency 4. Interval Frequency 5. Interval Frequency 91–100 81–90 71–80 61–70 51–60 5 9 7 2 4 30–34 25–29 20–24 15–19 10–14 5–9 5 10 10 12 0 2 1–3 4–6 7–9 10–12 13–15 16–18 24 30 28 41 19 8 6. For the table of grouped data given in Exercise 5, answer the following questions: a. What is the total frequency in the table? b. What interval contains the greatest frequency? c. The number of data values reported for the interval 4–6 is what percent of the total number of data values? d. How many data values from 10 through 18 were reported? Applying Skills 7. Towering Ted McGurn is the star of the school’s basketball team. The number of points scored by Ted in his last 20 games are as follows: 36, 32, 28, 30, 33, 36, 24, 33, 29, 30, 30, 25, 34, 36, 34, 31, 36, 29, 30, 34 a. Copy and complete the table to find the frequency for each interval. b. Construct a frequency histogram based on the data found in part a. c. Which interval contains the greatest frequency? d. In how many games did Ted score 32 or more points? e. In what percent of these 20 games did Ted score fewer than 26 points? Interval Tally Frequency 35–37 32–34 29–31 26–28 23–25 680 Statistics 8. Thirty students on the track team were timed in the 200-meter dash. Each student’s time was recorded to the nearest tenth of a second. Their times are as follows: 29.3, 31.2, 28.5, 37.6, 30.9, 26.0, 32.4, 31.8, 36.6, 35.0, 38.0, 37.0, 22.8, 35.2, 35.8, 37.7, 38.1, 34.0, 34.1, 28.8, 29.6, 26.9, 36.9, 39.6, 29.9, 30.0, 36.0, 36.1, 38.2, 37.8 a. Copy and complete the table to find the frequency in each interval. b. Construct a frequency histogram for the given data. c. Determine the number of students who ran the 200- meter dash in under 29 seconds. d. If a student on the track team is chosen at random, what is the probability that he or she ran the 200meter dash in fewer than 29 seconds? Interval Tally Frequency 37.0–40.9 33.0–36.9 29.0–32.9 25.0–28.9 21.0–24.9 Hands-On-Activity Construct a histogram to display the data that you collected and organized in the Hands-On Activities for Sections 16-1 and 16-2. 1. Draw the histogram on graph paper. 2. Follow the steps in this section to display the histogram on a graphing calculator. 16-4 THE MEAN, THE MEDIAN, AND THE MODE In a statistical study, after we have collected the data, organized them, and presented them graphically, we then analyze the data and summarize our findings. To do this, we often look for a representative, or typical, score. Averages in Arithmetic In your previous study of arithmetic, you learned how to find the average of two or more numbers. For example, to find the average of 17, 25, and 30: STEP 1. Add these three numbers: 17 25 30 72. STEP 2. Divide this sum by 3 since there are three numbers: 72 3 24. The average of the three numbers is 24. Averages in Statistics The word average has many different meanings. For example, there is an average of test scores, a batting average, the average television viewer, an average intelligence, and the average size of a family. These averages are not found by The Mean, the Median, and the Mode 681 the same rule or procedure. Because of this confusion, in statistics we speak of measures of central tendency. These measures are numbers that usually fall somewhere in the center of a set of organized data. We will discuss three measures of central tendency: the mean, the median, and the mode. The Mean In statistics, the arithmetic average previously studied is called the mean of a set of numbers. It is also called the arithmetic mean or the numerical average. The mean is found in the same way as the arithmetic average is found. Procedure To find the mean of a set of n numbers, add the numbers and divide the sum by n.The symbol used for the mean is x–. For example, if Ralph’s grades on five tests in science during this marking period are 93, 80, 86, 72, and 94, he can find the mean of his test grades as follows: STEP 1. Add the five data values: 93 80 86 72 94 425. STEP 2. Divide this sum by 5, the number of tests: 425 5 85. The mean (arithmetic average) is 85. Let us consider another example. In a car wash, there are seven employees whose ages are 17, 19, 20, 17, 46, 17, and 18. What is the mean of the ages of these employees? Here, we add the seven ages to get a sum of 154. Then, 154 7 22. While the mean age of 22 is the correct answer, this measure does not truly represent the data. Only one person is older than 22, while six people are under 22. For this reason, we will look at another measure of central tendency that will eliminate the extreme case (the employee aged 46) that is distorting the data. The Median The median is the middle value for a set of data arranged in numerical order. For example, the median of the ages 17, 19, 20, 17, 46, 17, and 18 for the car-wash employees can be found in the following manner: STEP 1. Arrange the ages in numerical order: STEP 2. Find the middle number: 17, 17, 17, 18, 19, 20, 46 17, 17, 17, 18, 19, 20, 46 ↑ The median is 18 because there are three ages less than 18 and three ages greater than 18. The median, 18, is a better indication of the typical age of the 682 Statistics employees than the mean, 22, because there are so many younger people working at the car wash. Now, let us suppose that one of the car-wash employees has a birthday, and her age changes from 17 to 18. What is now the median age? STEP 1. Arrange the ages in numerical order: STEP 2. Find the middle number: 17, 17, 18, 18, 19, 20, 46 17, 17, 18, 18, 19, 20, 46 ↑ The median, or middle value, is again 18. We can no longer say that there are three ages less than 18 because one of the three youngest employees is now 18. We can say, however, that: 1. the median is 18 because there are three ages less than or equal to 18 and three ages greater than or equal to 18; or 2. the median is 18 because, when the data values are arranged in numerical order, there are three values below this median, or middle number, and three values above it. Recently, the car wash hired a new employee whose age is 21. The data now include eight ages, an even number, so there is no middle value. What is now the median age? STEP 1. Arrange the ages in numerical order: STEP 2. There is no single middle number. 17, 17, 18, 18, 19, 20, 21, 46 17, 17, 18, 18, 19, 20, 21, 46 ↑ ↑ 2 5 181 2 18 1 19 Find the two middle numbers: STEP 3. Find the mean (arithmetic average) of the two middle numbers: 181 The median is now 2 and four ages greater than . There are four ages less than this center value of 181 2 . 181 2 Procedure To find the median of a set of n numbers: 1. Arrange the numbers in numerical order. 2. If n is odd, find the middle number.This number is the median. 3. If n is even, find the mean (arithmetic average) of the two middle numbers. This average is the median. The Mode The mode is the data value that appears most often in a given set of data. It is usually best to arrange the data in numerical order before finding the mode. The Mean, the Median, and the Mode 683 Let us consider some examples of finding the mode: 1. The ages of employees in a car wash are 17, 17, 17, 18, 19, 20, 46. The mode, which is the number appearing most often, is 17. 2. The number of hours each of six students spent reading a book are 6, 6, 8, 11, 14, 21. The mode, or number appearing most frequently, is 6. In this case, however, the mode is not a useful measure of central tendency. A better indication is give
n by the mean or the median. 3. The number of photographs printed from each of Renee’s last six rolls of film are 8, 8, 9, 11, 11, and 12. Since 8 appears twice and 11 appears twice, we say that there are two modes: 8 and 11. We do not take the average of these two numbers since the mode tells us where most of the scores appear. We simply report both numbers. When two modes appear within a set of data, we say that the data are bimodal. 4. The number of people living in each house on Meryl’s street are 2, 2, 3, 3, 4, 5, 5, 6, 8. These data have three modes: 2, 3, and 5. 5. Ralph’s test scores in science are 72, 80, 86, 93, and 94. Here, every number appears the same number of times, once. Since no number appears more often than the others, we define such data as having no mode. Procedure To find the mode for a set of data, find the number or numbers that occur most often. 1. If one number appears most often in the data, that number is the mode. 2. If two or more numbers appear more often than all other data values, and these numbers appear with the same frequency, then each of these numbers is a mode. 3. If each number in a set of data occurs with the same frequency, there is no mode. KEEP IN MIND Three measures of central tendency are: 1. The mean, or mean average, found by adding n data values and then divid- ing the sum by n. 2. The median, or middle score, found when the data are arranged in numeri- cal order. 3. The mode, or the value that appears most often. A graphing calculator can be used to arrange the data in numerical order and to find the mean and the median. The calculator solution in the following example lists the keystrokes needed to do this. 684 Statistics EXAMPLE 1 The weights, in pounds, of five players on the basketball team are 195, 168, 174, 182, and 181. Find the average weight of a player on this team. Solution The word average, by itself, indicates the mean. Therefore: (1) Add the five weights: 195 168 174 182 181 900. (2) Divide the sum by 5, the number of players: 900 5 180. Calculator Solution Enter the data into list L1. Then use 1-Var Stats from the STAT CALC menu to display information about this set of data. ENTER: STAT ENTER ENTER DISPLAY< n = 5 The first value given is , the mean. x2 Answer 180 pounds The second value given is Σx 900. The symbol Σ represents a sum and Σx 900 can be read as “The sum of the values of x is 900.” The list shows other values related to this set of data. The arrow at the bottom of the display indicates that more entries follow what appears on the screen. These can be displayed by pressing the down arrow. One of these is the median (Med 181). The display also shows that there are 5 data values (n = 5). Others we will use in later sections in this chapter and in more advanced courses. EXAMPLE 2 Renaldo has marks of 75, 82, and 90 on three mathematics tests. What mark must he obtain on the next test to have an average of exactly 85 for the four math tests? The Mean, the Median, and the Mode 685 Solution The word average, by itself, indicates the mean. Let x Renaldo’s mark on the fourth test. The sum of the four test marks divided by 4 is 85. 75 1 82 1 90 1 x 4 5 85 247 1 x 4 5 85 247 1 x 5 340 x 5 93 Check 75 1 82 1 90 1 93 4 5? 85 340 4 5? 85 85 5? 85 ✔ Answer Renaldo must obtain a mark of 93 on his fourth math test. EXAMPLE 3 Find the median for each distribution. a. 4, 2, 5, 5, 1 b. 9, 8, 8, 7, 4, 3, 3, 2, 0, 0 Solution a. Arrange the data in numerical order: The median is the middle value: 1, 2, 4, 5, 5 1, 2, 4, 5, 5 ↑ Answer median 4 b. Since there is an even number of values, there are two middle values. Find the mean (average) of these two middle values: 9, 8, 8, 7, 4, 3, 3, 2, 0 31 2 Answer median or 3.5 31 2 EXAMPLE 4 Find the mode for each distribution. a. 2, 9, 3, 7, 3 b. 3, 4, 5, 4, 3, 7, 2 c. 1, 2, 3, 4, 5, 6, 7 Solution a. Arrange the data in numerical order: 2, 3, 3, 7, 9. The mode, or most frequent value, is 3. b. Arrange the data in numerical order: 2, 3, 3, 4, 4, 5, 7. Both 3 and 4 appear twice. There are two modes. c. Every value occurs the same number of times in the data set 1, 2, 3, 4, 5, 6, 7. There is no mode. Answers a. The mode is 3. b. The modes are 3 and 4. c. There is no mode. 686 Statistics Linear Transformations of Data Multiplying each data value by the same constant or adding the same constant to each data value is an example of a linear transformation of a set of data. Let us start by examining additive transformations. For instance, consider the data 2, 2, 3, 4, 5. If 10 is added to each data value, the data set becomes: 12, 12, 13, 14, 15 Notice that every measure of central tendency has been shifted to the right by 10 units old mean old median 3 old mode 2 5 3.2 12 1 12 1 13 1 14 1 15 5 new mean new median 13 new mode 12 5 13.2 In fact, this result is valid for any additive transformation of a data set. In general: If x–, d, and o are the mean, median, and mode of a set of data and the constant c is added to each data value, then x– c, d c, and o c are the mean, median, and mode of the transformed data. It can be also shown that a similar result holds for multiplicative transfor- mations, that is: If x–, d, and o are the mean, median, and mode of a set of data and each data value is multiplied by the nonzero constant c, then cx–, cd, and co are the mean, median, and mode of the transformed data. EXAMPLE 5 In Ms. Huan’s Algebra class, the average score on the most recent quiz was 65. Being in a generous mood, Ms. Huan decided to curve the quiz by adding 10 points to each quiz score. What will be the new average score for the class? Answer 65 10 75 points EXERCISES Writing About Mathematics 1. On her first two math tests, Rene received grades of 67 and 79. Her mean (average) grade for these two tests was 73. On her third test she received a grade of 91. Rene found the mean of 73 and 91 and said that her mean for the three tests was 82. Do you agree with Rene? Explain why or why not. The Mean, the Median, and the Mode 687 2. Carlos said that when there are n numbers in a set of data and n is an odd number, the th number when the data are arranged in order. Do you agree with n 1 1 median is the 2 Carlos? Explain why or why not. Developing Skills 3. For each set of data, find the mean. a. 7, 3, 5, 11, 9 23 53 4 4 71 2 c. , , , 51 2 41 2 4. Find the median for each set of data. , b. 22, 38, 18, 14, 22, 30 d. 1.00, 0.01, 1.10, 0.12, 1.00, 1.03 a. 1, 2, 5, 3, 4 c. 3, 8, 12, 7, 1, 0, 4 e. 3.2, 8.7, 1.4 b. 2, 9, 2, 9, 7 d. 80, 83, 97, 79, 25 f. 2.00, 0.20, 2.20, 0.02, 2.02 g. 21, 24, 23, 22, 20, 24, 23, 21, 22, 23 h. 5, 7, 9, 3, 8, 7, 5, 6 5. What is the median for the digits 1, 2, 3, . . . , 9? 6. What is the median for the counting numbers from 1 through 100? 7. Find the mode for each distribution. a. 2, 2, 3, 4, 8 c. 2, 2, 8, 8, 8 e. 2, 2, 3, 8, 8, 9, 9 g. 1, 2, 3, 2, 1, 2, 3, 2, 1 b. 2, 2, 3, 8, 8 d. 2, 3, 4, 7, 8 f. 1, 2, 1, 2, 1, 2, 1 h. 3, 19, 21, 75, 0, 6 i. 3, 2, 7, 6, 2, 7, 3, 1, 4, 2, 7, 5 j. 19, 21, 18, 23, 19, 22, 18, 19, 20 8. A set of data consists of six numbers: 7, 8, 8, 9, 9, and x. Find the mode for these six numbers when: a. x 9 b. x 8 c. x 7 d. x 6 9. A set of data consists of the values 2, 4, 5, x, 5, 4. Find a possible value of x such that: a. there is no mode because all scores appear an equal number of times b. there is only one mode c. there are two modes 10. For the set of data 5, 5, 6, 7, 7, which statement is true? (1) mean mode (2) median mode (3) mean median (4) mean median 11. For the set of data 8, 8, 9, 10, 15, which statement is true? (1) mean median (2) mean mode (3) median mode (4) mean median 688 Statistics 12. When the data consists of 3, 4, 5, 4, 3, 4, 5, which statement is true? (1) mean median (2) mean mode (3) median mode (4) mean median 13. For which set of data is there no mode? (1) 2, 1, 3, 1, 2 (2) 1, 2, 3, 3, 3 (3) 1, 2, 4, 3, 5 (4) 2, 2, 3, 3, 3 14. For which set of data is there more than one mode? (1) 8, 7, 7, 8, 7 (2) 8, 7, 4, 5, 6 (3) 8, 7, 5, 7, 6, 5 (4) 1, 2, 2, 3, 3, 3 15. For which set of data does the median equal the mode? (1) 3, 3, 4, 5, 6 (2) 3, 3, 4, 5 (3) 3, 3, 4 (4) 3, 4 16. For which set of data will the mean, median, and mode all be equal? (1) 1, 2, 5, 5, 7 (2) 1, 2, 5, 5, 8, 9 (3) 1, 1, 1, 2, 5 (4) 1, 1, 2 17. The median of the following data is 11: 2, 5, 9, 11, 40, 3, 4, 5, 10, 45, 32, 40, 67, 7, 11, 9, 20, 34, 5, 1, 8, 15, 16, 19, 39 a. If 4 is subtracted from each data value, what is the median of the transformed data set? b. If the largest data value is doubled and the smallest data value is halved, what is the median of the new data set? 18. The mean of the following data is 37.625: 3, 0, 1, 7, 8, 11, 31, 15, 99, 98, 92, 81, 85, 87, 55, 54, 34, 27, 26, 21, 14, 17, 19, 18 If each data value is multiplied by 2 and increased by 5, what is the mean of the transformed data set? 19. Three consecutive integers can be represented by x, x 1, and x 2. The average of these consecutive integers is 32. What are the three integers? 20. Three consecutive even integers can be represented by x, x 2, and x 4. The average of these consecutive even integers is 20. Find the integers. 21. The mean of three numbers is 31. The second is 1 more than twice the first. The third is 4 less than 3 times the first. Find the numbers. Applying Skills 22. Sid received grades of 92, 84, and 70 on three tests. Find his test average. 23. Sarah’s grades were 80 on each of two of her tests and 90 on each of three other tests. Find her test average. 24. Louise received a grade of x on each of two of her tests and of y on each of three other tests. Represent her average for all the tests in terms of x and y. The Mean, the Median, and the Mode 689 25. Andy has grades of 84, 65, and 76 on three social studies tests. What grade must he obtain on the next test to have an average of exactly 80 for the four tests? 26. Rosemary has grades of 90, 90, 92, and 78 on four English tests. What grade must sh
e obtain on the next test so that her average for the five tests will be 90? 27. The first three test scores are shown below for each of four students. A fourth test will be given and averages taken for all four tests. Each student hopes to maintain an average of 85. Find the score needed by each student on the fourth test to have an 85 average, or explain why such an average is not possible. a. Pat: 78, 80, 100 c. Helen: 90, 92, 95 b. Bernice: 79, 80, 81 d. Al: 65, 80, 80 28. The average weight of Sue, Pam, and Nancy is 55 kilograms. a. What is the total weight of the three girls? b. Agnes weighs 60 kilograms. What is the average weight of the four girls: Sue, Pam, Nancy, and Agnes? 29. For the first 6 days of a week, the average rainfall in Chicago was 1.2 inches. On the last day of the week, 1.9 inches of rain fell. What was the average rainfall for the week? 30. If the heights, in centimeters, of a group of students are 180, 180, 173, 170, and 167, what is the mean height of these students? 31. What is the median age of a family whose members are 42, 38, 14, 13, 10, and 8 years old? 32. What is the median age of a class in which 14 students are 14 years old and 16 students are 15 years old? 33. In a charity collection, ten people gave amounts of $1, $2, $1, $1, $3, $1, $2, $1, $1, and $1.50. What was the median donation? 34. The test scores for an examination were 62, 67, 67, 70, 90, 93, and 98. What is the median test score? 35. The weekly salaries of six employees in a small firm are $440, $445, $445, $450, $450, and $620. a. For these six salaries, find: (1) the mean (2) the median (3) the mode b. If negotiations for new salaries are in session and you represent management, which measure of central tendency will you use as the average salary? Explain your answer. c. If negotiations are in session and you represent the labor union, which measure of cen- tral tendency will you use as an average salary? Explain your answer. 36. In a certain school district, bus service is provided for students living at least miles from 11 2 school. The distances, rounded to the nearest half mile, from school to home for ten students are 0, , and 10 miles. , 1, 1, 1, 1, , , 1 2 1 2 11 2 31 2 a. For these data, find: (1) the mean (2) the median (3) the mode b. How many of the ten students are entitled to bus service? c. Explain why the mean is not a good measure of central tendency to describe the aver- age distance between home and school for these students. 690 Statistics 37. Last month, a carpenter used 12 boxes of nails each of which contained nails of only one size. The sizes marked on the boxes were: 4 in., a. For these data, find: (1) the mean (2) the median (3) the mode 4 in., 3 4 in., 3 4 in., 3 4 in., 3 4 in., 3 4 in., 3 3 4 in., 3 1 in., 1 in., 2 in., 2 in. b. Describe the average-size nail used by the carpenter, using at least one of these mea- sures of central tendency. Explain your answer. Hands-On Activity Find the mean, the median, and the mode for the data that you collected in the Hands-On Activity for Section 16-1. It may be necessary to go back to your original data to do this. 16-5 MEASURES OF CENTRAL TENDENCY AND GROUPED DATA Intervals of Length 1 In a statistical study, when the range is small, we can use intervals of length 1 to group the data. For example, each member of a class of 25 students reported the number of books he or she read during the first half of the school year. The data are as follows: 5, 3, 5, 3, 1, 8, 2, 4, 2, 6, 3, 8, 8, 5, 3, 4, 5, 8, 5, 3, 3, 5, 6, 2, 3 These data, for which the values range from 1 to 8, can be organized into a table such as the one shown at the right, with each value representing an interval. Since 25 students were included in this study, the total frequency, N, is 25. We can use this table, with intervals of length 1, to find the mode, median, and mean for these data. Interval Frequency 25 Mode of a Set of Grouped Data Since the greatest frequency, 7, appears for interval 3, the mode for the data is 3. In general: For a set of grouped data, the mode is the value of the interval that contains the greatest frequency. Measures of Central Tendency and Grouped Data 691 Median of a Set of Grouped Data We have learned that the median for a set of data in numerical order is the middle value. For these 25 numbers, there are 12 numbers greater than or equal to the median, and 12 numbers less than or equal to the median. Therefore, when the numbers are written in numerical order, the median is the 13th number from either end. 1, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 4, 4, 5, 5, 5, 5, 5, 5, 6, 6, 8, 8, 8, 8 ↑– The median is 4. When the data are grouped in the table shown earlier, a simple counting procedure can be used to find the median, the 13th number. When we add the frequencies of the first four intervals, starting at the top, we find that these intervals include data for: 4 0 2 6 12 students Therefore, the next lower interval (with frequency greater than 0) must include the median, the value for the 13th student. This is the interval for the data value 4. When we add the frequencies of the first three intervals, starting at the bot- tom, we find that these intervals include data for: 1 3 7 11 students The next higher interval contains two scores, one for the 12th student and one that is the median, or the value for the 13th student. Again this is the interval for the data value 4. In general: For a set of grouped data, the median is the value of the interval that con- tains the middle data value. Mean of a Set of Grouped Data By adding the four 8’s in the ungrouped data, we see that four students, reading eight books each, have read 8 8 8 8 or 32 books. We can arrive at this same number by using the grouped intervals in the table: we multiply the four 8’s by the frequency 4. Thus, . Applying this multiplication shortcut to each row of the table, we obtain the third column of the following table: (4)(8) 5 32 692 Statistics Interval Frequency (Interval) (Frequency = 25 8 4 32 7 0 0 6 2 12 5 6 30 4 2 8 3 7 21 2 3 6 1 1 1 Total 110 The total (110) represents the sum of all 25 pieces of data. We can check this by adding the 25 scores in the unorganized data. Finally, to find the mean, we divide the total number, 110, by the number of items, 25. Thus, the mean for the data is: 110 25 4.4. Procedure To find the mean for N values in a table of grouped data when the length of each interval is 1: 1. For each interval, multiply the interval value by its corresponding frequency. 2. Find the sum of these products. 3. Divide this sum by the total frequency, N. Calculator Solution for Grouped Data The calculator can be used to find the mean and median for the grouped data shown above. Enter the number of books read by each student into L1 and the frequency for each number of books into L2. Then use the 1-Var Stats from the STAT CALC menu to display information about the data. ENTER: STAT ENTER 2nd L1 , 2nd L2 ENTER DISPLAY< n = 2 5 Measures of Central Tendency and Grouped Data 693 The display shows that the mean, x–, is 4.4, the sum of the number of books read is 110, and the number of students, the total frequency, N, is 25. Use the down arrow to display the median, Med 4. Intervals Other Than Length 1 There are specific mathematical procedures to find the mean, median, and mode for grouped data with intervals other than length 1, but we will not study them at this time. Instead, we will simply identify the intervals that contain some of these measures of central tendency. For example, a small industrial plant surveyed 50 workers to find the number of miles each person commuted to work. The commuting distances were reported, to the nearest mile, as follows: 0, 0, 1, 1, 2, 2, 2, 3, 3, 4, 4, 4, 5, 5, 6, 6, 6, 7, 7, 7, 9, 10, 10, 10, 10, 10, 10, 10, 10, 12, 12, 14, 15, 17, 17, 18, 22, 23, 25, 28, 30, 32, 32, 33, 34, 34, 36, 37, 37, 52 These data are organized into a table with intervals of length 10, as follows: Interval (commuting distance) Frequency (number of workers) 50–59 40–49 30–39 20–29 10–19 0–9 1 0 9 4 15 21 N 50 Modal Interval In the table, interval 0–9 contains the greatest frequency, 21. We say that interval 0–9 is the group mode, or modal interval, because this group of numbers has the greatest frequency. The modal interval is not the same as the mode. The modal interval is a group of numbers; the mode is usually a single number. For this example, the original data (before being placed into the table) show that the number appearing most often is 10. Hence, the mode is 10. The modal interval, which is 0–9, tells us that, of the six intervals in the table, the most frequently occurring commuting distance is 0 to 9 miles. Both the mode and the modal interval depend on the concept of greatest frequency. For the mode, we look for a single number that has the greatest frequency. For the modal interval, we look for the interval that has the greatest frequency. 694 Statistics Interval Containing the Median To find the interval containing the median, we follow the procedure described earlier in this section. For 50 numbers, the median, or middle number, will be at a point where 25 numbers are at or above the median and 25 are at or below it. Count the frequencies in the table from the uppermost interval and move downward. We add 1 0 9 4 14. Since there are 15 numbers in the next lower interval, and , we see that the 25th number will be reached somewhere in that interval, 10–19. 14 1 15 5 29 Count from the bottom interval and move up. We have 21 numbers in the first interval. Since there are 15 numbers in the next higher interval, and 21 15 36, we see that 25th number will be reached somewhere in that interval, 10–19. This is the same result that we obtained when we moved downward. The interval containing the median for this grouping is 10–19. In this course, we will not deal with problems in which the median is not found in any interval. Interval Containing the Mean When data are gro
uped using intervals of length other than 1, there is no simple procedure to identify the interval containing the mean. However, the mean can be approximated by assuming that the data are equally distributed throughout each interval. The mean is then found by using the midpoint of each interval as the value of each entry in the interval. This problem is studied in higher-level courses. EXAMPLE 1 In the table, the data indicate the heights, in inches, of 17 basketball players. For these data find: a. the mode b. the median c. the mean Height (inches) Frequency (number) Solution a. The greatest frequency, 5, occurs for the height of 75 inches. The mode, or height appearing most often, is 75. 77 76 75 2 0 5 74 3 73 b. For 17 players, the median is the 9th number, so there are 8 heights greater than or equal to the median and 8 heights less than or equal to the median. Counting the frequencies going down, we have 2 0 5 7. Since the frequency of the next interval is 3, the 8th, 9th, and 10th heights are in this interval, 74. Counting the frequencies going up, we have 1 2 4 7. Again, the frequency of the next interval is 3, and the 8th, 9th, and 10th heights are in this interval. The 9th height, the median, is 74. 71 72 2 1 4 Measures of Central Tendency and Grouped Data 695 c. (1) Multiply each height by its corresponding frequency: 75 5 375 76 0 0 71 1 71 72 2 144 77 2 154 73 4 292 (2) Find the total of these products: 154 0 375 222 292 144 71 1,258 74 3 222 (3) Divide this total, 1,258, by the total frequency, 17 to obtain the mean: 1258 17 74 Calculator Solution Clear any previous data that may be stored in L1 and L2. Enter the heights of the players into L1 and the frequencies into L2. Then use 1-Var Stats from the STAT CALC menu to display information about the data. The screen will show the mean, x–. Press the down arrow key to display the median. ENTER: STAT ENTER 2nd L1 , 2nd L2 ENTER DISPLAY Answers a. mode 75 b. median 74 c. mean 74 EXERCISES Writing About Mathematics 1. The median for a set of 50 data values is the average of the 25th and 26th data values when the data is in numerical order. What must be true if the median is equal to one of the data values? Explain your answer. 2. What must be true about a set of data if the median is not one of the data values? Explain your answer. 696 Statistics Developing Skills In 3–5, the data are grouped in each table in intervals of length 1. Find: a. the total frequency b. the mean c. the median d. the mode 3. Interval Frequency 4. Interval Frequency 5. Interval Frequency 10 15 16 17 18 19 20 3 2 4 1 5 6 25 24 23 22 21 20 19 4 0 3 2 4 5 2 In 6–8, the data are grouped in each table in intervals other than length 1. Find: a. the total frequency b. the interval that contains the median c. the modal interval 6. Interval Frequency 7. Interval Frequency 8. Interval Frequency 55–64 45–54 35–44 25–34 15–24 3 8 7 6 2 4–9 10–15 16–21 22–27 28–33 34–39 12 13 9 12 15 10 126–150 101–125 76–100 51–75 26–50 1–25 4 6 6 3 7 2 Applying Skills 9. On a test consisting of 20 questions, 15 students received the following scores: 17, 14, 16, 18, 17, 19, 15, 15, 16, 13, 17, 12, 18, 16, 17 a. Make a frequency table for these students listing scores from 12 to 20. b. Find the median score. c. Find the mode. d. Find the mean. Measures of Central Tendency and Grouped Data 697 10. A questionnaire was distributed to 100 people. The table shows the time taken, in minutes, to complete the questionnaire. a. For this set of data, find: (1) the mean (2) the Interval Frequency median (3) the mode b. How are the three measures found in part a related for these data? 6 5 4 3 2 12 20 36 20 12 11. A storeowner kept a tally of the sizes of suits purchased in the store, as shown in the table. a. For this set of data, find: (1) the total frequency (3) the median (2) the mean (4) the mode b. Which measure of central tendency should the store- owner use to describe the average suit sold? Size of Suit Number Sold (interval) (frequency) 48 46 44 42 40 38 36 34 1 1 3 5 3 8 2 2 12. Test scores for a class of 20 students are as follows: 93, 84, 97, 98, 100, 78, 86, 100, 85, 92, 72, 55, 91, 90, 75, 94, 83, 60, 81, 95 a. Organize the data in a table using 51–60 as the smallest interval. b. Find the modal interval. c. Find the interval that contains the median. 13. The following data consist of the weights, in pounds, of 35 adults: 176, 154, 161, 125, 138, 142, 108, 115, 187, 158, 168, 162 135, 120, 134, 190, 195, 117, 142, 133, 138, 151, 150, 168 172, 115, 148, 112, 123, 137, 186, 171, 166, 166, 179 a. Organize the data in a table, using 100–119 as the smallest interval. b. Construct a frequency histogram based on the grouped data. c. In what interval is the median for these grouped data? d. What is the modal interval? 698 Statistics 16-6 QUARTILES, PERCENTILES, AND CUMULATIVE FREQUENCY Quartiles When the values in a set of data are listed in numerical order, the median separates the values into two equal parts. The numbers that separate the set into four equal parts are called quartiles. To find the quartile values, we first divide the set of data into two equal parts and then divide each of these parts into two equal parts. The heights, in inches, of 20 students are shown in the following list. The median, which is the average of the 10th and 11th data values, is shown here enclosed in a box. Lower half \|______________________________\| 53, 60, 61, 63, 64, 65, 65, 65, 65, 66, Upper half \|_______________________________\| 66, 67, 67, 68, 69, 70, 70, 71, 71, 73 ↑ 66 Median Ten heights are listed in the lower half, 53–66. The middle value for these 10 heights is the average of the 5th and 6th values from the lower end, or 64.5. This value separates the lower half into two equal parts. Ten heights are also listed in the upper half, 66–73. The middle value for these 10 heights is the average of the 5th and 6th values from the upper end, or 69.5. This value separates the upper half into two equal parts. The 20 data values are now separated into four equal parts, or quarters. \|_______________\| 53, 60, 61, 63, 64, \|_______________\| 65, 65, 65, 65, 66, ↑ 64.5 \|_______________\| 66, 67, 67, 68, 69, \|_______________\| 70, 70, 71, 71, 73 ↑ 69.5 ↑ 66 First quartile Median Second quartile Third quartile The numbers that separate the data into four equal parts are the quartiles. For this set of data: 1. Since one quarter of the heights are less than or equal to 64.5 inches, 64.5 is the lower quartile, or first quartile. 2. Since two quarters of the heights are less than or equal to 66 inches, 66 is the second quartile. The second quartile is always the same as the median. 3. Since three quarters of the heights are less than or equal to 69.5 inches, 69.5 is the upper quartile, or third quartile. Note: The quartiles are sometimes denoted Q1, Q2, and Q3. Quartiles, Percentiles, and Cumulative Frequency 699 Procedure To find the quartile values for a set of data: 1. Arrange the data in ascending order from left to right. 2. Find the median for the set of data.The median is the second quartile value. 3. Find the middle value for the lower half of the data.This number is the first, or lower, quartile value. 4. Find the middle value for the upper half of the data.This number is the third, or upper, quartile value. Note that when finding the first quartile, use all of the data values less than or equal to the median, but do not include the median in the calculation. Similarly, when finding the third quartile, use all of the data values greater than or equal to the median, but do not include the median in the calculation. Constructing a Box-and-Whisker Plot A box-and-whisker plot is a diagram that uses the quartile values, together with the maximum and minimum values, to display information about a set of data. To draw a box-and-whisker plot, we use the following steps. STEP 1. Draw a scale with numbers from the minimum to the maximum value of a set of data. For example, for the set of heights of the 20 students, the scale should include the numbers from 53 to 73. STEP 2. Above the scale, place dots to represent the five numbers that are the statistical summary for this set of data: the minimum value, the first quartile, the median, the third quartile, and the maximum value. For the heights of the 20 students, these numbers are 53, 64.5, 66, 69.5 and 73. 50 55 60 65 70 75 STEP 3. Draw a box between the dots that represent the lower and upper quartiles, and a vertical line in the box through the point that represents the median. 50 55 60 65 70 75 700 Statistics STEP 4. Add the whiskers by drawing a line segment joining the dots that represent the minimum data value and the lower quartile, and a second line segment joining the dots that represent the maximum data value and the upper quartile. 50 55 60 65 70 75 The box indicates the ranges of the middle half of the set of data. The long whisker at the left shows us that the data are more scattered at the lower than at the higher end. A graphing calculator can display a boxand-whisker plot. Enter the data in L1, then go to the STAT PLOT menu to select the type of graph to draw. ENTER: 2nd STAT PLOT 1 ENTER ENTER 2nd L1 ALPHA = Ty Now display the box-and-whisker plot by P 1 : L 1 entering ZOOM 9 . We can press TRACE and the right and left arrow keys to display the minimum value, first quartile, median, third quartile, and maximum value. The five statistical summary can also be displayed in 1-Var Stats. Scroll down to the last five values. ENTER: STAT ENTER ENTER EXAMPLE 1 Find the five statistical summary for the following set of data: 8, 5, 12, 9, 6, 2, 14, 7, 10, 17, 11, 8, 14, 5 Solution (1) Arrange the data in numerical order: Quartiles, Percentiles, and Cumulative Frequency 701 2, 5, 5, 6, 7, 8, 8, 9,10, 11, 12, 14, 14, 17 We can see that 2 is the minimum value and 17 is the maximum value. (2) Find the median. Since there are 14 data values in the set, the median is the average of the 7t
h and 8th values. Therefore, 8.5 is the second quartile. Median 8 1 9 2 8.5 (3) Find the first quartile. There are seven values less than 8.5. The middle value is the 4th value from the lower end of the set of data, 6. Therefore, 6 is the first, or lower, quartile. (4) Find the third quartile. There are seven values greater than 8.5. The middle value is the 4th value from the upper end of the set of data, 12. Therefore, 12 is the third, or upper, quartile. Answer The minimum is 2, first quartile is 6, the second quartile is 8.5, the third quartile is 12, and the maximum is 17. Note: The quartiles 6, 8.5, and 12 separate the data values into four equal parts even though the original number of data values, 14, is not divisible by 4: \|_____\| 2, 5, 5, 6, \|_____\| \|________\| 7, 8, 8, 9, 10, 11, ↑ 8.5 \|_________\| 14, 14, 17 12, The first and third quartile values, 6 and 12, are data values. If we think of each of these as a half data value in the groups that they separate, each group contains 3 data values, which is 25% of the total. 1 2 Percentiles A percentile is a number that tells us what percent of the total number of data values lies at or below a given measure. Let us consider again the set of data values representing the heights of 20 students. What is the percentile rank of 65? To find out, we separate the data into the values that are less than or equal to 65 and those that are greater than or equal to 65, so that the four 65’s in the set are divided equally between the two groups: 53, 60, 61, 63, 64, 65, 65, 65, 65, 66, 66, 67, 67, 68, 69, 70, 70, 71, 71, 73 Half of 4, or 2, of the 65’s are in the lower group and half are in the upper group. 702 Statistics Since there are seven data values in the lower group, we find what percent 7 is of 20, the total number of values: 7 20 5 0.35 5 35% Therefore, 65 is at the 35th percentile. To find the percentile rank of 69, we separate the data into the values that are less than or equal to 69 and those that are greater or equal to 69: 53, 60, 61, 63, 64, 65, 65, 65, 65, 66, 66, 67, 67, 68, 69, 70, 70, 71, 71, 73 Because 69 occurs only once, we will include it as half of a data value in the 141 lower group and half of a data value in the upper group. Therefore, there are 2 or 14.5 data values in the lower group. 14.5 20 5 0.725 5 72.5% Because percentiles are usually not written using fractions, we say that 69 is at the 73rd percentile. EXAMPLE 2 Find the percentile rank of 87 in the following set of 30 marks: 56, 65, 65, 67, 72, 73, 75, 77, 77, 78, 78, 78, 80, 80, 80, 82, 83, 85, 85, 85, 86, 87, 87, 87, 88, 90, 92, 93, 95, 98 Solution (1) Find the sum of the number of marks less than 87 and half of the number of 87’s: Number of marks less than 87 21 Half of the number of 87’s (0.5 3) 1.5 22.5 (2) Divide the sum by the total number of marks: (3) Change the decimal value to a percent: 0.75 75%. 22.5 30 5 0.75 Answer: A mark of 87 is at the 75th percentile. Note: 87 is also the upper quartile mark. Cumulative Frequency In a school, a final examination was given to all 240 students taking biology. The test grades of these students were then grouped into a table. At the same time, a histogram of the results was constructed, as shown below. Interval (test scores) Frequency (number) 91–100 81–90 71–80 61–70 51–60 45 60 75 40 20 y c n e u q e r F 80 60 40 20 0 Quartiles, Percentiles, and Cumulative Frequency 703 HISTOGRAM 51–60 61–70 71–80 81–90 91–100 Test scores From the table and the histogram, we can see that 20 students scored in the interval 51–60, 40 students scored in the interval 61–70, and so forth. We can use these data to construct a new type of histogram that will answer the question, “How many students scored below a certain grade?” By answering the following questions, we will gather some information before constructing the new histogram: 1. How many students scored 60 or less on the test? From the lowest interval, 51–60, we know that 20 students scored 60 or less. 2. How many students scored 70 or less on the test? By adding the frequencies for the two lowest intervals, 51–60 and 61–70, we see that 20 40, or 60, students scored 70 or less. 3. How many students scored 80 or less on the test? By adding the frequencies for the three lowest intervals, 51–60, 61–70, and 71–80, we see that 20 40 75, or 135, students scored 80 or less. 4. How many students scored 90 or less on the test? Here, we add the frequencies in the four lowest intervals. Thus, 20 40 75 60, or 195, students scored 90 or less. 5. How many students scored 100 or less on the test? By adding the five lowest frequencies, 20 40 75 60 45, we see that 240 students scored 100 or less. This result makes sense because 240 students took the test and all of them scored 100 or less. Constructing a Cumulative Frequency Histogram The answers to the five questions we have just asked were found by adding, or accumulating, the frequencies for the intervals in the grouped data to find the cumulative frequency. The accumulation of data starts with the lowest interval of data values, in this case, the lowest test scores. The histogram that displays these accumulated figures is called a cumulative frequency histogram. 704 Statistics Interval (test scores) Frequency (number) Cumulative Frequency CUMULATIVE FREQUENCY HISTOGRAM 240 91–100 81–90 71–80 61–70 51–60 45 60 75 40 20 240 195 135 60 20 CUMULATIVE FREQUENCY HISTOGRAM 240 100% 210 180 150 120 90 60 30 75% 50% 25% 51–60 51–70 51–80 51–90 51–100 Test scores 210 180 150 120 90 60 30 0 51–60 51–70 51–80 51–90 51–100 Test scores To find the cumulative frequency for each interval, we add the frequency for that interval to the frequencies for the intervals with lower values. To draw a cumulative frequency histogram, we use the cumulative frequencies to determine the heights of the bars. For our example of the 240 biology students and their scores, the frequency scale for the cumulative frequency histogram goes from 0 to 240 (the total frequency for all of the data). We can replace the scale of the cumulative frequency histogram shown above with a different one that expresses the cumulative frequency in percents. Since 240 students represent 100% of the students taking the biology test, we write 100% to correspond to a cumulative frequency of 240. Similarly, since 0 students represent 0% of the students taking the biology test, we write 0% to correspond to a cumulative frequency of 0. If we divide the percent scale into four equal parts, we can label the three added divisions as 25%, 50%, and 75%. Quartiles, Percentiles, and Cumulative Frequency 705 Thus the graph relates each cumulative frequency to a percent of the total number of biology students. For example, 120 students (half of the total number) corresponds to 50%. Let us use the percent scale to answer the question, “What percent of the students scored 70 or below on the test?” The height of each bar represents both the number of students and the percent of the students who had scores at or below the largest number in the interval represented by that bar. Since 25%, or a quarter, of the scores were 70 or below, we say that 70 is an approximate value for the lower quartile, or the 25th percentile. CUMULATIVE FREQUENCY HISTOGRAM 240 100 210 180 150 120 90 60 30 0 75% 56% 50% 25% 8% 0% 51–60 51–70 51–80 51–90 51–100 Test scores From the histogram, we can see that about 56% of the students had scores at or below 80. Thus, the second quartile, the median, is in the 51–80 interval. For these data, the upper quartile is in the 51–90 interval. From the histogram, we can also conveniently read the approximate percentiles for the scores that are the end values of the intervals. For example, to find the percentile for a score of 60, the right-end score of the first interval, we draw a horizontal line segment from the height of the first interval to the percent scale, as shown by the dashed line in the histogram above. The fact that the horizontal line crosses the percent scale at about one-third the distance between 0% and 25% tells us that approximately 8% of the students scored 60 or below 60. Thus, the 8th percentile is a good estimate for a score of 60. 706 Statistics EXAMPLE 3 A reporter for the local newspaper is preparing an article on the ice cream stores in the area. She listed the following prices for a two-scoop cone at 15 stores. $2.48, $2.57, $2.30, $2.79, $2.25, $3.00, $2.82, $2.75, $2.55, $2.98, $2.53, $2.40, $2.80, $2.50, $2.65 a. List the data in a stem-and-leaf diagram. b. Find the median. c. Find the first and third quartiles. d. Construct a box-and-whisker plot. e. Draw a cumulative frequency histogram. f. Find the percentile rank of a price of $2.75. Solution a. The first two digits in each price will be the stem. The lowest price is $2.25 and the highest price is $3.00. b. Since there are 15 prices, the median is the 8th from the top or from the bottom. The median is $2.57. c. The middle value of the set of numbers below the median is the first quartile. That price is $2.48. The middle value of the set of numbers above the median is the third quartile. That price is $2.80. Stem Leaf 3.0 2.9 2.8 2.7 2.6 2.5 2.4 2.3 2. Key: 2.9 8 $2.98 d. Use a scale from $2.25 to $3.00. Place dots at $2.48, $2.57, and $2.80 for the first quartile, the median, and the third quartile. Draw the box around the quartiles with a vertical line through the median. Add the whiskers. 2.25 2.50 2.75 3.00 Interval Frequency Cumulative Frequency 3.00–3.09 2.90–2.99 2.80–2.89 2.70–2.79 2.60–2.69 2.50–2.59 2.40–2.49 2.30–2.39 2.20–2.29 1 1 2 2 1 4 2 1 1 15 14 13 11 9 8 4 2 1 Quartiles, Percentiles, and Cumulative Frequency 707 CUMULATIVE FREQUENCY HISTOGRAM 15 12 .20–2.29 2.20–2.39 2.20–2.49 2.20–2.59 2.20–2.69 2.20–2.79 2.20–2.89 2.20–2.99 2.20–3.09 e. Make a cumulative frequency table and draw the histogram. Use 2.20–2.29 Price of a Two-Scoop Cone as the smallest interval. f. There are 9 data values below $2.75. Add 1 2 Percentile
rank: 0.63 192 15 for the data value $2.75. – 63% Answers a. Diagram b. median $2.57 c. first quartile $2.48; third quartile $2.80 d. Diagram e. Diagram f. 63rd percentile Note: A cumulative frequency histogram can be drawn on a calculator just like a regular histogram. In list L2, where we previously entered the frequencies for each individual interval, we now enter each cumulative frequency. EXERCISES Writing About Mathematics 1. a. Is it possible to determine the percentile rank of a given score if the set of scores is arranged in a stem-and-leaf diagram? Explain why or why not. b. Is it possible to determine the percentile rank of a given score if the set of scores is shown on a cumulative frequency histogram? Explain why or why not. 2. A set of data consisting of 23 consecutive numbers is written in numerical order from left to right. a. The number that is the first quartile is in which position from the left? b. The number that is the third quartile is in which position from the left? 708 Statistics Developing Skills In 3–6, for each set of data: a. Find the five numbers of the statistical summary b. Draw a box-andwhisker plot. 3. 12, 17, 20, 21, 25, 27, 29, 30, 32, 33, 33, 37, 40, 42, 44 4. 67, 70, 72, 77, 78, 78, 80, 84, 86, 88, 90, 92 5. 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 5, 7, 9, 9 6. 3.6, 4.0, 4.2, 4.3, 4.5, 4.8, 4.9, 5.0 In 7–9, data are grouped into tables. For each set of data: a. Construct a cumulative frequency histogram. b. Find the interval in which the lower quartile lies. c. Find the interval in which the median lies. d. Find the interval in which the upper quartile lies. 7. Interval Frequency 8. Interval Frequency 9. Interval Frequency 41–50 31–40 21–30 11–20 1–10 8 5 2 5 4 25–29 20–24 15–19 10–14 5–9 3 1 3 9 9 1–4 5–8 9–12 13–16 17–20 4 3 7 2 2 10. For the data given in the table: a. Construct a cumulative frequency his- togram. b. In what interval is the median? c. The value 10 occurs twice in the data. What is the percentile rank of 10? 11. For the data given in the table: a. Construct a cumulative frequency his- togram. b. In what interval is the median? c. In what interval is the upper quartile? d. What percent of scores are 17 or less? e. In what interval is the 25th percentile? Interval Frequency 21–25 16–20 11–15 6–10 1–5 5 4 6 3 2 Interval Frequency 33–37 28–32 23–27 18–22 13–17 8–12 3–7 4 3 7 12 8 5 1 Applying Skills 12. A group of 400 students were asked to state the number of minutes that each spends watching television in 1 day. The cumulative frequency histogram shown below summarizes the responses as percents. a. What percent of the students questioned watch television for 90 minutes or less each day? b. How many of the students watch television for 90 minutes or less each day? c. In what interval is the upper quartile? d. In what interval is the lower quar- tile? e. If one of these students is picked at random, what is the probability that he or she watches 30 minutes or less of television each day? Quartiles, Percentiles, and Cumulative Frequency 709 CUMULATIVE FREQUENCY HISTOGRAM 100% 90% 80% 70% 60% 50% 40% 30% 20% 10% 1–30 1–60 1–90 1–120 1–150 1–180 Number of minutes 13. A journalism student was doing a study of the readability of the daily newspaper. She chose several paragraphs at random and listed the number of letters in each of 88 words. She prepared the following chart. a. Copy the chart, adding a column that lists the cumulative frequency b. Find the median. c. Find the first and third quartiles. d. Construct a box-and whisker plot. e. Draw a cumulative frequency his- togram. f. Find the percentile rank of a word with 7 letters. Number of letters Frequency 1 2 3 4 5 6 7 8 9 10 4 14 20 20 3 18 5 2 1 1 14. Cecilia’s average for 4 years is 86. Her average is the upper quartile for her class of 250 students. At most, how many students in her class have averages that are less than Cecilia’s? 710 Statistics 15. In the table at the right, data are given for the heights, in inches, of 22 football players. a. Copy and complete the table. b. Draw a cumulative frequency histogram. c. Find the height that is the lower quartile. d. Find the height that is the upper quartile. Height (inches) Frequency Cumulative Frequency 77 76 75 74 73 72 71 2 2 7 5 3 2 1 16. The lower quartile for a set of data was 40. These data consisted of the heights, in inches, of 680 children. At most, how many of these children measured more than 40 inches? In 17 and 18, select, in each case, the numeral preceding the correct answer. 17. On a standardized test, Sally scored at the 80th percentile. This means that (1) Sally answered 80 questions correctly. (2) Sally answered 80% of the questions correctly. (3) Of the students who took the test, about 80% had the same score as Sally. (4) Of the students who took the test, at least 80% had scores that were less than or equal to Sally’s score. 18. For a set of data consisting of test scores, the 50th percentile is 87. Which of the following could be false? (1) 50% of the scores are 87. (2) 50% of the scores are 87 or less. (3) Half of the scores are at least 87. (4) The median is 87. 16-7 BIVARIATE STATISTICS We have been studying univariate statistics or statistics that involve a single set of numbers. Statistics are often used to study the relationship between two different sets of values. For example, a dietician may want to study the relationship between the number of calories from fat in a person’s diet and the level of cholesterol in that person’s blood, or a merchant may want to study the relationship between the amount spent on advertising and gross sales. Although these examples involving two-valued statistics or bivariate statistics require complex statistical methods, we can investigate some of the properties of similar but simpler problems by looking at graphs and by using a graphing calculator. A graph that shows the pairs of values in the data as points in the plane is called a scatter plot. Bivariate Statistics 711 Correlation We will consider five cases of two-valued statistics to investigate the relationship or correlation between the variables based on their scatter plots. CASE 1 approximate a straight line that has a positive slope. The data has positive linear correlation. The points in the scatter plot A driver recorded the number of gallons of gasoline used and the number of miles driven each time she filled the tank. In this example, there is both correlation and causation since the increase in the number of miles driven causes the number of gallons of gasoline needed to increase. Gallons 7.2 5.8 7.0 5.5 5.6 7.1 6.0 4.4 5.0 6.2 4.7 5.7 Miles 240 188 226 193 187 235 202 145 167 212 154 188 This scatter plot can be duplicated on your graphing calculator. Enter the number of miles as L1 and the number of gallons of gasoline as corresponding entries in L2. The miles will be graphed as x-values and the gallons of gasoline as y-values. First, turn on Plot 1: 250 225 s e l i M 200 175 150 3 4 5 6 7 8 9 Gallons of Gasoline ENTER: 2nd STAT PLOT 1 ENTER ENTER 2nd L1 2nd L2 DISPLAY = Ty : + . 712 Statistics Now use ZoomStat from the ZOOM menu to construct a window that will include all values of x and y. ENTER: ZOOM 9 DISPLAY: The data has moderate positive correlation. The points in a scatter plot CASE 2 do not lie in a straight line but there is a general tendency for the values of y to increase as the values of x increase. Last month, each student in an English class was required to choose a book to read. The teacher recorded, for each student in the class, the number of days spent reading the book and the number of pages in the book. Days 8 14 12 26 9 17 28 13 15 30 18 20 Pages 225 300 298 356 200 412 205 215 310 357 209 250 Days 29 22 17 14 11 14 22 19 16 7 18 30 Pages 314 288 256 225 232 256 300 305 276 172 318 480 While the books with more pages may have required more time, some students read more rapidly and some spent more time each day reading. The graph shows that, in general, as the number of days needed to read a book increased, the number of pages that were read also increased. 5 7 9 11 13 15 17 19 21 23 25 27 29 31 Days 500 475 450 425 400 375 350 325 300 275 250 225 200 175 s e g a P Bivariate Statistics 713 CASE 3 The data has no correlation. Before giving a test, a teacher asked each student how many minutes each had spent the night before preparing for the test. After correcting the test, she prepared the table below which compares the number of minutes of study to the number of correct answers. Minutes of Study Correct Answers 20 15 40 5 10 25 30 12 15 10 3 19 16 6 12 3 5 5 20 35 40 8 16 14 The graph shows that there is no correlation between the time spent studying just before the test and the number of correct answers on the test 20 18 16 14 12 10 8 6 4 2 0 5 10 15 20 25 30 35 40 45 Minutes of Study The data has moderate negative correlation. The points in a scatter plot CASE 4 do not lie in a straight line but there is a general tendency for the values of y to decrease as the values of x increase. A group of children go to an after-school program at a local youth club. The director of the program keeps a record, shown below, of the time, in minutes, each student spends playing video games and doing homework. Games Homework 20 50 30 60 90 10 60 40 30 40 50 35 70 15 40 30 80 30 60 10 In this instance, the unit of measure, minutes, is found in the problem rather than in the table. To create meaningful graphs, always include a unit of measure on the horizontal and vertical axes. 714 Statistics 70 60 50 40 30 20 10 10 20 30 40 50 60 70 80 90 100 Minutes Playing Games The graph shows that, in general, as the number of minutes spent playing video games increases, the number of minutes spent doing homework decreases. CASE 5 approximate a straight line that has a negative slope. The data has negative linear correlation. The points in the scatter plot A long-distance truck driver travels 500 miles each day. As he passes t
hrough different areas on his trip, his average speed and the length of time he drives each day vary. The chart below shows a record of average speed and time for a 10-day period. Speed Time 50 10 64 68 60 54 66 70 62 64 58 7.9 7.5 8.5 9.0 7.0 7.1 8.0 8.2 9.0 11 10.5 10 9.5 9 8.5 8 7.5 7 6. 50 55 60 65 70 75 Speed in Miles per Hour In this case, the increase in the average speed causes the time required to drive a fixed distance to decrease. This example indicates both negative correlation and causation. It is important to note that correlation is not the same as causation. Correlation is an indication of the strength of the linear relationship or association between the variables, but it does not mean that changes in one variable are the cause of changes in the other. For example, suppose a study found there was a strong positive correlation between the number of pages in the daily newspaper and the number of voters who turn out for an election. One would not be correct in concluding that a greater number of pages causes a greater turnout. Rather, it is likely that the urgency of the issues is reflected in the increase of both the size of the newspaper and the size of the turnout. Bivariate Statistics 715 Another example where there is no causation occurs in time series or data that is collected at regular intervals over time. For instance, the population of the U.S. recorded every ten years is an example of a time series. In this case, we cannot say that time causes a change in the population. All we can do is note a general trend, if any. Line of Best Fit When it makes sense to consider one variable as the independent variable and the other as the dependent variable, and the data has a linear correlation (even if it is only moderate correlation), the data can be represented by a line of best fit. For example, we can write an equation for the data in Case 1. Enter the data into L1 and L2 if it is not already there. Find the mean values for x, the number of miles driven, and for y, the number of gallons of gasoline used. Then use 2-Var Stats from the STAT CALC menu: ENTER: STAT 2 ENTER DISPLAY< n = 1 2 The calculator gives x– 5.85 and, by pressing the down arrow key, y– 194.75. We will use these mean values, (5.85, 194.75), as one of the points on our line. We will choose one other data point, for example (7.1, 235), as a second point and write the equation of the line using the slope-intercept form y mx b. First find the slope: m y2 2 y1 x2 2 x1 5 194.75 2 235 5.85 2 7.1 5 240.25 21.25 5 32.2 Now use one of the points to find the y-intercept: 194.75 32.2(5.85) b 194.75 188.37 b 6.38 b Round the values to three significant digits. A possible equation for a line of best fit is y 32.2x 6.38. The calculator can also be used to find a line called the regression line to fit a bivariate set of data. Use the LinReg(ax+b) function in the STAT CALC menu. 716 Statistics ENTER: STAT 4 ENTER DISPLAY If we round the values to three significant digits, the equation of the regression line is y 32.7x 3.59. In this case, the difference between these two equations is negligible. However, this is not always the case. The regression line is a special line of best fit that minimizes the square of the vertical distances to each data point. We can compare these two equations with the actual data. Graph the scatter plot of the data using ZoomStat. Then write the two equations in the Y= menu. ENTER: Y 32.2 X,T,,n 6.38 ENTER 32.7 X,T,,n 3.59 GRAPH DISPLAY: Notice that the lines are very close and do approximate the data. Note 1: The equation of the line of best fit is very sensitive to rounding. Try to round the coefficients of the line of best fit to at least three significant digits or to whatever the test question asks. Note 2: A line of best fit is appropriate only for data that exhibit a linear pattern. In more advanced courses, you will learn how to deal with nonlinear patterns. These equations can be used to predict values. For example, if the driver has driven 250 miles before filling the tank, how many gallons of gasoline should be needed? We will use the equation from the calculator. y 32.7x 3.59 250 32.7x 3.59 246.41 32.7x 7.535474006 x Bivariate Statistics 717 It is reasonable to say that the driver can expect to need about 7.5 gallons of gasoline. What we just did is called extrapolation, that is, using the line of best fit to make a prediction outside of the range of data values. Using the line of best fit to make a prediction within the given range of data values is called interpolation. In general, interpolation is usually safe, while care should be taken when extrapolating. The observed correlation pattern may not be valid outside of the given range of data values. For example, consider the scatter plot of the population of a town shown below. The population grew at a constant rate during the years in which the data was gathered. However, we do not expect the population to continue to grow forever, and thus, it may not be possible to extrapolate far into the future. 280 270 260 250 240 230 220 210 200 190 180 170 160 ) 1960 1965 1970 1975 1980 1985 1990 1995 2000 2005 Year Keep in Mind In general, when a given relationship involves two sets of data: 1. In some cases a straight line, a line of best fit, can be drawn to approxi- mate the relationship between the data sets. 2. If a line of best fit has a positive slope, the data has positive linear correla- tion. 3. If the line of best fit has a negative slope, the data has negative linear cor- relation. 4. A line of best fit can be drawn through (x–, y–), the point whose coordinates are the means of the given data. Any data point that appears to lie on or near the line of best fit can be used as a second point to write the equation. 718 Statistics 5. A calculator can be used to find the regression line as the line of best fit. 6. When the graphed data points are so scattered that it is not possible to draw a straight line that approximates the given relationship, the data has no correlation. To study bivariate data without using a graphing calculator: 1. Make a table that lists the data. 2. Plot the data as points on a graph. 3. Find the mean of each set of data and locate the point (x–, y–) on the graph. 4. Draw a line that best approximates the data. 5. Choose the point (x–, y–) and one other point or any two points that are on or close to the line that you drew. Use these points to write an equation of the line. 6. Use the equation of the line to predict related outcomes. To study bivariate data using a graphing calculator: 1. Enter the data into L1 and L2 or any two lists in the memory of the calcu- lator. 2. Use STAT PLOT to turn on a plot and to choose the type of plot needed. Enter the names of the lists in which the data is stored and choose the mark to be used for each data point. 3. Use ZoomStat to choose a viewing window that shows all of the data points. 4. Find the regression line using LinReg(ax+b) from the STAT CALC menu. 5. Enter the equation of the regression line in the Y= menu and use to show the relationship between the data and the regression line. GRAPH 6. Use the equation of the line to predict related outcomes. In this course, we have found a line of best fit by finding a line that seems to represent the data or by using a calculator. In more advanced courses in statistics, you will learn detailed methods for finding the line of best fit. The table below shows the number of calories and the number of grams of carbohydrates in a half-cup serving of ten different canned or frozen vegetables. Carbohydrates 9 23 Calories 45 100 4 20 5 19 25 110 8 35 12 50 7 30 13 70 17 80 EXAMPLE 1 Bivariate Statistics 719 a. Draw a scatter plot on graph paper. Let the horizontal axis represent grams of carbohydrates and the vertical axis represent the number of calories. b. Find the mean number of grams of carbohydrates in a serving of vegetables and the mean number of calories in a serving of vegetables. c. On the graph, draw a line that approximates the data in the table, and deter- mine its equation. d. Enter the data in L1 and L2 on your calculator and find the linear regression equation, LinReg(ax+b). e. Use each equation to find the expected number of calories in a serving of vegetables with 20 grams of carbohydrates. Compare the answers. Solution a. s e i r o l a C 120 110 100 90 80 70 60 50 40 30 20 0 5 10 15 20 25 Grams of Carbohydrates b. Enter the number of grams of carbohydrates in L1 and the number of calo- ries in L2. Find x– and y–, using 2-Var Stats. ENTER: STAT 2 ENTER The means of the x- and y-coordinates are x– 11.7 and y– 56.5. Locate the point (11.7, 56.5) on the graph. 720 Statistics c. 120 110 100 90 80 70 60 50 40 30 20 x, y) 10 20 5 Grams of Carbohydrates 15 25 The line we have drawn seems to go through the point (4, 20). We will use this point and the point with the mean values, (11.7, 56.5), to write an equation of a line of best fit. m 5 56.5 2 20 11.7 2 4 5 36.5 7.7 < 4.74 y 5 mx 1 b 20 5 36.5 7.7 (4) 1 b An equation of a best fit line is y 4.74x 1.05. 1.05 < b d. The data are in L1 and L2. ENTER: STAT 4 ENTER DISPLAY The equation of the regression line is y = 4.89x 0.660. e. Let x 20. Use the equation from c. y 4.74x 1.05 y 4.74(20) 1.05 y 95.85 Use the equation from d. y = 4.89x 0.660 y 4.89(20) 0.660 y 97.14 The two equations give very similar results. It would be reasonable to say that we could expect the number of calories to be about 96 or close to 100. Bivariate Statistics 721 EXERCISES Writing About Mathematics 1. a. Give an example of a set of bivariate data that has negative correlation. b. Do you think that the change in the independent variable in your example causes the change in the dependent variable? 2. Explain the purpose of finding a line of best fit. Applying Skills 3. When Gina bought a new car, she decided to keep a record of how much gas she uses. Each time she puts gas in the car, she records the number of gallons
of gas purchased and the number of miles driven since the last fill-up. Her record for the first 2 months is as follows: Gallons of gas 10 12 9 6 11 10 8 12 10 7 Miles driven 324 375 290 190 345 336 250 375 330 225 a. Draw a scatter plot of the data. Let the horizontal axis represent the number of gallons of gas and the vertical axis represent the number of miles driven. b. Does the data have positive, negative, or no correlation? c. Is this a causal relationship? d. Find the mean number of gallons of gasoline per fill-up. e. Find the mean number of miles driven between fill-ups. f. Locate the point that represents the mean number of gallons of gasoline and the mean number of miles driven. Use (0, 0) as a second point. Draw a line through these two points to approximate the data in the table. g. Use the line drawn in part d to approximate the number of miles Gina could drive on 3 gallons of gasoline. 4. Gemma made a record of the cost and length of each of the 14 long-distance telephone calls that she made in the past month. Her record is given below. Minutes 3.7 1.0 19.6 0.8 4.3 34.8 2.9 Cost $0.35 $0.11 $2.12 $0.09 $0.47 $3.78 $0.24 Minutes 2.5 7.1 10.9 5.8 1.5 1.4 8.0 Cost $0.27 $0.79 $1.21 $0.65 $0.20 $0.17 $0.89 a. Draw a scatter plot of the data on graph paper. Let the horizontal axis represent the number of minutes, and the vertical axis represent the cost of the call. b. Does the data have positive, negative, or no correlation? c. Is this a causal relationship? 722 Statistics d. Find the mean number of minutes per call. e. Find the mean cost of the calls. f. On the graph, draw a line of best fit for the data in the table and write its equation. g. Use a calculator to find the equation of the regression line. h. Approximate the cost of a call that lasted 14 minutes using the equation written in d. i. Approximate the cost of a call that lasted 14 minutes using the equation written in e. 5. A local store did a study comparing the cost of a head of lettuce with the number of heads sold in one day. Each week, for five weeks, the price was changed and the average number of heads of lettuce sold per day was recorded. The data is shown in the chart below. Cost per Head of Lettuce $1.50 $1.25 $0.90 $1.75 $0.50 No. of Heads Sold 48 52 70 42 88 a. Draw a scatter plot of the data. Let the horizontal axis represent the cost of a head of lettuce and the vertical axis represent the number of heads sold. b. Does the data have positive, negative, or no correlation? c. Is this a causal relationship? d. Find the mean cost per head. e. Find the mean number of heads sold. f. On the graph, draw a line that approximates the data in the table. g. What appears to be the result of raising the price of a head of lettuce? 6. The chart below shows the recorded heights in inches and weights in pounds for the last 24 persons who enrolled in a health club. Height Weight Height Weight Height Weight Height Weight 69 67 63 73 71 79 160 160 135 185 215 225 75 76 70 73 68 74 180 155 175 170 190 190 66 66 68 68 72 77 145 130 160 140 170 195 71 66 67 78 72 69 165 155 140 210 160 145 a. Draw a scatter plot on graph paper to display the data. b. Does the data have positive, negative, or no linear correlation? c. Is this a causal relationship? d. Draw and find the equation of a line of best fit. Use (77, 195) as a second point. e. Use a calculator to find the linear regression line. Bivariate Statistics 723 f. According to the equation written in d., if the next person who enrolls in the health club is 62 inches tall, what would be the expected weight of that person? g. According to the equation written in e., if the next person who enrolls in the health club weighs 200 pounds, what would be the expected height of that person? 7. The chart below shows the number of millions of cellular telephones in use in the United States by year from 1994 to 2003. Year Phones ’94 24.1 ’95 33.8 ’96 44.0 ’97 55.3 ’98 69.2 ’99 86.0 ’00 ’01 ’02 ’03 109.5 128.3 140.8 158.7 Let L1 be the number of years after 1990: 4, 5, 6, 7, 8, 9, 10, 11, 12, 13. a. Draw a scatter plot on graph paper to display the data. b. Does the data have positive, negative, or no linear correlation? c. Is this a causal relationship? d. Draw and find the equation of a line of best fit. e. On the graph, draw a line that approximates the data in the table, and determine its equation. Use (6, 44.0) as a second point. f. If the line of best fit is approximately correct for years beyond 2003, estimate how many cellular phones will be in use in 2007. 8. The chart below shows, for the last 20 Supreme Court Justices to have left the court before 2000, the age at which the judge was nominated and the number of years as a Supreme Court judge. Age Years Age Years 47 15 50 33 64 16 56 16 62 24 62 16 62 17 59 7 59 24 50 18 55 4 56 7 54 3 57 13 45 31 49 6 43 23 49 12 56 5 62 1 a. Draw a scatter plot on graph paper to display the data. b. Does the data have positive, negative, or no linear correlation? c. Is there a causal relationship? d. Draw and find an equation of a line of best fit. e. Use a calculator to find the linear regression line. f. Do you think that the data, the line of best fit, the regression line, or none of these could be used to approximate the number of years as a Supreme Court justice for the next person to retire from that office? 724 Statistics 9. A cook was trying different recipes for potato salad and comparing the amount of dressing with the number of potatoes given in the recipe. The following data was recorded. Number of Potatoes Cups of Dressing 7 11 2 4 7 8 2 3 4 8 11 4 6 1 7 13 4 5 11 8 4 3 4 a. Draw a scatter plot on graph paper to display the data. b. Does the data have positive, negative, or no linear correlation? c. Draw and find the equation of a line of best fit. Use d. Use a calculator to find the linear regression line. as a second point. 4, 7 8 A B e. According to the equation written in c., if the cook needs to use 10 potatoes to have enough salad, approximately how many cups of dressing are needed? CHAPTER SUMMARY Statistics is the study of numerical data. In a statistical study, data are col- lected, organized into tables and graphs, and analyzed to draw conclusions. Data can either be quantitative or qualitative. Quantitative data represents counts or measurements. Qualitative data represents categories or qualities. In an experiment, a researcher imposes a treatment on one or more groups. The treatment group receives the treatment, while the control group does not. Tables and stem-and-leaf diagrams are used to organize data. A table should have between five and fifteen intervals that include all data values, are of equal size, and do not overlap. A histogram is a bar graph in which the height of a bar represents the fre- quency of the data values represented by that bar. A cumulative frequency histogram is a bar graph in which the height of the bar represents the total frequency of the data values that are less than or equal to the upper endpoint of that bar. The mean, median, and mode are three measures of central tendency. The mean is the sum of the data values divided by the total frequency. The median is the middle value when the data values are placed in numerical order. The mode is the data value that has the largest frequency. Quartile values separate the data into four equal parts. A box-and-whisker plot displays a set of data values using the minimum, the first quartile, the median, the third quartile, and the maximum as significant measures. The percentile rank tells what percent of the data values lie at or below a given measure. In two-valued statistics or bivariate statistics, a relation between two different sets of data is studied. The data can be graphed on a scatter plot. The data may have positive, negative, or no correlation. Data that has positive or negative linear correlation can be represented by a line of best fit. The line of best fit can be used to predict values not in the included data set. Interpolation is predicting within the given data range. Extrapolation is predicting outside of the given data range. Review Exercises 725 VOCABULARY 16-1 Data • Statistics • Descriptive statistics • Qualitative data • Quantitative data • Census • Sample • Bias • Experiment • Treatment group • Control group • Placebo effect • Placebo • Blinding • Singleblind experiment • Double-blind experiment 16-2 Tally • Frequency • Total frequency • Frequency distribution table • Group • Interval • Grouped data • Range • Stem-and-leaf diagram • 16-3 Histogram • Frequency histogram 16-4 Average • Measures of central tendency • Mean • Arithmetic mean • Numerical average • Median • Mode • Bimodal • Linear transformation of a data set 16-5 Group mode • Modal interval 16-6 Quartile • Lower quartile • First quartile • Second quartile • Upper quartile • Third quartile • Box-and-whisker plot • Five statistical summary • Percentile • Cumulative frequency • Cumulative frequency histogram 16-7 Univariate statistics • Two-valued statistics • Bivariate statistics • Scatter plot • Correlation • Causation • Time series • Line of best fit • Regression line • Extrapolation • Interpolation REVIEW EXERCISES 1. Courtney said that the mean of a set of consecutive integers is the same as the median and that the mean can be found by adding the smallest and the largest numbers and dividing the sum by 2. Do you agree with Courtney? Explain why or why not. 2. A set of data contains N numbers arranged in numerical order. a. When is the median one of the numbers in the set of data? b. When is the median not one of the numbers in the set of data? 3. For each of the following sets of data, find: a. the mean b. the median c. the mode (if one exists) (1) 3, 4, 3, 4, 3, 5 (3) 9, 3, 2, 8, 3, 3 (2) 1, 3, 5, 7, 1, 2, 4 (4) 9, 3, 2, 3, 8, 2, 7 4. Express, in terms of y, the mean of 3y 2 and 7y 18. 726 Statistics 5. For the following data: 78, 91, 60, 65, 81, 72, 78, 80, 65, 63, 59, 78, 78, 54,
87, 75, 77 a. Use a stem-and-leaf diagram to organize the data. b. Draw a histogram, using 50–59 as the lowest interval. c. Draw a cumulative frequency histogram. d. Draw a box-and-whisker plot. 6. The weights, in kilograms, of five adults are 53, 72, 68, 70, and 72. a. Find: (1) the mean (2) the median (3) the mode b. If each of the adults lost 5 kilograms, find, for the new set of weights: (1) the mean (2) the median (3) the mode 7. Steve’s test scores are 82, 94, and 91. What grade must Steve earn on a fourth test so that the mean of his four scores will be exactly 90? 8. From Monday to Saturday of a week in May, the recorded high temperature readings were 72°, 75°, 79°, 83°, 83°, and 88°. For these data, find: a. the mean b. the median c. the mode 9. Paul worked the following numbers of hours each week over a 20-week period: 15, 3, 7, 6, 2, 14, 9, 25, 8, 12, 8, 8, 15, 0, 8, 12, 28, 10, 14, 10 a. Organize the data in a frequency table, using 0–5 as the lowest interval. b. Draw a frequency histogram of the data. c. In what interval does the median lie? d. Which interval contains the lower quartile? Score Frequency Cumulative Frequency 10. The table shows the scores of 25 test papers. a. Is the data univariate or bivariate? b. Find the mean score. c. Find the median score. d. Find the mode. e. Copy and complete the table. 100 f. Draw a cumulative frequency histogram. g. Find the percentile rank of 90. 60 70 80 90 1 9 8 2 5 h. What is the probability that a paper chosen at random has a score of 80? Review Exercises 727 11. The electoral votes cast for the winning presidential candidate in elections from 1900 to 2004 are as follows: 292, 336, 321, 435, 277, 404, 382, 444, 472, 523, 449, 432, 303, 442, 457, 303, 486, 301, 520, 297, 489, 525, 426, 370, 379, 271, 286 a. Organize the data in a stem-and-leaf diagram. (Use the first digit as the stems, and the last two digits as the leaves.) b. Find the median number of electoral votes cast for the winning candi- date. c. Find the first-quartile and third-quartile values. d. Draw a box-and-whisker plot to display the data. 12. The ages of 21 high school students are shown in the table at the right. a. What is the median age? b. What is the percentile rank of age 15? c. When the ages of these 21 students are combined with the ages of 20 additional students, the median age remains unchanged. What is the smallest possible number of students under 16 in the second group? 13. For each variable, determine if it is qualitative or Age Frequency 18 17 16 15 14 13 1 4 2 7 2 5 quantitative. a. Major in college b. GPA in college c. Wind speed of a hurricane d. Temperature of a rodent e. Yearly profit of a corporation f. Number of students late to class g. Zip code h. Employment status 14. Researchers looked into a possible relationship between alcoholism and pneumonia. They conducted a study of 100 current alcoholics, 50 former alcoholics, and 1,000 non-alcoholics who were hospitalized for a mild form of pneumonia. The researchers found that 30% of alcoholics and 30% of former alcoholics, versus only 15% of the non-alcoholics developed a more dangerous form of pneumonia. The researchers concluded that alcoholism raises the risk for developing pneumonia. Discuss possible problems with this study. 728 Statistics 15. Aurora buys oranges every week. The accompanying table lists the weights and the costs of her last 10 purchases of oranges. Weight (lb) 2.2 1.2 3.6 4.5 1.0 2.5 1.8 5.0 3.5 1.7 Cost ($) 1.22 0.60 1.04 1.58 0.50 0.89 0.95 1.88 1.46 0.70 a. Is the data univariate or bivariate? b. Draw a scatter plot of the data on graph paper. Let the horizontal axis represent the weights of the oranges and the vertical axis the costs. c. Is there a correlation between the weight and the cost of the oranges? If so, is it positive or negative? d. If the price is determined by the number of oranges purchased, do the variables have a causal relationship? Explain your answer. e. On the graph, draw a line of best fit that approximates the data in the table and write its equation. f. Use the equation written in d to approximate the cost of 4 pounds of oranges. 16. Explain why the graph on the right is misleading. (Hint: In accounting, numbers enclosed by parentheses denote negative numbers.) NET INCOME OF XYZ COMPANY (in thousands) $12,000 $8,050 $7,100 $5,123 $(4,000) 2002 2003 2004 2005 2006 Exploration a. Marny took the SAT in 2004 and scored a 1370. She was in the 94th percentile. Jordan took the SAT in 2000 and scored 1370. He was in the 95th percentile. Explain how this is possible. b. Taylor’s class rank stayed the same even though he had a cumulative grade point average of 3.4 one semester and 3.8 the next semester. Explain how this is possible. CUMULATIVE REVIEW Part I Cumulative Review 729 CHAPTERS 1–16 Answer all questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. 1. When the domain is the set of integers, the solution set of the inequality 0 , 0.1x 2 0.4 # 0.2 (1) { } is (2) {4, 5} 2. The product (2a 3)(2a 3) can be written as (3) {4, 5, 6} (4) {5, 6} (1) 2a2 9 (2) 4a2 9 (3) 4a2 9 (4) 4a2 12a 9 3. When 0.00034 is written in the form 3.4 10n, the value of n is (4) 4 (1) 3 (2) 4 (3. When (1) 5 , x equals (2) 1 (3) 1 2 5. The mean of the set of even integers from 2 to 100 is (1) 49 (2) 50 (3) 51 (4) 1 (4) 52 6. The probability that 9 is the sum of the numbers that appear when two dice are rolled is (1) 4 6 (2) 2 36 (3) 2 6 (4) 4 36 7. If the circumference of a circle is 12 centimeters, then the area of the circle is (1) 36 square centimeters (2) 144 square centimeters (3) (4) 36 p 144 p square centimeters square centimeters 8. Which of the following is not an equation of a function? (1) y 3x 2 (2) y x2 3x 1 9. The value of 10P8 is (1) 80 (2) 90 (3) y2 x (4) y x (3) 1,814,400 (4) 3,628,800 10. Which of the following is an equation of a line parallel to the line whose equation is (1) 2x y 7 y 5 22x 1 4 ? (2) y 2x 7 (3) 2x y 7 (4) y 2x 7 Part II Answer all questions in this part. Each correct answer will receive 2 credits. Clearly indicate the necessary steps, including appropriate formula substitu- 730 Statistics tions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 11. In a bridge club, there are three more women than men. How many persons are members of the club if the probability that a member chosen at random, is a woman is ? 3 5 12. Find to the nearest degree the measure of the smallest angle in a right tri- angle whose sides measure 12, 35, and 37 inches. Part III Answer all questions in this part. Each correct answer will receive 3 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 13. The lengths of the sides of a triangle are in the ratio 3 : 5 : 6. The perimeter of the triangle is 49.0 meters. What is the length of each side of the triangle? 14. Huy worked on an assignment for four days. Each day he worked half as long as he worked the day before and spent a total of 3.75 hours on the assignment. a. How long did Huy work on the assignment each day? b. Find the mean number of hours that Huy worked each day. Part IV Answer all questions in this part. Each correct answer will receive 4 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 15. The perimeter of a garden is 16 feet. Let x represent the width of the garden. a. Write an equation for the area of the land, y, in terms of x. b. Sketch the graph of the equation that you wrote in a. c. What is the maximum area of the land? 16. Each morning, Malcolm leaves for school at 8:00 o’clock. His brother Marvin leaves for the same school at 8:15. Malcolm walks at 2 miles an hour and Marvin rides his bicycle at 8 miles an hour. They follow the same route to school and arrive at the same time. a. At what time do Malcolm and Marvin arrive at school? b. How far is the school from their home? INDEX A Abscissa, 76 Absolute value, 6–7, 55 Absolute value function, 382–386 transformations of, reflection, 386 scaling, 386 translation, 385–386 Accuracy, 31 Acute angle, 249 Acute triangle, 263 Addition, 38 of algebraic expressions, 168–171 of algebraic fractions, 550–554 associative property of, 47–48 commutative property of, 47 distributive property of multiplication over, 48–49 of radicals, 487–489 of like radicals, 487 of unlike radicals, 488–489 of signed numbers, 54–58 with opposite signs, 56–58 with same signs, 54–55 verbal phrases involving, 89 Addition method in solving systems of linear equations, 416–420 Addition property of equality, 118 of inequality, 147–148 of zero, 49–50 Additive identity, 49–50 Additive inverse (opposite), 50 Adjacent angles, 250 Adjacent side of an angle, 307 Algebraic equation, 565 Algebraic expression(s), 89 addition of, 168–171 evaluating, 100–102 subtraction of, 168–171 terms in, 95–96, 168 like, 123, 168 unlike, 123, 168 translating verbal phrases into, 89–90, 92–93 writing in words, 98–99, Algebraic fraction(s), 540 addition of, 550–554 division of, 548–549 multiplication of, 545–547 subtraction of, 550–554 Algebraic solution of quadratic-linear Area system, 529–531 Alternate exterior angle(s), 259 parallel lines and, 260 Alternate interior angle(s), 258 parallel lines and, 259 Angle(s), 248–249 acute, 249 adjacent, 250 alternate exterior, 259 alternate interior, 258 base, 265 classifying triangles according to, 263 complementary, 250–251 congruent, 252 consecutive, 272 corresponding, 259 cosine of, 317 definition of, 248 of depression, 313 of elevation, 313 exterior, 258 interior, 258 on the same side of transversal, 2
59 linear pair of, 252 measuring, 248 obtuse, 249 opposite, 272 pairs of, 250–254 perpendicular, 249 parallel lines and, 258–261 parallelograms and, informal proofs, 274–275 polygons and, 275 of quadrilaterals, 273 right, 248 sides of an, 248 sine of, 317 straight, 249 sum of the measures of, for polygons, 275 for triangles, 263–265 supplementary, 251 tangent of, 308 types of, 248–249 vertical, 252–253 Angle bisector, 271 Approximation, 20–21 of numbers, 469 rational, 20, 477–478 of irregular polygons, 279–280 surface, 282–284 Arithmetic averages in, 680 order of operations in, 38–43 Arithmetic mean, 681 Ascending order, polynomials in, 170 Associative property of addition, 47–48 of multiplication, 48 Average(s). See also Central tendency in arithmetic, 680 numerical, 681 in statistics, 680–681 Axiom (postulate), 246 Axis, graphing line parallel to, 352–353 Axis of symmetry of the parabola, 510 B Bar (fractional line), 41 Base(s), in geometry, angles of an isosceles triangle, 265 of a prism, 282 of a trapezoid, 275 of an isosceles triangle, 265 in number theory, division of powers with the same, 186–187 of exponents, 39–40 multiplication of powers with the same, 173–174 in percent, 227 powers of, 95–96 Bias, 662 Biased object, 579 Bimodal data, 683 Binary operation, 38 Binomial(s), 169 division of polynomial by, 200–201 multiplication of, 454–455 Bisector angle, 271 perpendicular, 271 Bivariate statistics, 710–720 causation, 711–715 correlation, 711–715 line of best fit, 715–720 extrapolation, 717 interpolation, 717 732 Index Bivariate statistics, line of best fit cont. regression line, 715 scatter plot, 710–715 time series, 715 Blinding, 663 single-blind experiments, 663 double-blind experiments, 664 Box-and-whisker plot, 699–701 Brackets, as grouping symbol, 41 C Calculated probability, 585. See also Theoretical probability Calculator, algebraic expressions, entering variables, 101–102, 130 equations, intersect feature, 526–527 solving systems of equations, 414, 526–527 testing for equality, 129–130, 218 graphing functions on, 349–350, 352 absolute value, 384 CALC, 513–514 exponential, 390 quadratic, 512–515 TBLSET, 515 TRACE, 349–350, 414 vertex, finding, 512–514 ZOOM, 349–350 number theory, additive inverses (opposites), 50 decimals, 13–14 converting to a fraction, 52, 209–210 FRAC, 52, 209–210 exponents, 40, 67 fractions, comparing, 12 converting to a decimal, 13–14 mixed, 57–58 grouping symbols, 42–43 multiplicative inverses (reciprocals), 51 negative symbol, 61 numerical expressions, 3–4 reciprocals, 51 roots, general, 472–475 square roots, 18–19 scientific notation, 193–195 squaring numbers, 18, 470–472 probability, combinations, 642 factorials, 628–629 permutations, 631–634 statistics, 1-Var Stats, 684, 695 box-and-whisker plot, 700 five statistical summary, 700 frequency histogram, 676–678 line of best fit, 715–716 mean, 684 median, 684 scatter plot, 711–712 trigonometry, cosine ratios, 319–320 degree mode, 309 sine ratios, 319–320 tangent ratios, 308–311 Cancellation, 542 Cancellation method, 545 Cards, standard deck of, 580 Causation, 711 Check, in solving equations, 119 Census, 662 Center of a sphere, 286 Central tendency (average), 690–695 and linear transformations, 686 mean, 681 for grouped data, 691–692 for intervals, 694–695 median, 681–682 for grouped data, 691 for intervals, 694 mode, 682–683 for grouped data, 690 for intervals (modal interval), 693 Certainty, 591 Closure, property of, 45–47 Coefficient, 95 Combination(s), 639–644, 646 comparing permutations and, 639–641 relationships involving, 641–644 Commas in verbal phrases, 90 Common denominator, 551 Common fraction, 13 Common monomial factor, 447 Commutative property of addition, 47, 170 of multiplication, 47 Compass, 270 Complement of a set, 73 Complementary angles, 250–251 Completeness property of real numbers, 25 Composite number, 39, 443 Compound event, 609 Compound interest, 389 Computations with more than one operation, 41 Conditional equation, 117 Conditional probability, 619–623 Cone, volume of, 288 Congruent line segments, 265 Congruent angles, 252 Consecutive angles, 272 Consistent system of equations, 410–411 Constant of variation, 222 Constant ratio, 222 Construction, geometric, 270 angle bisector, 271 congruent angles, 270 congruent line segments, 270 perpendicular bisector, 271 Continued ratio, 209–210 Control group, 663 Convergence, 579 Coordinate axes, 75 Coordinate plane, 75 locating a point on, 76–77 Coordinates of a point, 76 finding on a plane, 77–78 Correlation, 711–715 Corresponding angles, 259 parallel lines and, 259–260 Cosine of an angle, 317 Cosine ratio, 318 applications of, 323–324 finding, on calculator, 319–320 Counting numbers, 2, 11 Counting principle, 610–611, 612 Cross-multiplication, 217 Cross-product, 217 Cube root, 472 Cumulative frequency, 702–703 Cumulative frequency histogram, 703–707 Cumulative relative frequency, 578–579 Cylinder(s) right circular, 283 surface area of, 283 D Data, 661 collection of, 661–665 bias, 662 sampling, 662 techniques of, 662–663 grouped, 668–670, 690–695 organization of, 667–672 frequency distribution table, 668 grouped data, 668–670 cumulative frequency, 702–703 types of, univariate, 710 bivariate, 710 qualitative, 661 quantitative, 661 visualization of data, interpreting graphs of data, 664–665 types of, box-and-whisker plot, 699–701 cumulative frequency histogram, 703–707 scatter plot, 711–714 stem-and-leaf diagram, 670–672 histogram, 675–678 Decimal(s) equivalent, 586–587 expressing as rational numbers, 15 expressing rational numbers as, 13–14 in ordering real numbers, 26 periodic, 14 repeating, 14, 17, 587 terminating, 14, 586 Decimal fraction, 13 Decimal notation, changing scientific notation to, 193–195 Decrease, percent of, 230 Degree, 248 Denominator common, 551 least common (LCD), 552 Dependent system of equations, 412–414 Dependent events, 617 Dependent variable, 341 Depression, angle of, 313 Descending order, polynomials in, 170 Descriptive statistics, 661 Diagram stem-and-leaf, 670–672 tree, 609–611, 627 Diameter of a sphere, 286 Die, 580, 584 rolling of fair, 596 Difference of two perfect squares, factoring, 452–453 Digits, set of, 3 significant, 29–32, 144–145, 195, 280, 284, 289, 304 Dimensional analysis, 234 Direct measurement, 301 Direct variation, 222–224 constant of, 222 equation of, 375 graphing, 374–376 Directly proportional, 222 Disjoint sets, 72 Distance formula, 136 Distributive property, 48–49 of multiplication over addition or subtraction, 168, 178, 183 Division, 4, 38 of fractions, 548–549 of a monomial by a monomial, 197–198 of a polynomial by a binomial, 200–201 by a monomial, 198–199 of powers that have the same base, 186–187 of signed numbers, 68–70 of square-root radicals, 494–495 verbal phrases involving, 90 Division property of equality, 118 of a fraction, 541 Domain, 89, 151 of a function, 341 of an inequality, 151 of an open sentence, 104–105 of a relation, 339 Double-blind experiment, 664 Double root, 506 E Elevation, angle of, 313 Empirical probability, 576–581 Empirical study, 577, 579 Empty set, 3, 72 Endpoint(s), 247–248 of a ray, 247 of a segment, 247 Equality addition property of, 118 division property of, 118 multiplication property of, 118 properties of, 117–119 subtraction property of, 118 Equally likely outcomes, 586 Equation(s), 117 conditional, 117 equivalent, 117 left side of, 117 right side of, 117 root of, 117 simplifying each side of, 122–126 solving linear, with more than one operation, 117–121 with the variable in both sides, 128–132 solution set of, 117 systems of, linear, 416–420, 422–424 quadratic-linear, 529–531 writing linear given slope and one point, 402 given the intercepts, 407–409 given two points, 404–405 Equiangular triangle, 263 Equilateral triangle, 265, 266 Equivalent decimals, 586–587 Equivalent equations, 117 Equivalent fractions, 541 Equivalent inequalities, 151 Equivalent ratios, 208–209 Error, 227 in geometric calculations, 289–290 percent of, 227–229 relative, 228 Estimation, radicals and, 477–478 Evaluating an algebraic expression, 100–102 Event(s), 584 compound, 609 dependent, 617 favorable, 584 independent, 611–612, 618 mutually exclusive, 600–603 probability of an, 592 singleton, 585, 590 unfavorable, 584 Everywhere dense, 12 Experimental design, 663–664 Experiment(s), 579, 663 double-blind, 664 in probability, 579–581 single-blind, 663 Exponent(s), 39, 40, 95–96 negative integral, 189–190 zero, 188–189 Exponential decay, 389 Exponential function, 387–391 exponential decay, 389–391 exponential growth, 388–391 Exponential growth, 388 Expression. See Algebraic expression(s); numerical expressions Exterior angle(s), 258 alternate, 259 Extraneous solutions, 565 Extrapolation, 717 Extremes, of a proportion, 216 F Face, of a solid, 282 Factorial n, 628 Factorial symbol (!), 628 Factoring, 443–445 a number, 443 over the set of integers, 443–445 a polynomial, 447, completely, 461–463 the difference of two perfect squares, 452–453 trinomials, 457–460 solving quadratic equations by, 503–507 Factor(s), 39, 443 common monomial, 447 Index 733 greatest common, 444–445 greatest common monomial, 447 of a term, 95 Fair and unbiased objects, 579 Favorable event, 584 Finite set, 3, 339–340 First-degree equations in one variable, 122 graphing, 346–350, 370–373 parallel to axes, 352–353 solving with more than one operation, 117–121 with the variable in both sides, 128–132 systems of, consistent, 410–411 dependent, 412–414 inconsistent, 411 independent, 411 solving, graphically, 410–414 using substitution, 422–424 using the addition method, 416–420 writing given slope and one point, 402 given the intercepts, 407–409 given two points, 404–405 First-degree inequalities in two variables, graphing, 378–381 First quartile, 698 Five statistical summary, 699 FOIL method, 183 in multiplication of binomials, 454 in multiplication of polynomials, 183 Formula(s), 107, 143 distance, 136 in problem solving, 134–137 for sur
face area, 293 transforming, 143–145 for volume, 293 writing, 107–108 Fractional coefficients solving equations with, 556–559 solving inequalities with, 562–563 Fractional equation(s), 565 solving, 565–567 Fractional expression, 540 Fraction line, as grouping symbol, 41 Fraction(s) addition of, 550–554 algebraic, 540 common, 13 decimal, 13 division of, 548–549 division property of, 541 equivalent, 13–14, 541, 216 lowest terms, 541 multiplication of, 545–547 multiplication property of a, 542 reducing to lowest terms, 541–543 subtraction of, 550–554 writing probability of events as, 586 Frequency distribution table, 668 Frequency histogram, 676 cumulative, 703–707 734 Index Function, 341. See also Absolute value function; Exponential function; Linear function; Quadratic function domain of a, 341 range of a, 341 G Geometric calculations, error in, 289–290 Geometry angles in, 248–249, 258–261 areas of irregular polygons in, 279–280 half-lines in, 247 line segments in, 247 lines in, 246–247 parallel lines in, 258–261 perpendicularity in, 249 planes in, 246 points in, 246 quadrilaterals in, 272–276 rays, 247–248 surface areas of solids in, 282–284 triangles in, 262–267 undefined terms in, 246 volumes of solids in, 286–290 Graph(s), interpreting, 664–665 of ordered pairs, 76, 610 of a point, 6 of a polygon, 78–79 and root finding, 522–524 of sets, intersection, 153–154 union, 154–155 and solving systems of equations, linear, 410–414 quadratic-linear, 525–527 types of, absolute value, 382–386 direct variation, 374–376 exponential, 387–391 inequalities, 378–381 linear, 346–350, 352–353, 370–373 quadratic, 508–516 Graphic solution of a quadratic-linear system, 525–527 Histogram, 675–678 cumulative frequency, 703–707 frequency, 676 Horizontal number line, 6 Hypotenuse, 263, 301, 307 I Identity, 117 additive, 49–50 multiplicative, 50, 179 Identity element of addition, 49–50 of multiplication, 50 Impossibility, 591 Inconsistent system of equations, 411 Increase, percent of, 230 Independent events, 611–612, 618 Independent system of equations, 411 Independent variable, 341 Index, of a radical, 473 Indirect measurement, 301 Inequalities domain (replacement set) of, 151 equivalent, 151 finding and graphing the solution of, 151–155 with fractional coefficients, 562–563 graphing first-degree, in one variable, 151–155 in two variables, 378–381 systems of, 431–434 in problem solving, 157–159 properties of, 146–149 addition, 147–148 multiplication, 148–149 order, 146–147 transitive property, 147 solution set of, 151 symbols of, 7–8 verbal phrases involving, 157–159 Infinite set, 3, 340–344 Integer(s), 2–8 ordering, 6 set of, 4–5 subsets of, 5 Graphing calculator. See also Calculator Greatest common factor (GCF), Intercept form of a linear equation, 367 Intercept(s) 444–445 Greatest common monomial factor, 447 Greatest possible error (GPE), 29 Grouped data, 668–670 calculator solution for, 692–693 mean of a set of, 691–692 measures of central tendency and, 690–695 median of a set of, 691 mode of a set of, 690 Grouping symbols expressions with, 41–43 multiplication and, 179–180 Group mode, 693 Group, 669 H Half-line, 247 Half-plane, 378 of a line, 366–367 writing equation given, 407–409 Interest, compound, 389 Interior angle(s), 258 alternate, 258 on the same side of the transversal, 258 Interpolation, 717 Intersection of sets, 71–72, 597 graphing, 153–154 Interval, 669 containing the mean, 694–695 containing the median, 694 containing the mode (modal interval), 693 Interval notation, 151–152 Inverse operation, using, in dividing signed numbers, 68–69 Inverse additive, 50 multiplicative, 51–52, 548 Irrational numbers, 17–23 basic rules for radicals that are, 478–481 radicals and, 476–481 set of, 17–18 Irregular polygons, area of, 279–280 Isosceles trapezoid, 276 Isosceles triangle, 265–266 base angles of, 265 base of, 265 legs of, 265 vertex angle of, 265 L Latitude, 75 Leaf, in a stem-and-leaf diagram, 670 Least common denominator (LCD), 552 Left member of an equation, 117 Left side of an equation, 117 Legs of isosceles triangles, 265 of right triangles, 263, 307 Letters, using, to represent numbers, 89–90 Length of a line segment, 247 Like radicals, 487 addition of, 487 subtraction of, 487 Like terms (similar terms), 123, 168 Line, in geometry, 246–247. See also Linear function Line of best fit, 715–720 Line segment, 247 Linear equation, 347. See also Firstdegree equation Linear function, graphing, using slope, 370–373 using solutions, 346–350 intercepts of, 366–367 x-intercept, 366–369 y-intercept, 366–369 slope of, 355–360 finding, 355–357 parallel, 258–261, 352–353, 363 perpendicular, 249, 364–365 transformations of, reflection, 372 scaling, 372 translation, 372 Linear growth, 387 Linear pair of angles, 252 Linear regression. See Bivariate statistics Linear system of equations, 410 solving, addition method, 416–420 graphically, 410–415 substitution method, 422–424 types of, consistent, 411 dependent, 412 inconsistent, 411 independent, 411 using to solve verbal problems, 426–428 Linear transformation of a set of data, 686 Line segment(s), 247 congruent, 265 length of, 247 List of ordered pairs, 609 Longitude, 75 Lower quartile, 698 Lowest terms, 212 reducing fractions to, 541–543 Lowest terms fraction, 541 M Mathematics, defined, 2 Maximum, 511 Mean, 681 arithmetic, 681 interval containing, 694–695 of a set of grouped data, 691–692 Measure of a line segment, 247 Measurement(s) accuracy of, 31 changing units of, 234–236 direct, 301 indirect, 301 numbers as, 28–32 precision of, 30–31 Measures of central tendency. See Central tendency Median, 681–682 interval containing, 694 of set of grouped data, 691 Member of a set (∈), symbol for, 338 Minimum, 509 Minus symbol, 61–62 Modal interval, 693 Mode, 682–685 of set of grouped data, 690 Monomial, 169 division of a, by a monomial, 197–198 division of a polynomial by a, 198–199 finding the principal square root of a, 482–483 multiplication of a monomial by a, 177–178 multiplication of a polynomial by a, 178–179 square of a, 449 Monomial square roots, division of, 494–495 Multiplication, 3–4, 38 associative property of, 48 of binomials, 454–455 commutative property of, 47 distributive property of, over addition, 49, 168, 178, 183 of fractions, 545–547 grouping symbols and, 179–180 of monomial by a monomial, 177–178 of polynomials, 183–184 by a monomial, 178–179 of powers that have the same base, 173–174 of signed numbers, 64–67 of square-root radicals, 491–492 of sum and difference of two terms, 450–451 verbal phrases involving, 89–90 Multiplication property of equality, 118 of a fraction, 542 of inequality, 148–149 of one, 50 of zero, 52 Multiplicative identity, 50, 179 Multiplicative inverse (reciprocal), 51–52, 548 Mutually exclusive events, 600–603 N n factorial, 628 Natural numbers, 2 Negative integral exponent, 189–190 Negative numbers, 4 absolute value of, 7 Negative slope, 358 Nonlinear function, 388 No slope, 358 Not a member of a set (∉), symbol for, 338 Notation decimal, 193–195 interval, 151–152 scientific, 191–195 set-builder, 338–344 Null set, 3 Numerical coefficient, 95 Number line, 2, 6 ordering real numbers on, 25 rational, 11–12 real, 25 standard, 6 Number pairs, graphing, 75–79 Number(s), 2 approximations of, 469 composite, 39, 443 counting, 2, 4, 5, 11 factoring, 443 irrational, 17–20, 469, 470 as measurements, 28–32 natural, 2 negative, 4 positive, 4 prime, 39, 443 rational, 11–15, 470–475 real, 2, 25–26 representing two, with the same variable, 125–126 symbols for, 2 using letters to represent, 89–90 whole, 3 writing, in scientific notation, 192–193 Number systems, 1–36 integers in, 2–8 irrational numbers in, 17–20 measurements and, 28–32 rational numbers in, 11–15 real numbers in, 2, 25–26 Numeral, 2 Numerical average, 681 Numerical expression(s), 3–4 with grouping symbols, 41–43 simplifying, 3, 42–43 O Obtuse angle, 249 Obtuse triangle, 263 One, multiplication property of, 50 Open sentence, 104–105 Index 735 Operation(s) binary, 38 computations with more than one, 41 order of, 38–43 properties of, 45–52 with sets, 71–73 solving equations using more than one, 117–121 Operational symbols, 3 Opposite angles, 272 Opposite rays, 248 Opposite (additive inverse), 4–5, 50 Opposite side of an angle, 307 Ordered pair(s), 38, 75 graph of, 610 list of, 609 Order of operations, 38–43 Order property of real numbers, 146–147 Ordinate, 75 Origin, 75 Outcome(s), 584 equally likely, 586 P Palindrome, 37 Parabola, 509 axis of symmetry of the, 510 Parallel lines, 258–261 alternate exterior angles and, 260 alternate interior angles and, 259 corresponding angles and, 259–260 slope of, 363 to the x-axis, 352 to the y-axis, 353 Parallelograms, 272 family of, 273–274 informal proofs for statements about angles in, 274–275 Parentheses as grouping symbol, 41 Percent, 227 of decrease, 230 of error, 227–229 of increase, 230 Percentage, 227 Percentiles, 701–702 Perfect squares, 449, 471 factoring the difference of two, 452–453 square root of, 471 Perfect square trinomial, 455 Perimeter, 134 Periodic decimals, 14 Permutations, 627–634, 646 calculator and, 631–634 comparing with combinations, 639–641 with repetition, 636–637 representing, 629–630 symbols for, 631 that use some of the elements, 630–631 Perpendicular bisector, 271 Perpendicularity, 249 Perpendicular lines, 249 slope of, 364–365 Pi (p), 20 Placebo, 663 Placebo effect, 663 736 Index Placeholder, 89 Plane, 246 points on, 75–76 finding the coordinates of, 77–78 Plane divider, 378 Plot box-and-whisker, 699–701 scatter, 710, 711 Point(s), 246 coordinates of, 75 finding the coordinates of, on a plane, 77–78 locating, on the coordinate plane, 76–77 on a plane, 75–76 writing equation given slope and one, 402 writing equations given two, 404–405 Polygon(s), 78, 262 angles and, 275 area of irregular, 279–280 graphing, 78–79 sides of, 262 vertices of, 262 permutations and, 627–634 as sums, 605–606 theoretical, 584–587 with two or more activiti
es with replacement, 617–618 without replacement, 617 uniform, 586 writing as fractions, 586 Problem solving formulas in, 134–137 inequalities in, 157–159 tangent ratio in, 313–315 trigonometric ratios in, 327–328 Product, 443 Properties of equality, 117–119 addition property of equality, 118 division property of equality, 118 multiplication property of equality, 118 substitution principle, 118 subtraction property of equality, 118 Properties of inequalities, 146–149 addition property of inequality, Polynomial equation of degree two, 147–148 503 multiplication property of inequality, Polynomial function, second-degree, 148–149 509 Polynomial(s), 169–170 in ascending order, 170 in descending order, 170 division of by a binomial, 200 by a monomial, 198–199 factoring, 447 factoring completely, 461–463 multiplication of, 183–184 by a monomial, 178–179 prime, 447 Positive numbers, 4 absolute value of, 7 Positive slope, 357 Postulate, 246 Power(s), 39, 40, 95–96 division of, 186–187 finding a power of a, 174–176 finding the product of, 173–174 Powers of 10, 192 Precision, 30–31 Price, unit, 212 Prime number, 39, 443 Prime polynomial, 447 over the set of integers, 461 Principal square root, 471–472 finding, of a monomial, 482–483 Prism, 282–283 right, 282–283 Probabilities, 575–659 of (A and B), 596–597 of (A or B), 599–603 of (not A), 605–606 of any event, 592 calculated, 585 combinations and, 639–644 conditional, 619–623 defined, 576 empirical, 576–581 evaluating simple, 590–594 experiments in, 579–581 order property of real numbers, 146–147 transitive property of inequality, 147 Properties of operations, 45–52 addition property of zero, 49–50 additive inverse, 50 associative of addition, 47–48 of multiplication, 48 closure, 45–47 commutative of addition, 47 of multiplication, 47 distributive, 48–49 multiplication property of one, 50 multiplication property of zero, 52 multiplicative inverses, 51–52 Proportion, 216–220 cross products in, 217 extremes of, 216 general form of, 216 inner terms of, 216 outer terms of, 216 products of the means and extremes in, 217 Pyramid(s), 288 volume of, 288 Pythagoras, 301 Pythagorean Theorem, 227, 301–304 statements of, 302–304 Pythagorean triple, 305 Q Quadrant, 76 Quadratic equation(s), 503 finding roots from a graph, 522–524 roots of, 504, 522 solving, by factoring, 503–507 standard form of, 503 Quadratic function, 509–518 axis of symmetry, 510–511 leading coefficient, 509–511 transformations of, reflection, 517–518 scaling, 517–518 translation, 517–518 turning point, 509 vertex, 509 Quadratic-linear system of equations, 525 solving, algebraically, 529–531 graphically, 525–527 Quadrilateral(s), 78, 272–276 angles of, 273 consecutive angles in, 272 opposite angles in, 272 special, 272–273 Qualitative data, 661 Quantitative data, 661 Quartile, 698–699 first, 698 lower, 698 second, 698 third, 698 upper, 698 R Radical(s), 470, 470–475 addition of like, 487 addition of unlike, 488–489 basic rules for irrational numbers as, 478–481 estimation and, 477–478 index of, 473 irrational numbers and, 476–481 simplifying square-root, 484–485 subtraction of like, 487 subtraction of unlike, 488–489 that are square roots, 471–472 Radical sign, 18, 470 Radicand, 470 Radius of a sphere, 286 Random selection, 586–588 Range, of data, 669 of a function, 341 of a relation, 339 Rate, 212, 227 unit, 212 using ratio to express, 212 Rational approximation, 20, 477–478 Rational expression, 540 Rational number line, 11–12 Rational numbers, 11–15, 470–475 expressing, as decimals, 13–14 expressing decimals as, 15 properties of, 12–13 set of, 11–15 Ratio(s), 208 constant, 222 continued, 209–210 equivalent, 208–209 expression of, in simplest form, 208–209 using, to express a rate, 212 verbal problems involving, 214–215 Ray(s), 247–248 opposite, 248 Real number line, 25 Real numbers, 2, 25–26 completeness property of, 25 ordering, 25–26 order property of, 146–147 set of, 25 using properties of, to multiply signed numbers, 65–67 Reciprocal(s), 51–52 in dividing signed numbers, 69–70 Rectangle, 272, 273 Rectangular solid, 283 Reduced to lowest terms, 541–543 Reflection rule for absolute value functions, 386 for linear functions, 372 for quadratic functions, 517 Regression line, 715–716. See also Linear regression Relation, 339 domain of, 339 range of, 339 Relative error, 228 Relative frequency, 577 cumulative, 578–579 Repeating decimals, 14, 17, 587 Repetition, permutations with, 636– 637 Replacement probability with, 617–618 probability without, 617 Replacement set, 89 of inequalities, 151 Rhombus, 273 Right angle, 248 Right circular cylinder, 283 Right member of an equation, 117 Right prism, 282–283 Right side of an equation, 117 Right triangle, 263, 301 hypotenuse of, 263, 301, 307 legs of, 263, 307 Root(s), 117 calculators and, 473–475 cube, 472 double, 505 of an equation, 504 finding, from a graph, 522–524 square, 470 Roster form, 338 Rounding, 21 Round-off error, 31 S Sample, 662 Sample space(s), 584 subscripts in, 592–594 Sampling, 662 techniques of, 662–663 Scalene triangle, 265 Scaling rules, for absolute value functions, 386 for linear functions, 372 for quadratic functions, 517 Scatter plot, 710–711 Scientific notation, 191–195 changing to ordinary decimal notation, 193–195 Index 737 writing numbers in, 192–193 Second-degree polynomial function, 509 Second quartile, 698 Segment of a line, 247 Set(s), 3, 71 in finding equations, 402 Solids surface area of, 282–284 volumes of, 286–290 Solutions disjoint, 72 empty, 3, 72 finite, 3, 339–340 infinite, 3, 340–344 of integers, 4–5 of irrational numbers, 17–18 null, 3 operations with, 71–73 intersection of, 71–72, 597 graphing, 153–154 complement of, 73 union of, 72–73 of rational numbers, 11–15 of real numbers, 25 types of, 3 graphing, 154–155 universal, 71 of whole numbers, 3 Set-builder notation, 338–339 Sides, of an angle, 248 classifying triangles according to, 265–266 of polygons, 262 Signed numbers, addition of, with opposite signs, 56–58 with same signs, 54–55 division of, 68–70 multiplication of, 64–67 subtraction of, 59–62 Significant digits, 29–32, 144–145, 195, 280, 284, 289, 304 rules for determining, 29–30 Similar terms, 123, 168 Similar triangles, 307 Simplest form, of a polynomial, 170 expression of ratio in, 208–209 of square-root radical, 485 Simplify a numerical expression, 3 Sine of an angle, 317 Sine ratio, 317–318 applications of, 323–324 finding, on a calculator, 319–320 Single-blind experiment, 663 Singleton event, 585, 590 Slope, 355–360 applications of, 360 graphing linear functions using, 370–373 of parallel lines, 363 of perpendicular lines, 364–365 slope-intercept form, 367–368 types of, negative, 358 positive, 357 undefined, 358 zero, 358 as unit rate of change, 356 writing equations given one point and, 402 Slope-intercept form of a linear equation, 368 extraneous, 565 graphing linear functions using their, 346–350 Solution set(s), 104 graphing, for system of inequalities, 431–434 of inequalities, 151–153 of open sentences, 104–105, 117 Sphere, 286 center of, 286 diameter of, 286 radius of, 286 volume of, 286 Square(s), 18, 273, 470 of monomials, 449 perfect, 449, 471 Square root(s), 18, 470 calculators and, 472 finding the principal, of a monomial, 482–483 of a perfect square, 471 principal, 471–472 radicals that are, 471–472 Square-root radicals division of, 494–495 multiplication of, 491–492 simplest form of, 485 simplifying, 484–485 Standard deck of cards, 580 Standard form, 503 first-degree equation in, 347 Standard number line, 6 Statistical summary (five statistical summary), 699 Statistics, 660–730 bivariate data, 710–720 correlation, 711–715 causation, 711–715 line of best fit, 715–720 extrapolation, 717 interpolation, 717 regression line, 715–720 central tendency (average), 690–695 and linear transformations, 686 mean, 681 for grouped data, 691–692 for intervals, 694–695 median, 681–682 for grouped data, 691 for intervals, 694 mode, 682–683 for grouped data, 690 for intervals (modal interval), 693 data collection, 661–665 bias, 662 sampling, 662 techniques of, 662–663 defined, 661 experimental design, 663–664 organization of data, 667–672 frequency distribution table, 668 grouped data, 668–670 cumulative frequency, 702–703 percentiles, 701–702 738 Index Statistics cont. quartiles, 698–699 five statistical summary, 699 sampling in, 662–663 visualization of data, types of, box-and-whisker plot, 699–701 cumulative frequency histogram, 703–707 histogram, 675–678 scatter plot, 711–714 stem-and-leaf diagram, 670–672 interpreting graphs of data, 664–665 univariate data, 710 Stem-and-leaf diagram, 670–672 Straight angle, 249 Straight line(s), 246–247 facts about, 246–247 Straightedge, 270 Subscripts, 592 in sample spaces, 592–594 Subsets of the integers, 5 Substitution method for solving a system of linear equations, 422–424 Substitution principle, 118, 423 Subtraction, 3, 38 of algebraic expressions, 168–171 of algebraic fractions, 550–554 distributive property of multiplication over, 168, 178, 183 of like radicals, 487 of signed numbers, 59–62 of unlike radicals, 488–489 verbal phrases involving, 89 Subtraction property of equality, 118 Successor, 2 Sums, probabilities as, 605–606 Supplementary angles, 251 Surface area of cylinders, 283 formulas for, 293 of rectangular solids, 283 of solids, 282–284 Symbol(s) approximately equal to (), 20 factorial (!), 628 grouping, 41–43 of inequality, 7–8 infinity (∞), 151–152 is an element of (∈), 338 is not an element of (∉), 338 minus (), 61–62 for numbers, 2 operational, 3 for permutations, 631 translating verbal phrases into, 91–93 vertical bar (), 338 System of equations. See Linear system of equations; Quadratic-linear system of equations System of inequalities, graphing the solution set of, 431–434 System of simultaneous equations, 410 T Table(s) frequency distribution, 668 preparing, 668 Tally mark, 668 Tangent of an angle, 308 Tangent ratio(s), 307–311 applications of, 313–317 an
gle of depression, 313 angle of elevation, 313 in problem solving, 313–315 finding, on a calculator, 308–311 Terminating decimals, 14, 586 Term(s), 95, 122, 168 factors of, 95 like, 123, 168 lowest, 212 multiplication of the sum and difference of two, 450–451 reducing fractions to lowest, 541–543 similar, 123, 168 undefined, 246 unlike, 123, 168 Theorem, 253 Theoretical probability, 584–587 Third quartile, 698 Time series, 715 Total frequency, 668 Transitive property of inequality, 147 Translation rules for absolute value functions, 385–386 for linear functions, 372 for quadratic functions, 517 Transversal, 258 Trapezoid, 272, 275–276 isosceles, 276 Treatment group, 663 Tree diagram, 609–611, 627 Trial, 579 Triangle(s), 262–267 classifying according to angles, 263 according to sides, 265–266 properties of special, 266–267 similar, 307 sides of, 262 sum of the measures of the angles of a, 263–265 types of, acute, 263 equiangular, 263 equilateral, 265–266 isosceles, 265–266 obtuse, 263 right, 263, 301 scalene, 265 vertices of, 262 Trigonometry, 300 cosine ratio in, 318–320, 323–324 problem solving using ratios, 327–328 Pythagorean Theorem in, 301–304 sine ratio in, 317–320, 323–324 tangent ratio in, 307–317 Trinomial(s), 170 factoring, 457–460 Turning point, 509 Two-valued statistics, 710. See also Bivariate statistics U Undefined term, 246 Unfavorable event, 584 Uniform probability, 586 Union of sets, 72–73 graphing, 154–155 Unit measure, 6 Unit price, 212 Unit rate, 212 of change, 356 Units of measure, changing, 234–236 Univariate statistics, 710 Universal set, 71 Unlike radicals addition of, 488–489 subtraction of, 488–489 Unlike terms, 123, 168 Upper quartile, 698 V Variable(s), 89 dependent, 341 first-degree equations in one, 122 graphing first-degree inequalities in two, 378–381 independent, 341 representing two numbers with the same, 125–126 solving, for in terms of another variable, 142–143 solving equations that have, in both sides, 128–132 writing algebraic expressions involving, 92 Verbal phrases commas in, 90 involving addition, 89 involving division, 90 involving multiplication, 89–90 involving subtraction, 89 translating, into symbols, 91–93 Verbal problems involving ratio, 214–215 using systems of equations in solving, 426–428 Vertex angle of an isosceles triangle, 265 Vertex of a parabola, 509 Vertex of an angle, 248 Vertical angles, 252–254 Vertical bar (\|), 338 Vertical number line, 6 Vertices, 78 of a polygon, 262 Volume of cones, 288 formulas for, 293 of pyramids, 286 of solids, 286–290 of spheres, 286 W Whole numbers, 3 set of, 3 With replacement, probability problems, 617–618 Without replacement, probability problems, 617 Words, writing algebraic expressions in, 98–99 X x-axis, 75 lines parallel to, 352 x-coordinate, 76 x-intercept, 366 writing equation given, 407–409 Y y-axis, 75 lines parallel to, 353 y-coordinate, 76 y-intercept, 366 slope and, 367–368 writing equation given, 407–409 Z Zero, 3 addition property of, 49–50 division of, by nonzero number, 69 multiplication property of, 52 Zero exponent, 188–189 Zero slope, 358 Index 739
iew Note: Persian mathematician Omar Khayyam would solve algebraic problems geometrically by intersecting graphs rather than solving them algebraically. 35 1.2 Practice - Two-Step Problems Solve each equation. 1) 5 + n 4 = 4 3) 102 = 7r + 4 − 8n + 3 = 5) − 7) 0 = 6v − 8 = x 9) − 11 − 13) 12 + 3x = 0 − 15) 24 = 2n 8 − 12 + 2r 17) 2 = − 3 + 7 = 10 19) b 77 − 6) 8) − − 4 − 2 + x 2) 2 = 2m + 12 − − 4) 27 = 21 3x − b = 8 2 = 4 10) 12) 14) 16) − − − − 18) − 20) x 5 = a 1 4 − 6 = 15 + 3p 5m + 2 = 27 37 = 8 + 3x 8 + n 12 = 7 − 8 8 = − 1 − 21) 152 = 8n + 64 23) − 16 = 8a + 64 25) 56 + 8k = 64 27) 29) − − 2x + 4 = 22 20 = 4p + 4 31) − 33 = − 40 = 4n 35) − 37) 87 = 3 7v − x + 1 = 39) − 32 − 11 − 8 + v 2 − 3 = − 3n = 22) 24) − − 11 = 2x 26) 4 − − 28) 67 = 5m − − 8 29 16 − 30) 9 = 8 + x 6 32) m 1 = 4 − − 80 = 4x 2 28 − 34) − 36) 33 = 3b + 3 3 = 38) 3x − 40) 4 + a 3 = 1 3 − 36 1.3 Solving Linear Equations - General Equations Objective: Solve general linear equations with variables on both sides. Often as we are solving linear equations we will need to do some work to set them up into a form we are familiar with solving. This section will focus on manipulating an equation we are asked to solve in such a way that we can use our pattern for solving two-step equations to ultimately arrive at the solution. One such issue that needs to be addressed is parenthesis. Often the parenthesis can get in the way of solving an otherwise easy problem. As you might expect we can get rid of the unwanted parenthesis by using the distributive property. This is shown in the following example. Notice the ﬁrst step is distributing, then it is solved like any other two-step equation. Example 59. 4(2x 8x − − 6) = 16 Distribute 4 through parenthesis 24 = 16 Focus on the subtraction ﬁrst + 24 + 24 Add 24 to both sides 8x = 40 Now focus on the multiply by 8 8 Divide both sides by 8 8 x = 5 Our Solution! Often after we distribute there will be some like terms on one side of the equation. Example 2 shows distributing to clear the parenthesis and then combining like terms next. Notice we only combine like terms on the same side of the equation. Once we have done this, our next example solves just like any other two-step equation. Example 60. 3(2x 6x − − 4) + 9 = 15 Distribute the 3 through the parenthesis 12 + 9 = 15 Combine like terms, 6x 12 + 9 Focus on the subtraction ﬁrst 3 = 15 − − + 3 + 3 Add 3 to both sides 6x = 18 Now focus on multiply by 6 37 6 6 Divide both sides by 6 x = 3 Our Solution A second type of problem that becomes a two-step equation after a bit of work is one where we see the variable on both sides. This is shown in the following example. Example 61. 4x − 6 = 2x + 10 Notice here the x is on both the left and right sides of the equation. This can make it diﬃcult to decide which side to work with. We ﬁx this by moving one of the terms with x to the other side, much like we moved a constant term. It doesn’t matter which term gets moved, 4x or 2x, it would be the author’s suggestion to move the smaller term (to avoid negative coeﬃcients). For this reason we begin this problem by clearing the positive 2x by subtracting 2x from both sides. however, 6 = 2x + 10 Notice the variable on both sides − 4x 2x 2x − 2x − 6 = 10 − + 6 + 6 2x = 16 2 2 x = 8 Subtract 2x from both sides Focus on the subtraction ﬁrst Add 6 to both sides Focus on the multiplication by 2 Divide both sides by 2 Our Solution! The previous example shows the check on this solution. Here the solution is plugged into the x on both the left and right sides before simplifying. Example 62. 4(8) 32 6 = 2(8) + 10 Multiply 4(8) and 2(8) ﬁrst − 6 = 16 + 10 − 26 = 26 Add and Subtract True! The next example illustrates the same process with negative coeﬃcients. Notice ﬁrst the smaller term with the variable is moved to the other side, this time by adding because the coeﬃcient is negative. 38 Example 63. 3x + 9 = 6x − + 3x + 3x 9 = 9x − − 27 Notice the variable on both sides, Add 3x to both sides Focus on the subtraction by 27 27 3x is smaller − + 27 + 27 Add 27 to both sides 36 = 9x 9 9 4 = x Focus on the mutiplication by 9 Divide both sides by 9 Our Solution Linear equations can become particularly intersting when the two processes are combined. In the following problems we have parenthesis and the variable on both sides. Notice in each of the following examples we distribute, then combine like terms, then move the variable to one side of the equation. Example 64. 2(x 2x − − 5) + 3x = x + 18 Distribute the 2 through parenthesis 10 + 3x = x + 18 Combine like terms 2x + 3x 5x x 4x 10 = x + 18 Notice the variable is on both sides − x − − 10 = 18 − + 10 + 10 4x = 28 4 4 x = 7 Subtract x from both sides Focus on the subtraction of 10 Add 10 to both sides Focus on multiplication by 4 Divide both sides by 4 Our Solution Sometimes we may have to distribute more than once to clear several parenthesis. Remember to combine like terms after you distribute! Example 65. 3(4x − 12x 5) − − 15 4(2x + 1) = 5 Distribute 3 and 8x 4 = 5 Combine like terms 12x − 4x − 19 = 5 − Focus on subtraction of 19 − + 19 + 19 Add 19 to both sides − 4 through parenthesis 8x and 4 15 − − 4x = 24 Focus on multiplication by 4 39 4 4 Divide both sides by 4 x = 6 Our Solution This leads to a 5-step process to solve any linear equation. While all ﬁve steps aren’t always needed, this can serve as a guide to solving equations. 1. Distribute through any parentheses. 2. Combine like terms on each side of the equation. 3. Get the variables on one side by adding or subtracting 4. Solve the remaining 2-step equation (add or subtract then multiply or divide) 5. Check your answer by plugging it back in for x to ﬁnd a true statement. The order of these steps is very important. World View Note: The Chinese developed a method for solving equations that involved ﬁnding each digit one at a time about 2000 years ago! We can see each of the above ﬁve steps worked through our next example. Example 66. 7) + 8x Distribute 4 and 3 through parenthesis 21 + 8x Combine like terms 24 + 9 and 3x + 8x 4(2x 8x 6) + 9 = 3(x − 24 + 9 = 3x − 8x 8x − − 15 = 11x 8x − 15 = 3x − − − + 21 21 − 21 − + 21 6 = 3x 3 3 2 = x − Notice the variable is on both sides Subtract 8x from both sides Focus on subtraction of 21 Add 21 to both sides Focus on multiplication by 3 Divide both sides by 3 Our Solution Check: 4[2(2) 4[4 − − 6] + 9 = 3[(2) 6] + 9 = 3[ − 5] + 8(2) − 7] + 8(2) Plug 2 in for each x. Multiply inside parenthesis Finish parentesis on left, multiply on right 40 4[ − − 2 15 + 8(2) 15 + 16 Finish multiplication on both sides Add True! When we check our solution of x = 2 we found a true statement, 1 = 1. Therefore, we know our solution x = 2 is the correct solution for the problem. There are two special cases that can come up as we are solving these linear equations. The ﬁrst is illustrated in the next two examples. Notice we start by distributing and moving the variables all to the same side. Example 67. 3(2x 6x 6x − − − − − − 5) = 6x 15 = 6x 6x − 15 = 15 − 15 Distribute 3 through parenthesis 15 Notice the variable on both sides Subtract 6x from both sides Variable is gone! True! 15 = Here the variable subtracted out completely! We are left with a true statement, 15. If the variables subtract out completely and we are left with a true − statement, this indicates that the equation is always true, no matter what x is. Thus, for our solution we say all real numbers or R. − Example 68. 2(3x 6x − − − 5) 10 2x 2x − − − 4x = 2x + 7 Distribute 2 through parenthesis 4x = 2x + 7 Combine like terms 6x 10 = 2x + 7 Notice the variable is on both sides 4x − 2x − 10 7 Subtract 2x from both sides Variable is gone! False! − Again, the variable subtracted out completely! However, this time we are left with a false statement, this indicates that the equation is never true, no matter what x is. Thus, for our solution we say no solution or ∅. 41 1.3 Practice - General Linear Equations Solve each equation. 1) 2 3) ( − 5( 3a 8) = 1 − − 4 + 2v) = − − 5) 66 = 6(6 + 5x) 50 − 7) 0 = 8(p 5) − − 2 + 2(8x 9) 11) − − 13) − 15) 1 − 17) 20 − 7) = 16 − 6 3x − − 8m + 7 21x + 12 = 1 7m = − 12r = 29 − 8r − 12b + 30 2v − − 32 2 7b = − 24v = 34 − 5(2 − − 19) 21) 23) 25) 29) 31) 33) − − − − − − 37) 39) 41) 43) 45) 47) 49) − − − − − − − − 27) 2(4x − 4m) = 33 + 5m 4n + 11 = 2(1 8n) + 3n 6v 29 = 5(v + 1) − 4v − − 4x 4) = 20 − − 1) = 39 5(8a 7a − − p + 1) + 2(6 + 8p) ( − a − 57 = − − − 2(m 2) + 7(m 8) = 67 − 35) 50 = 8 (7 + 7r) (4r + 6) − − − − − 61 = 2(8n − 3( − 7(x 5(5r 4) + 4(3r 4) − 4) = 8(1 n) − 7v + 3) + 8v = 5v − 4(1 − − 6v) 2) = 4 6(x 1) − − 6(8k + 4) = − − 8(6k + 3) − 7p) = 8(p 7) − 2(1 − 2 − 2) 2( 3n + 8) = 20 − 4 + 3x) = 34 − 4) 2 8( − 6) 32 = 2 − − 8) 10) 12) (3 3n − − − 14) 56p − 5( 4n + 6) − 55 = 8 + 7(k − 5) − 5n) = 12 27 = − 48 = 6p + 2 − 27 3n − 16) 4 + 3x = 12x + 4 − 16n + 12 = 39 7n 18) − 20) 17 2x = 35 − 8x − 7x = 6(2x − 25 − 7(1 + b) = 5 − − 8(8r 2) = 3r + 16 1) − 5b − 19 = − 8n − 4 + 4k = 4(8k − 8) − 22) 24) 26) 28) 30) 36) 38) 40) 42) 44) 46(8n 3) + 3n − 32) 16 = − 34) 7 = 4(n 5(1 6x) + 3(6x + 7) − 7) + 5(7n + 7) − 8(6 + 6x) + 4( 3 + 6x) = − − 3) 4 6(x 8) 4(x − − 4(1 + a) = 2a 2) = − 8(5 + 3a) − − 3) + 5 = 6(x − (n + 8) + n = 2 5(x − − − 8n + 2(4n 12 − 5) 4) − 48) − 50) 8( 5(x + 7) = 4( − 8n + 4) = 4( 8x 2) − 7n + 8) − − − 8(n 7) + 3(3n 3) = 41 76 = 5(1 + 3b) + 3(3b 42 1.4 Solving Linear Equations - Fractions Objective: Solve linear equations with rational coeﬃcients by multiplying by the least common denominator to clear the fractions. Often when solving linear equations we will need to work with an equation with fraction coeﬃcients. We can solve these problems as we have in the past. This is demonstrated in our next example. Example 69 Focus on subtraction Add 7 2 to both sides Notice we will need to get a common denominator to add 5 6 + 7 common denominator of 6. So we build up the denominator, 7 2 can now add the fractions: 2. Notice we have a = 21 6 , and we 3 3 3 4 x − 21 6
= 5 6 Same problem, with common denominator 6 + 21 6 + 21 6 Add 21 6 to both sides 3 4 3 4 x = x = 26 6 13 3 Reduce 26 6 to 13 3 Focus on multiplication by 3 4 We can get rid of 3 4 by dividing both sides by 3 same as multiplying by the reciprocal, so we will multiply both sides by 4 3. 4. Dividing by a fraction is the 4 3 3 4 4 3 x = x = 13 3 52 9 Multiply by reciprocal Our solution! While this process does help us arrive at the correct solution, the fractions can make the process quite diﬃcult. This is why we have an alternate method for dealing with fractions - clearing fractions. Clearing fractions is nice as it gets rid of the fractions for the majority of the problem. We can easily clear the fractions 43 by ﬁnding the LCD and multiplying each term by the LCD. This is shown in the next example, the same problem as our ﬁrst example, but this time we will solve by clearing fractions. Example 70 LCD = 12, multiply each term by 12 (12)3 4 x − (12)7 2 = (12)5 6 Reduce each 12 with denominators (3)3x − (6)7 = (2)5 Multiply out each term 9x Focus on subtraction by 42 42 = 10 − + 42 + 42 Add 42 to both sides 9x = 52 9 9 52 9 x = Focus on multiplication by 9 Divide both sides by 9 Our Solution The next example illustrates this as well. Notice the 2 isn’t a fraction in the origional equation, but to solve it we put the 2 over 1 to make it a fraction. Example 71 LCD = 6, multiply each term by 6 (6)2 3 x − (6)2 1 = (6)3 2 x + (6)1 6 Reduce 6 with each denominator (2)2x − (6)2 = (3)3x + (1)1 Multiply out each term 4x 4x − − − 12 = 9x + 1 Notice variable on both sides Subtract 4x from both sides Focus on addition of 1 Subtract 1 from both sides Focus on multiplication of 5 Divide both sides by 5 4x 12 = 5x + 1 1 1 − 13 = 5x 5 5 13 5 = x Our Solution − − − − We can use this same process if there are parenthesis in the problem. We will ﬁrst distribute the coeﬃcient in front of the parenthesis, then clear the fractions. This is seen in the following example. 44 Example 72. 5 9 x + 4 27 3 2 5 6 (18)18)3 9 (18)5 6 x + = 3 Distribute 3 2 through parenthesis, reducing if possible LCD = 18, multiply each term by 18 Reduce 18 with each denominator (3)5x + (2)2 = (18)3 Multiply out each term − 15x + 4 = 54 4 4 − 15x = 50 15 10 3 . 15 x = Focus on addition of 4 Subtract 4 from both sides Focus on multiplication by 15 Divide both sides by 15. Reduce on right side. Our Solution While the problem can take many diﬀerent forms, the pattern to clear the fraction is the same, after distributing through any parentheses we multiply each term by the LCD and reduce. This will give us a problem with no fractions that is much easier to solve. The following example again illustrates this process. Example 73 − − Distribute 1 3 , reduce if possible LCD = 4, multiply each term by 4. (4)3 4 x − (4)1 2 = (4)1 4 x + (4)2 1 − (4)7 2 Reduce 4 with each denominator (1)3x − (2)1 = (1)1x + (4)2 3x − 2 = x + 8 − 2 = x 3x − x x 2x (2)7 Multiply out each term 14 Combine like terms 8 6 Notice variable on both sides Subtract x from both sides Focus on subtraction by 2 − 2 = 14 − − 6 − − − + 2 + 2 Add 2 to both sides − 2x = 2 x = Focus on multiplication by 2 Divide both sides by 2 4 − 2 2 Our Solution − World View Note: The Egyptians were among the ﬁrst to study fractions and linear equations. The most famous mathematical document from Ancient Egypt is the Rhind Papyrus where the unknown variable was called “heap” 45 1.4 Practice - Fractions Solve each equation. 1) 3 5(1 + p) = 21 20 6 5 ) 5 4(x − − m = 113 24 5 4 3) 0 = 5) 3 4 − 7) 635 72 = 5 2( − − 11 4 + x) 9) 2b + 9 5 = 11 5 − n + 1) = 3 2 11) 3 2( 7 3 5 4( a − 8 3 a + 1) = 19 4 − 5 3) 4 3) − − 4 3 n 13) − 15) 55 17) 16 r 19) 21 11 3 + 3 2 2(b 5 3 ) − 5 2 x 23) ( − − 25) 45 16 + 19 16 n − 27) 3 2(v + 3 2) = 29) 47 − 19 6 x + 1) − 4) 3 2 n 8 3 = 29 12 − 4 + 3 4 − r = 163 32 6) 11 16 9 = − 4 3( 5 3 + n) − 10) 3 12) 41 14) 1 3( 13 8 − 83 24 − 2( 2 1 3 x 16) − − 3(m + 9 18) 2 4) 7 4 k + 1) 10 3 k = − 3 4) 7 2 x = − 10 3 = − 53 18 − x + 5 3 (x − 7 4) 20) 1 12 = 4 3 − n + 2(n + 3 2 ) 3 2 − 11 3 r= 22) 7 6 − 4 3 n = 24) 26) − − 149 16 − 2 ( 5 7 3 − 3 ) = 11 4 − a + 25 8 a + 1 7 4 r 5 4( 4 3 − r + 1) 1 2 28) 8 3 − − n + 29 30( 4 − n + 2 3) 3 2 3 ( 13 4 x + 1) − 46 1.5 Solving Linear Equations - Formulas Objective: Solve linear formulas for a given variable. Solving formulas is much like solving general linear equations. The only diﬀerence is we will have several varaibles in the problem and we will be attempting to solve for one speciﬁc variable. For example, we may have a formula such as A = πr2 + πrs (formula for surface area of a right circular cone) and we may be interested in solving for the varaible s. This means we want to isolate the s so the equation has s on one side, and everything else on the other. So a solution might look like s = A . This second equation gives the same information as the ﬁrst, they are algebraically equivalent, however, one is solved for the area, while the other is solved for s (slant height of the cone). In this section we will discuss how we can move from the ﬁrst equation to the second. − πs πr2 When solving formulas for a variable we need to focus on the one varaible we are trying to solve for, all the others are treated just like numbers. This is shown in the following example. Two parallel problems are shown, the ﬁrst is a normal onestep equation, the second is a formula that we are solving for x Example 74. 3x = 12 3 3 x = 4 In both problems, x is multiplied by something wx = z w w To isolate the x we divide by 3 or w. x = Our Solution z w We use the same process to solve 3x = 12 for x as we use to solve w x = z for x. Because we are solving for x we treat all the other variables the same way we would treat numbers. Thus, to get rid of the multiplication we divided by w. This same idea is seen in the following example. Example 75. m + n = p for n Solving for n, treat all other variables like numbers m n = p Subtract m from both sides Our Solution m m − − − As p and m are not like terms, they cannot be combined. For this reason we leave the expression as p m. This same one-step process can be used with grouping symbols. − 47 Example 76. y) = b − y) (x a(x (x − y) − for a Solving for a, treat (x y) like a number − Divide both sides by (x y) − a = x b − y Our Solution − Because (x y) is in parenthesis, if we are not searching for what is inside the parenthesis, we can keep them together as a group and divide by that group. However, if we are searching for what is inside the parenthesis, we will have to break up the parenthesis by distributing. The following example is the same formula, but this time we will solve for x. Example 77. − − y) = b for x Solving for x, we need to distribute to clear parenthesis a(x ax ay = b + ay + ay ax = b + ay a This is a two Add ay to both sides The x is multipied by a Divide both sides by a − step equation, ay is subtracted from our x term a b + ay a x = Our Solution Be very careful as we isolate x that we do not try and cancel the a on top and bottom of the fraction. This is not allowed if there is any adding or subtracting in the fraction. There is no reducing possible in this problem, so our ﬁnal reduced answer remains x = b + ay . The next example is another two-step problem a Example 78. y = mx + b for m Solving for m, focus on addition ﬁrst b − y Subtract b from both sides m is multipied by x. Divide both sides by x b − b = mx x b = m Our Solution − x y − x It is important to note that we know we are done with the problem when the variable we are solving for is isolated or alone on one side of the equation and it does not appear anywhere on the other side of the equation. The next example is also a two-step equation, it is the problem we started with at the beginning of the lesson. 48 Example 79. A = πr2 + πrs for s Solving for s, focus on what is added to the term with s − πr2 A − − πr A πr2 πr2 = πrs πr πr2 − πr Subtract πr2 from both sides s is multipied by πr Divide both sides by πr = s Our Solution Again, we cannot reduce the πr in the numerator and denominator because of the subtraction in the problem. Formulas often have fractions in them and can be solved in much the same way we solved with fractions before. First identify the LCD and then multiply each term by the LCD. After we reduce there will be no more fractions in the problem so we can solve like any general equation from there. Example 80. h = 2m n (n)2m n (n)h = nh = 2m 2 2 nh 2 = m for m To clear the fraction we use LCD = n Multiply each term by n Reduce n with denominators Divide both sides by 2 Our Solution The same pattern can be seen when we have several fractions in our problem. Example 81. (b)a b + + c b = e (b) a b (b)c b a + c = eb c c − c a = eb − − = e for a To clear the fraction we use LCD = b Multiply each term by b Reduce b with denominators Subtract c from both sides Our Solution Depending on the context of the problem we may ﬁnd a formula that uses the same letter, one capital, one lowercase. These represent diﬀerent values and we must be careful not to combine a capital variable with a lower case variable. Example 82. a = A − b 2 for b Use LCD (2 b) as a group − 49 (2 − b)a = b)A b (2 − 2 − Multiply each term by (2 b) − − − (2 2a 2a ab = A a b)a = A ab = A 2a − 2a − a − 2a − ) with denominator reduce (2 Distribute through parenthesis Subtract 2a from both sides The b is multipied by Divide both sides by a − a − Our Solution Notice the A and a were not combined as like terms. This is because a formula will often use a capital letter and lower case letter to represent diﬀerent variables. Often with formulas there is more than one way to solve for a variable. The next example solves the same problem in a slightly diﬀerent manner. After clearing the denominator, we divide by a to move it to the other side, rather than distributing. Example 83. a = for b Use LCD = (2 b) as a group − Multiply each term by (2 b) − Reduce (2 − Divide both sides by a b) with denom
inator Focus on the positive 2 Subtract 2 from both sides 2 Still need to clear the negative (2 − b)a = (2 − A 2 b − b)A (2 − b 2 − b))( ( − − b) = ( 1) − 1) Multiply (or divide) each term by 1 − Our Solution Both answers to the last two examples are correct, they are just written in a different form because we solved them in diﬀerent ways. This is very common with formulas, there may be more than one way to solve for a varaible, yet both are equivalent and correct. World View Note: The father of algebra, Persian mathematician Muhammad ibn Musa Khwarizmi, introduced the fundamental idea of blancing by subtracting the same term to the other side of the equation. He called this process al-jabr which later became the world algebra. 50 1.5 Practice - Formulas Solve each of the following equations for the indicated variable. 1) ab = c for b x = b for x 3) f g 5) 3x = a b for x 7) E = mc2 for m 9) V = 4 3 πr3 for π 11) a + c = b for c 13) c = 4y m + n for y 15) V = πDn 12 for D 17) P = n(p c) for n 19) T = D − d L for D − 21) L = Lo(1 + at) for Lo 23) 2m + p = 4m + q for m 25) k m − r = q for k 16t2 for v − 27) h = vt 29) Q1 = P (Q2 − 31) R = kA(T1 + T2) d 2) g = h i for h 4) p = 3y q for y 6) ym b = c d for y 8) DS = ds for D 10) E = mv2 for m 2 12) x f = g for x − rs 3 = k for r a 14) − 16) F = k(R L) for k − 18) S = L + 2B for L 20) I = Ea Eq − R for Ea 22) ax + b = c for x 24) q = 6(L p) for L − 26) R = aT + b for T 28) S = πrh + πr2 for h Q1) for Q2 for T1 30) L = π(r1 + r2) + 2d for r1 32) P = V1(V2 − g V1) for V2 33) ax + b = c for a 35) lwh = V for w a + b = c a for a 37) 1 39) at − bw = s for t 41) ax + bx = c for a 43) x + 5y = 3 for y 45) 3x + 2y = 7 for y 47) 5a 49) 4x − − 7b = 4 for b 5y = 8 for y 34) rt = d for r 36) V = πr2h a + b = c 38) 1 3 a for b for h 40) at − bw = s for w 42) x + 5y = 3 for x 44) 3x + 2y = 7 for x 46) 5a − 7b = 4 for a 5y = 8 for x 48) 4x − 50) C = 5 9 (F 32) for F − 51 1.6 Solving Linear Equations - Absolute Value Objective: Solve linear absolute value equations. When solving equations with absolute value we can end up with more than one possible answer. This is because what is in the absolute value can be either negative or positive and we must account for both possibilities when solving equations. This is illustrated in the following example. Example 84. x \| x = 7 or x = \| = 7 Absolute value can be positive or negative 7 Our Solution − Notice that we have considered two possibilities, both the positive and negative. Either way, the absolute value of our number will be positive 7. World View Note: The ﬁrst set of rules for working with negatives came from 7th century India. However, in 1758, almost a thousand years later, British mathematician Francis Maseres claimed that negatives “Darken the very whole doctrines of the equations and make dark of the things which are in their nature excessively obvious and simple.” When we have absolute values in our problem it is important to ﬁrst isolate the absolute value, then remove the absolute value by considering both the positive and negative solutions. Notice in the next two examples, all the numbers outside of the absolute value are moved to the other side ﬁrst before we remove the absolute value bars and consider both positive and negative solutions. Example 85. x \| 5 + 5 = 8 Notice absolute value is not alone Subtract 5 from both sides \| − x = 3 Absolute value can be positive or negative \| x = 3 or x = 3 Our Solution − 5 \| − Example 86. = x \| 4 − 4 \| − − − 20 Notice absolute value is not alone 4 Divide both sides by 4 − 52 x = 5 \| x = 5 or x = \| − Absolute value can be positive or negative 5 Our Solution Notice we never combine what is inside the absolute value with what is outside the absolute value. This is very important as it will often change the ﬁnal result to an incorrect solution. The next example requires two steps to isolate the absolute value. The idea is the same as a two-step equation, add or subtract, then multiply or divide. Example 87. x 5 \| 4 = 26 Notice the absolute value is not alone \| − + 4 + 4 Add 4 to both sides 5 \| 5 = 30 Absolute value still not alone Divide both sides by 5 x \| 5 = 6 Absolute value can be positive or negative x \| x = 6 or x = \| 6 Our Solution − Again we see the same process, get the absolute value alone ﬁrst, then consider the positive and negative solutions. Often the absolute value will have more than just a variable in it. In this case we will have to solve the resulting equations when we consider the positive and negative possibilities. This is shown in the next example. Example 88. \| 1 = 7 or 2x 2x 1 \| − 1 = = 7 Absolute value can be positive or negative 7 Two equations to solve − − 2x − Now notice we have two equations to solve, each equation will give us a diﬀerent solution. Both equations solve like any other two-step equation. 2x 1 = 7 − + 1 + 1 2x = 8 2 2 x = 4 2x or 53 − 1 = 7 − + 1 + 1 2x = 6 2 x = − 2 3 − Thus, from our previous example we have two solutions, x = 4 or x = 3. − Again, it is important to remember that the absolute value must be alone ﬁrst before we consider the positive and negative possibilities. This is illustrated in below. Example 89. 2 4 2x + 3 \| \| − = − 18 4. Notice we cannot combine the 2 and To get the absolute value alone we ﬁrst need to get rid of the 2 by subtracting, 4 becuase they are then divide by − not like terms, the 4 has the absolute value connected to it. Also notice we do not distribute the 4 into the absolute value. This is because the numbers outside cannot be combined with the numbers inside the absolute value. Thus we get the absolute value alone in the following way: − − − 4 \| − 2 2 2x + 3 = \| − \| 4 = − 2x + 3 4 2x + 3 \| − = 5 \| 2x + 3 = 5 or 2x + 3 = \| − − − − − 18 Notice absolute value is not alone Subtract 2 from both sides 2 20 Absolute value still not alone 4 Divide both sides by 4 − Absoloute value can be positive or negative 5 Two equations to solve Now we just solve these two remaining equations to ﬁnd our solutions. − 2x + 3 = 5 3 3 − 2x = 2 2 2 x = 1 or 5 − 3 − 8 − 2 4 − 2x + 3 = 3 − 2x = 2 x = 4. − We now have our two solutions, x = 1 and x = As we are solving absolute value equations it is important to be aware of special cases. Remember the result of an absolute value must always be positive. Notice what happens in the next example. 54 Example 90. − 2x \| − 5 7 + 7 2x \| 5 \| − = − \| − = 4 Notice absolute value is not alone Subtract 7 from both sides 7 3 Result of absolute value is negative! Notice the absolute value equals a negative number! This is impossible with absolute value. When this occurs we say there is no solution or ∅. One other type of absolute value problem is when two absolute values are equal to eachother. We still will consider both the positive and negative result, the diﬀerence here will be that we will have to distribute a negative into the second absolute value for the negative possibility. Example 91. 2x = \| 7 = 4x + 6 or 2x − 7 \| 2x − \| − 4x + 6 7 = \| − Absolute value can be positive or negative (4x + 6) make second part of second equation negative Notice the ﬁrst equation is the positive possibility and has no signiﬁcant diﬀerence other than the missing absolute value bars. The second equation considers the negative possibility. For this reason we have a negative in front of the expression which will be distributed through the equation on the ﬁrst step of solving. So we solve both these equations as follows: 2x − − 2x 7 = 4x + 6 2x 7 = 2x + 6 6 6 − − − − 13 = 2x − 2 2 13 − 2 = x or 6 − − − − − (4x + 6) 2x 7 = 2x 4x 7 = + 4x + 4x 6x 6 7 = − + 7 + 7 6x = 1 6 6 1 6 x = − This gives us our two solutions, x = − 13 2 or x = 1 6. 55 1.6 Practice - Absolute Value Equations Solve each equation + 8a = 53 \| 3k + 8 \| 9 + 7x \| = 2 = 30 1) 3) 5) 7) \| \| \| \| 9) \| 11) 13) = 50 \| = 24 \| \| 8 + 6m 6 2x − 7 \| 3 15) − 17) 7 \| − 7x − 3 3r = \| = 21 \| − 19) \| − 10 \| = 3 \| − 4b − 8 3 = 5 21) 8 x + 7 \| − \| 3 + 7m \| 25) 3 + 5 23) 5 8 \| 2x + 1 = 51 = 63 \| 2 − \| + 10 = 44 27) 29) 31) 33) 35) 6b \| \| − 7 + 8 − 5x + 3 \| − = 1 2x \| − 2x + 3 = \| 4 \| \| \| \| 3x 4x − 2 − 5 \| 6x + 3 2 \| = \| \| 7x 3 \| − = 73 2) 4) 6) 8) \| \| \| \| 10) 12) 14) 16) 21 − = 7 = 2 n x \| \| 9n + 8 = 46 \| 3 \| \| \| \| x = 6 \| − 5n + 7 = 23 \| 9p + 6 = 3 \| 2 = 7 \| 3n − 2 + 2b + 1 = 3 \| 18) \| − 3n 4 − 4 \| = 2 5 = 11 40 = − \| 20) 8 22) 3 24) 4 5p + 8 \| \| − 6n + 7 − \| r + 7 \| + 3 = 59 \| 26) 5 + 8 10n 2 \| − \| − = 101 28) 7 10v \| 2 9 = 5 \| − 30) 8 − 3n 3 \| − 2 + 3x = \| 2x 5 − 3 3x + 2 2 = = \| \| \| \| 5 = 91 \| − 4 − \| 3x + 4 2 2x 3 − 3 2x \| \| \| \| \| 32) 34) 36) \| \| \| 56 1.7 Solving Linear Equations - Variation Objective: Solve variation problems by creating variation equations and ﬁnding the variation constant. One application of solving linear equations is variation. Often diﬀerent events are related by what is called the constant of variation. For example, the time it takes to travel a certain distance is related to how fast you are traveling. The faster you travel, the less time it take to get there. This is one type of variation problem, we will look at three types of variation here. Variation problems have two or three variables and a constant in them. The constant, usually noted with a k, describes the relationship and does not change as the other variables in the problem change. There are two ways to set up a variation problem, the ﬁrst solves for one of the variables, a second method is to solve for the constant. Here we will use the second method. The greek letter pi (π) is used to represent the ratio of the circumference of a circle to its diameter. World View Note: In the 5th centure, Chinese mathematician Zu Chongzhi calculated the value of π to seven decimal places (3.1415926). This was the most accurate value of π for the next 1000 years! If you take any circle and divide the circumference of the circle by the diameter you will always get the same value, about 3.14159... If you have a bigger circumference you will also have a bigger diameter. This relationship is called
direct variation or directly proportional. If we see this phrase in the problem we know to divide to ﬁnd the constant of variation. Example 92. m is varies directly as n ′′Directly′′ tells us to divide = k Our formula for the relationship m n In kickboxing, one will ﬁnd that the longer the board, the easier it is to break. If you multiply the force required to break a board by the length of the board you will also get a constant. Here, we are multiplying the variables, which means as one variable increases, the other variable decreases. This relationship is called indirect variation or inversly proportional. If we see this phrase in the problem we know to multiply to ﬁnd the constant of variation. Example 93. y is inversely proportional to z ′′Inversely′′ tells us to multiply yz = k Our formula for the relationship 57 The formula for the area of a triangle has three variables in it. If we divide the area by the base times the height we will also get a constant, 1 . This relationship 2 is called joint variation or jointly proportional. If we see this phrase in the problem we know to divide the ﬁrst variable by the product of the other two to ﬁnd the constant of variation. Example 94. A varies jointly as x and y ′′Jointly′′ tells us to divide by the product A xy = k Our formula for the relationship Once we have our formula for the relationship in a variation problem, we use given or known information to calculate the constant of variation. This is shown for each type of variation in the next three examples. Example 95. Example 96. Example 97. w is directly proportional to y and w = 50 when y = 5 w y (50) (5) = k ′′directly′′ tells us to divide = k Substitute known values 10 = k Evaluate to ﬁnd our constant c varies indirectly as d and c = 4.5 when d = 6 cd = k (4.5)(6) = k ′′indirectly′′ tells us to multiply Substitute known values 27 = k Evaluate to ﬁnd our constant x is jointly proportional to y and z and x = 48 when y = 2 and z = 4 x yz (48) (2)(4) = k ′′Jointly′′ tells us to divide by the product = k Substitute known values 6 = k Evaluate to ﬁnd our constant 58 Once we have found the constant of variation we can use it to ﬁnd other combinations in the same relationship. Each of these problems we solve will have three important steps, none of which should be skipped. 1. Find the formula for the relationship using the type of variation 2. Find the constant of variation using known values 3. Answer the question using the constant of variation The next three examples show how this process is worked out for each type of variation. Example 98. The price of an item varies directly with the sales tax. If a S25 item has a sales tax of S2, what will the tax be on a S40 item? ′′Directly′′ tells us to divide price (p) and tax (t) = k p t (25) (2) 12.5 = k Evaluate to ﬁnd our constant = k Substitute known values for price and tax 40 t = 12.5 Using our constant, substitute 40 for price to ﬁnd the tax = 12.5(t) Multiply by LCD = t to clear fraction 40 = 12.5t Reduce the t with the denominator (t)40 t 12.5 12.5 Divide by 12.5 3.2 = t Our solution: Tax is S3.20 Example 99. The speed (or rate) Josiah travels to work is inversely proportional to time it takes to get there. If he travels 35 miles per hour it will take him 2.5 hours to get to work. How long will it take him if he travels 55 miles per hour? rt = k (35)(2.5) = k ′′Inversely′′ tells us to multiply the rate and time Substitute known values for rate and time 87.5 = k Evaluate to ﬁnd our constant 55t = 87.5 Using our constant, substitute 55 for rate to ﬁnd the time 55 t 55 1.59 Our solution: It takes him 1.59 hours to get to work Divide both sides by 55 ≈ Example 100. 59 The amount of simple interest earned on an investment varies jointly as the principle (amount invested) and the time it is invested. In an account, S150 invested for 2 years earned S12 in interest. How much interest would be earned on a S220 investment for 3 years? I Pt (12) (150)(2) = k ′′Jointly′′ divide Interest (I) by product of Principle (P )& time (t) = k Substitute known values for Interest, Principle and time 0.04 = k Evaluate to ﬁnd our constant I (220)(3) = 0.04 Using constant, substitute 220 for principle and 3 for time I 660 = 0.04 Evaluate denominator = 0.04(660) Multiply by 660 to isolate the variable I = 26.4 Our Solution: The investment earned S26.40 in interest (660)I 660 Sometimes a variation problem will ask us to do something to a variable as we set up the formula for the relationship. For example, π can be thought of as the ratio of the area and the radius squared. This is still direct variation, we say the area varies directly as the radius square and thus our variable is squared in our formula. This is shown in the next example. Example 101. The area of a circle is directly proportional to the square of the radius. A circle with a radius of 10 has an area of 314. What will the area be on a circle of radius 4? ′′Direct′′ tells us to divide, be sure we use r2 for the denominator A r2 = k (314) (10)2 = k (314) 100 3.14 = k Divide to ﬁnd our constant = k Exponents ﬁrst Substitute known values into our formula A (4)2 = 3.14 Using the constant, use 4 for r, don′t forget the squared! A 16 = 3.14 Evaluate the exponent = 3.14(16) Multiply both sides by 16 A = 50.24 Our Solution: Area is 50.24 (16)A 16 When solving variation problems it is important to take the time to clearly state the variation formula, ﬁnd the constant, and solve the ﬁnal equation. 60 1.7 Practice - Variation Write the formula that expresses the relationship described 1. c varies directly as a 2. x is jointly proportional to y and z 3. w varies inversely as x 4. r varies directly as the square of s 5. f varies jointly as x and y 6. j is inversely proportional to the cube of m 7. h is directly proportional to b 8. x is jointly proportional with the square of a and the square root of b 9. a is inversely proportional to b Find the constant of variation and write the formula to express the relationship using that constant 10. a varies directly as b and a = 15 when b = 5 11. p is jointly proportional to q and r and p = 12 when q = 8 and r = 3 12. c varies inversely as d and c = 7 when d = 4 13. t varies directly as the square of u and t = 6 when u = 3 14. e varies jointly as f and g and e = 24 when f = 3 and g = 2 15. w is inversely proportional to the cube of x and w is 54 when x = 3 16. h is directly proportional to j and h = 12 when j = 8 17. a is jointly proportional with the square of x and the square root of y and a = 25 when x = 5 and y = 9 18. m is inversely proportional to n and m = 1.8 when n = 2.1 Solve each of the following variation problems by setting up a formula to express the relationship, ﬁnding the constant, and then answering 61 the question. 19. The electrical current, in amperes, in a circuit varies directly as the voltage. When 15 volts are applied, the current is 5 amperes. What is the current when 18 volts are applied? 20. The current in an electrical conductor varies inversely as the resistance of the conductor. If the current is 12 ampere when the resistance is 240 ohms, what is the current when the resistance is 540 ohms? 21. Hooke’s law states that the distance that a spring is stretched by hanging object varies directly as the mass of the object. If the distance is 20 cm when the mass is 3 kg, what is the distance when the mass is 5 kg? 22. The volume of a gas varies inversely as the pressure upon it. The volume of a gas is 200 cm3 under a pressure of 32 kg/cm2. What will be its volume under a pressure of 40 kg/cm2? 23. The number of aluminum cans used each year varies directly as the number of people using the cans. If 250 people use 60,000 cans in one year, how many cans are used each year in Dallas, which has a population of 1,008,000? 24. The time required to do a job varies inversely as the number of peopel working. It takes 5hr for 7 bricklayers to build a park well. How long will it take 10 bricklayers to complete the job? 25. According to Fidelity Investment Vision Magazine, the average weekly allowance of children varies directly as their grade level. In a recent year, the average allowance of a 9th-grade student was 9.66 dollars per week. What was the average weekly allowance of a 4th-grade student? 26. The wavelength of a radio wave varies inversely as its frequency. A wave with a frequency of 1200 kilohertz has a length of 300 meters. What is the length of a wave with a frequency of 800 kilohertz? 27. The number of kilograms of water in a human body varies directly as the mass of the body. A 96-kg person contains 64 kg of water. How many kilo grams of water are in a 60-kg person? 28. The time required to drive a ﬁxed distance varies inversely as the speed. It takes 5 hr at a speed of 80 km/h to drive a ﬁxed distance. How long will it take to drive the same distance at a speed of 70 km/h? 29. The weight of an object on Mars varies directly as its weight on Earth. A person weighs 95lb on Earth weighs 38 lb on Mars. How much would a 100-lb person weigh on Mars? 30. At a constant temperature, the volume of a gas varies inversely as the pres- 62 sure. If the pressure of a certain gas is 40 newtons per square meter when the volume is 600 cubic meters what will the pressure be when the volume is reduced by 240 cubic meters? 31. The time required to empty a tank varies inversely as the rate of pumping. If a pump can empty a tank in 45 min at the rate of 600 kL/min, how long will it take the pump to empty the same tank at the rate of 1000 kL/min? 32. The weight of an object varies inversely as the square of the distance from the center of the earth. At sea level (6400 km from the center of the earth), an astronaut weighs 100 lb. How far above the earth must the astronaut be in order to weigh 64 lb? 33. The stopping distance of a car after the brakes have been applied varies directly as the square of the speed r. If a car, traveling 60 mph can stop in 200 ft, how fast can a car go and still
stop in 72 ft? 34. The drag force on a boat varies jointly as the wetted surface area and the square of the velocity of a boat. If a boat going 6.5 mph experiences a drag force of 86 N when the wetted surface area is 41.2 ft2, how fast must a boat with 28.5 ft2 of wetted surface area go in order to experience a drag force of 94N? 35. The intensity of a light from a light bulb varies inversely as the square of the distance from the bulb. Suppose intensity is 90 W/m2 (watts per square meter) when the distance is 5 m. How much further would it be to a point where the intesity is 40 W/m2? 36. The volume of a cone varies jointly as its height, and the square of its radius. If a cone with a height of 8 centimeters and a radius of 2 centimeters has a volume of 33.5 cm3, what is the volume of a cone with a height of 6 centimeters and a radius of 4 centimeters? 37. The intensity of a television signal varies inversely as the square of the dis- tance from the transmitter. If the intensity is 25 W/m2 at a distance of 2 km, how far from the trasmitter are you when the intensity is 2.56 W/m2? 38. The intensity of illumination falling on a surface from a given source of light is inversely proportional to the square of the distance from the source of light. The unit for measuring the intesity of illumination is usually the footcandle. If a given source of light gives an illumination of 1 foot-candle at a distance of 10 feet, what would the illumination be from the same source at a distance of 20 feet? 63 1.8 Linear Equations - Number and Geometry Objective: Solve number and geometry problems by creating and solving a linear equation. Word problems can be tricky. Often it takes a bit of practice to convert the English sentence into a mathematical sentence. This is what we will focus on here with some basic number problems, geometry problems, and parts problems. A few important phrases are described below that can give us clues for how to set up a problem. • • • • A number (or unknown, a value, etc) often becomes our variable Is (or other forms of is: was, will be, are, etc) often represents equals (=) x is 5 becomes x = 5 More than often represents addition and is usually built backwards, writing the second part plus the ﬁrst Three more than a number becomes x + 3 Less than often represents subtraction and is usually built backwards as well, writing the second part minus the ﬁrst Four less than a number becomes x 4 − Using these key phrases we can take a number problem and set up and equation and solve. Example 102. If 28 less than ﬁve times a certain number is 232. What is the number? 5x 5x 28 − 28 = 232 Subtraction is built backwards, multiply the unknown by 5 Is translates to equals − + 28 + 28 Add 28 to both sides 5x = 260 The variable is multiplied by 5 5 Divide both sides by 5 5 x = 52 The number is 52. This same idea can be extended to a more involved problem as shown in the next example. Example 103. 64 Fifteen more than three times a number is the same as ten less than six times the number. What is the number 3x + 15 6x 3x + 15 = 6x 3x 3x − 15 = 3x − − − First, addition is built backwards 10 Then, subtraction is also built backwards 10 Is between the parts tells us they must be equal Subtract 3x so variable is all on one side 10 Now we have a two − − + 10 Add 10 to both sides + 10 step equation 3 25 = 3x The variable is multiplied by 3 3 Divide both sides by 3 25 3 = x Our number is 25 3 Another type of number problem involves consecutive numbers. Consecutive numbers are numbers that come one after the other, such as 3, 4, 5. If we are looking for several consecutive numbers it is important to ﬁrst identify what they look like with variables before we set up the equation. This is shown in the following example. Example 104. The sum of three consecutive integers is 93. What are the integers? First x Make the ﬁrst number x Second x + 1 To get the next number we go up one or + 1 Third x + 2 Add another 1(2 total) to get the third F + S + T = 93 First (F ) plus Second (S) plus Third (T ) equals 93 (x) + (x + 1) + (x + 2) = 93 Replace F with x, S with x + 1, and T with = 93 Here the parenthesis aren′t needed. 3x + 3 = 93 Combine like terms x + x + x and 2 + 1 3 − − 3 Add 3 to both sides 3x = 90 The variable is multiplied by 3 3 Divide both sides by 3 x = 30 Our solution for x 3 First 30 Replace x in our origional list with 30 Second (30) + 1 = 31 The numbers are 30, 31, and 32 Third (30) + 2 = 32 Sometimes we will work consective even or odd integers, rather than just consecutive integers. When we had consecutive integers, we only had to add 1 to get to the next number so we had x, x + 1, and x + 2 for our ﬁrst, second, and third number respectively. With even or odd numbers they are spaced apart by two. So if we want three consecutive even numbers, if the ﬁrst is x, the next number would be x + 2, then ﬁnally add two more to get the third, x + 4. The same is 65 true for consecutive odd numbers, if the ﬁrst is x, the next will be x + 2, and the third would be x + 4. It is important to note that we are still adding 2 and 4 even when the numbers are odd. This is because the phrase “odd” is refering to our x, not to what is added to the numbers. Consider the next two examples. Example 105. The sum of three consecutive even integers is 246. What are the numbers? First x Make the ﬁrst x Second x + 2 Even numbers, so we add 2 to get the next Third x + 4 Add 2 more (4 total) to get the third F + S + T = 246 Sum means add First (F ) plus Second (S) plus Third (T ) (x) + (x + 2) + (x + 4) = 246 Replace each F , S , and T with what we labeled them x + x + 2 + x + 4 = 246 Here the parenthesis are not needed 3x + 6 = 246 Combine like terms x + x + x and 2 + 4 6 6 Subtract 6 from both sides − − 3x = 240 The variable is multiplied by 3 3 Divide both sides by 3 3 x = 80 Our solution for x First 80 Replace x in the origional list with 80. Second (80) + 2 = 82 The numbers are 80, 82, and 84. Third ( 80) + 4 = 84 Example 106. Find three consecutive odd integers so that the sum of twice the ﬁrst, the second and three times the third is 152. First x Make the ﬁrst x Second x + 2 Odd numbers so we add 2(same as even!) Third x + 4 Add 2 more (4 total) to get the third 2F + S + 3T = 152 Twice the ﬁrst gives 2F and three times the third gives 3T 2(x) + (x + 2) + 3(x + 4) = 152 Replace F , S , and T with what we labled them 2x + x + 2 + 3x + 12 = 152 Distribute through parenthesis 6x + 14 = 152 Combine like terms 2x + x + 3x and 2 + 14 Subtract 14 from both sides − − 14 14 6x = 138 Variable is multiplied by 6 6 Divide both sides by 6 6 x = 23 Our solution for x First 23 Replace x with 23 in the original list Second (23) + 2 = 25 The numbers are 23, 25, and 27 Third (23) + 4 = 27 66 When we started with our ﬁrst, second, and third numbers for both even and odd we had x, x + 2, and x + 4. The numbers added do not change with odd or even, it is our answer for x that will be odd or even. Another example of translating English sentences to mathematical sentences comes from geometry. A well known property of triangles is that all three angles will always add to 180. For example, the ﬁrst angle may be 50 degrees, the second 30 degrees, and the third 100 degrees. If you add these together, 50 + 30 + 100 = 180. We can use this property to ﬁnd angles of triangles. World View Note: German mathematician Bernhart Thibaut in 1809 tried to prove that the angles of a triangle add to 180 without using Euclid’s parallel postulate (a point of much debate in math history). He created a proof, but it was later shown to have an error in the proof. Example 107. The second angle of a triangle is double the ﬁrst. The third angle is 40 less than the ﬁrst. Find the three angles. First x With nothing given about the ﬁrst we make that x Second 2x The second is double the ﬁrst, Third x 40 The third is 40 less than the ﬁrst F + S + T = 180 All three angles add to 180 − (x) + (2x) + (x x + 2x + x 4x 40) = 180 Replace F , S , and T with the labeled values. 40 = 180 Here the parenthesis are not needed. 40 = 180 Combine like terms, x + 2x + x − − − + 40 + 40 Add 40 to both sides 4x = 220 The variable is multiplied by 4 4 x = 55 Our solution for x Divide both sides by 4 4 First 55 Replace x with 55 in the original list of angles Second 2(55) = 110 Our angles are 55, 110, and 15 Third (55) 40 = 15 − Another geometry problem involves perimeter or the distance around an object. For example, consider a rectangle has a length of 8 and a width of 3. There are two lengths and two widths in a rectangle (opposite sides) so we add 8 + 8 + 3 + 3 = 22. As there are two lengths and two widths in a rectangle an alternative to ﬁnd the perimeter of a rectangle is to use the formula P = 2L + 2W . So for the rectangle of length 8 and width 3 the formula would give, P = 2(8) + 2(3) = 16 + 6 = 22. With problems that we will consider here the formula P = 2L + 2W will be used. Example 108. 67 The perimeter of a rectangle is 44. The length is 5 less than double the width. Find the dimensions. Length x We will make the length x − 5 Width is ﬁve less than two times the length Width 2x P = 2L + 2W The formula for perimeter of a rectangle 5) Replace P , L, and W with labeled values 10 Distribute through parenthesis 10 Combine like terms 2x + 4x − − − + 10 Add 10 to both sides + 10 (44) = 2(x) + 2(2x 44 = 2x + 4x 44 = 6x 54 = 6x The variable is multiplied by 6 Divide both sides by 6 6 6 9 = x Our solution for x Length 9 Replace x with 9 in the origional list of sides 5 = 13 The dimensions of the rectangle are 9 by 13. Width 2(9) − We have seen that it is imortant to start by clearly labeling the variables in a short list before we begin to solve the problem. This is important in all word problems involving variables, not just consective numbers or geometry problems. This is shown in the following example. Example 109. A sofa and a love seat together costs S444. The sofa costs double the love seat. How much d
o they each cost? Love Seat x With no information about the love seat, this is our x Sofa 2x Sofa is double the love seat, so we multiply by 2 S + L = 444 Together they cost 444, so we add. (x) + (2x) = 444 Replace S and L with labeled values 3x = 444 Parenthesis are not needed, combine like terms x + 2x 3 x = 148 Our solution for x Divide both sides by 3 3 Love Seat 148 Replace x with 148 in the origional list Sofa 2(148) = 296 The love seat costs S148 and the sofa costs S296. Be careful on problems such as these. Many students see the phrase “double” and believe that means we only have to divide the 444 by 2 and get S222 for one or both of the prices. As you can see this will not work. By clearly labeling the variables in the original list we know exactly how to set up and solve these problems. 68 1.8 Practice - Number and Geometry Problems Solve. 1. When ﬁve is added to three more than a certain number, the result is 19. What is the number? 2. If ﬁve is subtracted from three times a certain number, the result is 10. What is the number? 3. When 18 is subtracted from six times a certain number, the result is What is the number? 42. − 4. A certain number added twice to itself equals 96. What is the number? 5. A number plus itself, plus twice itself, plus 4 times itself, is equal to What is the number? 104. − 6. Sixty more than nine times a number is the same as two less than ten times the number. What is the number? 7. Eleven less than seven times a number is ﬁve more than six times the number. Find the number. 8. Fourteen less than eight times a number is three more than four times the number. What is the number? 9. The sum of three consecutive integers is 108. What are the integers? 10. The sum of three consecutive integers is 126. What are the integers? − 11. Find three consecutive integers such that the sum of the ﬁrst, twice the second, and three times the third is 76. − 12. The sum of two consecutive even integers is 106. What are the integers? 13. The sum of three consecutive odd integers is 189. What are the integers? 69 14. The sum of three consecutive odd integers is 255. What are the integers? 15. Find three consecutive odd integers such that the sum of the ﬁrst, two times the second, and three times the third is 70. 16. The second angle of a triangle is the same size as the ﬁrst angle. The third angle is 12 degrees larger than the ﬁrst angle. How large are the angles? 17. Two angles of a triangle are the same size. The third angle is 12 degrees smaller than the ﬁrst angle. Find the measure the angles. 18. Two angles of a triangle are the same size. The third angle is 3 times as large as the ﬁrst. How large are the angles? 19. The third angle of a triangle is the same size as the ﬁrst. The second angle is 4 times the third. Find the measure of the angles. 20. The second angle of a triangle is 3 times as large as the ﬁrst angle. The third angle is 30 degrees more than the ﬁrst angle. Find the measure of the angles. 21. The second angle of a triangle is twice as large as the ﬁrst. The measure of the third angle is 20 degrees greater than the ﬁrst. How large are the angles? 22. The second angle of a triangle is three times as large as the ﬁrst. The measure of the third angle is 40 degrees greater than that of the ﬁrst angle. How large are the three angles? 23. The second angle of a triangle is ﬁve times as large as the ﬁrst. The measure of the third angle is 12 degrees greater than that of the ﬁrst angle. How large are the angles? 24. The second angle of a triangle is three times the ﬁrst, and the third is 12 degrees less than twice the ﬁrst. Find the measures of the angles. 25. The second angle of a triangle is four times the ﬁrst and the third is 5 degrees more than twice the ﬁrst. Find the measures of the angles. 26. The perimeter of a rectangle is 150 cm. The length is 15 cm greater than the width. Find the dimensions. 27. The perimeter of a rectangle is 304 cm. The length is 40 cm longer than the width. Find the length and width. 28. The perimeter of a rectangle is 152 meters. The width is 22 meters less than the length. Find the length and width. 29. The perimeter of a rectangle is 280 meters. The width is 26 meters less than the length. Find the length and width. 70 30. The perimeter of a college basketball court is 96 meters and the length is 14 meters more than the width. What are the dimensions? 31. A mountain cabin on 1 acre of land costs S30,000. If the land cost 4 times as much as the cabin, what was the cost of each? 32. A horse and a saddle cost S5000. If the horse cost 4 times as much as the saddle, what was the cost of each? 33. A bicycle and a bicycle helmet cost S240. How much did each cost, if the bicycle cost 5 times as much as the helmet? 34. Of 240 stamps that Harry and his sister collected, Harry collected 3 times as many as his sisters. How many did each collect? 35. If Mr. Brown and his son together had S220, and Mr. Brown had 10 times as much as his son, how much money had each? 36. In a room containing 45 students there were twice as many girls as boys. How many of each were there? 37. Aaron had 7 times as many sheep as Beth, and both together had 608. How many sheep had each? 38. A man bought a cow and a calf for S990, paying 8 times as much for the cow as for the calf. What was the cost of each? 39. Jamal and Moshe began a business with a capital of S7500. If Jamal furnished half as much capital as Moshe, how much did each furnish? 40. A lab technician cuts a 12 inch piece of tubing into two pieces in such a way that one piece is 2 times longer than the other. 41. A 6 ft board is cut into two pieces, one twice as long as the other. How long are the pieces? 42. An eight ft board is cut into two pieces. One piece is 2 ft longer than the other. How long are the pieces? 43. An electrician cuts a 30 ft piece of wire into two pieces. One piece is 2 ft longer than the other. How long are the pieces? 44. The total cost for tuition plus room and board at State University is S2,584. Tuition costs S704 more than room and board. What is the tuition fee? 45. The cost of a private pilot course is S1,275. The ﬂight portion costs S625 more than the groung school portion. What is the cost of each? 71 1.9 Solving Linear Equations - Age Problems Objective: Solve age problems by creating and solving a linear equation. An application of linear equations is what are called age problems. When we are solving age problems we generally will be comparing the age of two people both now and in the future (or past). Using the clues given in the problem we will be working to ﬁnd their current age. There can be a lot of information in these problems and we can easily get lost in all the information. To help us organize and solve our problem we will ﬁll out a three by three table for each problem. An example of the basic structure of the table is below Age Now Change Person 1 Person 2 Table 6. Structure of Age Table Normally where we see “Person 1” and “Person 2” we will use the name of the person we are talking about. We will use this table to set up the following example. Example 110. Adam is 20 years younger than Brian. In two years Brian will be twice as old as Adam. How old are they now? Age Now + 2 Adam Brian We use Adam and Brian for our persons We use + 2 for change because the second phrase is two years in the future Age Now + 2 20 Adam x Brain − x Consider the ′′Now′′ part, Adam is 20 years youger than Brian. We are given information about Adam, not Brian. So Brian is x now. To show Adam 20. is 20 years younger we subtract 20, Adam is x − Age Now Adam x Brian − x 20 x + 2 20 + 2 − x + 2 Now the + 2 column is ﬁlled in. This is done by adding 2 to both Adam′s and Brian′s now column as shown in the table. Age Now + 2 Adam x Brian − x 20 x 18 − x + 2 Combine like terms in Adam′s future age: 20 + 2 This table is now ﬁlled out and we are ready to try and solve. − 72 B = 2A (x + 2) = 2(x 18) − − − 36 x + 2 = 2x x x − 2 = x 36 + 36 + 36 38 = x − Age now Adam 38 Brian − 20 = 18 38 − Our equation comes from the future statement: Brian will be twice as old as Adam. This means the younger, Adam, needs to be multiplied by 2. Replace B and A with the information in their future cells, Adam (A) is replaced with x 18 and Brian (B) is replaced with (x + 2) This is the equation to solve! Distribute through parenthesis Subtract x from both sides to get variable on one side Need to clear the − Add 36 to both sides Our solution for x The ﬁrst column will help us answer the question. Replace the x′s with 38 and simplify. Adam is 18 and Brian is 38 36 Solving age problems can be summarized in the following ﬁve steps. These ﬁve steps are guidelines to help organize the problem we are trying to solve. 1. Fill in the now column. The person we know nothing about is x. 2. Fill in the future/past collumn by adding/subtracting the change to the now column. 3. Make an equation for the relationship in the future. This is independent of the table. 4. Replace variables in equation with information in future cells of table 5. Solve the equation for x, use the solution to answer the question These ﬁve steps can be seen illustrated in the following example. Example 111. Carmen is 12 years older than David. Five years ago the sum of their ages was 28. How old are they now? Age Now 5 − Carmen David Carmen David 5 − Age Now x + 12 x Five years ago is − 5 in the change column. Carmen is 12 years older than David. We don′t know about David so he is x, Carmen then is x + 12 Carmen David Age Now x + 12 x 5 − x + 12 x − − 5 5 Subtract 5 from now column to get the change 73 Carmen David Age Now x + 12 x 5 − x + 7 x 5 − Simplify by combining like terms 12 Our table is ready! 5 − (x + 7) + ( = 28 5) = 28 5 = 28 2x + 2 = 28 2 2 − 2x = 26 2 2 x = 13 − The sum of their ages will be 29. So we add C and D Replace C and D with the change cells. Remove parenthesis Combine like terms x + x and 7 Subtract 2 from both sides Notice x is multiplied by 2
Divide both sides by 2 Our solution for x − 5 Age Now Caremen 13 + 12 = 25 David 13 Replace x with 13 to answer the question Carmen is 25 and David is 13 Sometimes we are given the sum of their ages right now. These problems can be tricky. In this case we will write the sum above the now column and make the ﬁrst person’s age now x. The second person will then turn into the subtraction problem total x. This is shown in the next example. − Example 112. The sum of the ages of Nicole and Kristin is 32. In two years Nicole will be three times as old as Kristin. How old are they now? 32 Age Now + 2 Nicole Kristen x 32 x − Nicole Kristen Age Now x + 2 x + 2 32 x − x + 2 32 − The change is + 2 for two years in the future The total is placed above Age Now The ﬁrst person is x. The second becomes 32 x − Add 2 to each cell ﬁll in the change column Nicole Kristen Age Now + 2 x + 2 x 34 32 x x − − (x + 2) = 3(34 x + 2 = 102 + 3x N = 3K x) − 3x − + 3x Combine like terms 32 + 2, our table is done! Nicole is three times as old as Kristin. Replace variables with information in change cells Distribute through parenthesis Add 3x to both sides so variable is only on one side 74 − − 4x + 2 = 102 2 2 4x = 100 4 4 x = 25 Nicole Kristen 32 Age Now 25 25 = 7 − − step equation Solve the two Subtract 2 from both sides The variable is multiplied by 4 Divide both sides by 4 Our solution for x Plug 25 in for x in the now column Nicole is 25 and Kristin is 7 A slight variation on age problems is to ask not how old the people are, but rather ask how long until we have some relationship about their ages. In this case we alter our table slightly. In the change column because we don’t know the time to add or subtract we will use a variable, t, and add or subtract this from the now column. This is shown in the next example. Example 113. Louis is 26 years old. Her daughter is 4 years old. In how many years will Louis be double her daughter’s age? Age Now + t Louis Daughter 26 4 As we are given their ages now, these numbers go into the table. The change is unknown, so we write + t for the change Age Now + t Louis Daughter 26 4 26 + t 4 + t Fill in the change column by adding t to each person′s age. Our table is now complete. L = 2D (26 + t) = 2(4 + t) 26 + t = 8 + 2t t t 26 = 8 + t 8 8 − 18 = t − − − Louis will be double her daughter Replace variables with information in change cells Distribute through parenthesis Subtract t from both sides Now we have an 8 added to the t Subtract 8 from both sides In 18 years she will be double her daughter′s age Age problems have several steps to them. However, if we take the time to work through each of the steps carefully, keeping the information organized, the problems can be solved quite nicely. World View Note: The oldest man in the world was Shigechiyo Izumi from Japan who lived to be 120 years, 237 days. However, his exact age has been disputed. 75 1.9 Practice - Age Problems 1. A boy is 10 years older than his brother. In 4 years he will be twice as old as his brother. Find the present age of each. 2. A father is 4 times as old as his son. In 20 years the father will be twice as old as his son. Find the present age of each. 3. Pat is 20 years older than his son James. In two years Pat will be twice as old as James. How old are they now? 4. Diane is 23 years older than her daughter Amy. In 6 years Diane will be twice as old as Amy. How old are they now? 5. Fred is 4 years older than Barney. Five years ago the sum of their ages was 48. How old are they now? 6. John is four times as old as Martha. Five years ago the sum of their ages was 50. How old are they now? 7. Tim is 5 years older than JoAnn. Six years from now the sum of their ages will be 79. How old are they now? 8. Jack is twice as old as Lacy. In three years the sum of their ages will be 54. How old are they now? 9. The sum of the ages of John and Mary is 32. Four years ago, John was twice as old as Mary. Find the present age of each. 10. The sum of the ages of a father and son is 56. Four years ago the father was 3 times as old as the son. Find the present age of each. 11. The sum of the ages of a china plate and a glass plate is 16 years. Four years ago the china plate was three times the age of the glass plate. Find the present age of each plate. 12. The sum of the ages of a wood plaque and a bronze plaque is 20 years. Four 76 years ago, the bronze plaque was one-half the age of the wood plaque. Find the present age of each plaque. 13. A is now 34 years old, and B is 4 years old. In how many years will A be twice as old as B? 14. A man’s age is 36 and that of his daughter is 3 years. In how many years will the man be 4 times as old as his daughter? 15. An Oriental rug is 52 years old and a Persian rug is 16 years old. How many years ago was the Oriental rug four times as old as the Persian Rug? 16. A log cabin quilt is 24 years old and a friendship quilt is 6 years old. In how may years will the log cabin quilt be three times as old as the friendship quilt? 17. The age of the older of two boys is twice that of the younger; 5 years ago it was three times that of the younger. Find the age of each. 18. A pitcher is 30 years old, and a vase is 22 years old. How many years ago was the pitcher twice as old as the vase? 19. Marge is twice as old as Consuelo. The sum of their ages seven years ago was 13. How old are they now? 20. The sum of Jason and Mandy’s age is 35. Ten years ago Jason was double Mandy’s age. How old are they now? 21. A silver coin is 28 years older than a bronze coin. In 6 years, the silver coin will be twice as old as the bronze coin. Find the present age of each coin. 22. A sofa is 12 years old and a table is 36 years old. In how many years will the table be twice as old as the sofa? 23. A limestone statue is 56 years older than a marble statue. In 12 years, the limestone will be three times as old as the marble statue. Find the present age of the statues. 24. A pewter bowl is 8 years old, and a silver bowl is 22 years old. In how many years will the silver bowl be twice the age of the pewter bowl? 25. Brandon is 9 years older than Ronda. In four years the sum of their ages will be 91. How old are they now? 26. A kerosene lamp is 95 years old, and an electric lamp is 55 years old. How many years ago was the kerosene lamp twice the age of the electric lamp? 27. A father is three times as old as his son, and his daughter is 3 years younger 77 than the son. If the sum of their ages 3 years ago was 63 years, ﬁnd the present age of the father. 28. The sum of Clyde and Wendy’s age is 64. In four years, Wendy will be three times as old as Clyde. How old are they now? 29. The sum of the ages of two ships is 12 years. Two years ago, the age of the older ship was three times the age of the newer ship. Find the present age of each ship. 30. Chelsea’s age is double Daniel’s age. Eight years ago the sum of their ages was 32. How old are they now? 31. Ann is eighteen years older than her son. One year ago, she was three times as old as her son. How old are they now? 32. The sum of the ages of Kristen and Ben is 32. Four years ago Kristen was twice as old as Ben. How old are they both now? 33. A mosaic is 74 years older than the engraving. Thirty years ago, the mosaic was three times as old as the engraving. Find the present age of each. 34. The sum of the ages of Elli and Dan is 56. Four years ago Elli was 3 times as old as Dan. How old are they now? 35. A wool tapestry is 32 years older than a linen tapestry. Twenty years ago, the wool tapestry was twice as old as the linen tapestry. Find the present age of each. 36. Carolyn’s age is triple her daughter’s age. In eight years the sum of their ages will be 72. How old are they now? 37. Nicole is 26 years old. Emma is 2 years old. In how many years will Nicole be triple Emma’s age? 38. The sum of the ages of two children is 16 years. Four years ago, the age of the older child was three times the age of the younger child. Find the present age of each child. 39. Mike is 4 years older than Ron. In two years, the sum of their ages will be 84. How old are they now? 40. A marble bust is 25 years old, and a terra-cotta bust is 85 years old. In how many years will the terra-cotta bust be three times as old as the marble bust? 78 1.10 Solving Linear Equations - Distance, Rate and Time Objective: Solve distance problems by creating and solving a linear equation. An application of linear equations can be found in distance problems. When solving distance problems we will use the relationship rt = d or rate (speed) times time equals distance. For example, if a person were to travel 30 mph for 4 hours. To ﬁnd the total distance we would multiply rate times time or (30)(4) = 120. This person travel a distance of 120 miles. The problems we will be solving here will be a few more steps than described above. So to keep the information in the problem organized we will use a table. An example of the basic structure of the table is blow: Rate Time Distance Person 1 Person 2 Table 7. Structure of Distance Problem The third column, distance, will always be ﬁlled in by multiplying the rate and time columns together. If we are given a total distance of both persons or trips we will put this information below the distance column. We will now use this table to set up and solve the following example 79 Example 114. Two joggers start from opposite ends of an 8 mile course running towards each other. One jogger is running at a rate of 4 mph, and the other is running at a rate of 6 mph. After how long will the joggers meet? Rate Time Distance Jogger 1 Jogger 2 Rate Time Distance Jogger 1 Jogger 2 4 6 Rate Time Distance Jogger 1 Jogger 2 4 6 t t Rate Time Distance Jogger 1 Jogger 2 4 6 t t 4t 6t 8 4t + 6t = 8 10t = 8 10 10 4 5 t = The basic table for the joggers, one and two We are given the rates for each jogger. These are added to the table We only know they both start and end at the same time. We use the variable t for b
oth times The distance column is ﬁlled in by multiplying rate by time We have total distance, 8 miles, under distance The distance column gives equation by adding Combine like terms, 4t + 6t Divide both sides by 10 Our solution for t, 4 5 hour (48 minutes) As the example illustrates, once the table is ﬁlled in, the equation to solve is very easy to ﬁnd. This same process can be seen in the following example Example 115. Bob and Fred start from the same point and walk in opposite directions. Bob walks 2 miles per hour faster than Fred. After 3 hours they are 30 miles apart. How fast did each walk? Rate Time Distance Bob Fred 3 3 The basic table with given times ﬁlled in Both traveled 3 hours 80 Rate Time Distance Bob r + 2 Fred r 3 3 Bob walks 2 mph faster than Fred We know nothing about Fred, so use r for his rate Bob is r + 2, showing 2 mph faster Bob r + 2 Fred r Rate Time Distance 3r + 6 3r 30 3 3 3r + 6 + 3r = 30 6r + 6 = 30 6 6 − 6r = 24 6 6 r = 4 − Rate Bob 4 + 2 = 6 Fred 4 Distance column is ﬁlled in by multiplying rate by Time. Be sure to distribute the 3(r + 2) for Bob. Total distance is put under distance The distance columns is our equation, by adding Combine like terms 3r + 3r Subtract 6 from both sides The variable is multiplied by 6 Divide both sides by 6 Our solution for r To answer the question completely we plug 4 in for r in the table. Bob traveled 6 miles per hour and Fred traveled 4 mph Some problems will require us to do a bit of work before we can just ﬁll in the cells. One example of this is if we are given a total time, rather than the individual times like we had in the previous example. If we are given total time we will write this above the time column, use t for the ﬁrst person’s time, and make t, for the second person’s time. This is shown in a subtraction problem, Total the next example − Example 116. Two campers left their campsite by canoe and paddled downstream at an average speed of 12 mph. They turned around and paddled back upstream at an average rate of 4 mph. The total trip took 1 hour. After how much time did the campers turn around downstream? Rate Time Distance 12 4 Down Up 1 Rate Time Distance Down Up 12 4 t 1 t − Basic table for down and upstream Given rates are ﬁlled in Total time is put above time column As we have the total time, in the ﬁrst time we have t, the second time becomes the subtraction, total t − 81 Rate Time Distance Down Up 12 4 t 1 t − Distance column is found by multiplying rate by time. Be sure to distribute 4(1 upstream. As they cover the same distance, = is put after the down distance t) for − 4t With equal sign, distance colum is equation 12t = 4 4t − 12t = 4 − + 4t + 4t 16t = 4 16 16 1 4 t = Add 4t to both sides so variable is only on one side Variable is multiplied by 16 Divide both sides by 16 Our solution, turn around after 1 4 hr (15 min ) Another type of a distance problem where we do some work is when one person catches up with another. Here a slower person has a head start and the faster person is trying to catch up with him or her and we want to know how long it will take the fast person to do this. Our startegy for this problem will be to use t for the faster person’s time, and add amount of time the head start was to get the slower person’s time. This is shown in the next example. Example 117. Mike leaves his house traveling 2 miles per hour. Joy leaves 6 hours later to catch up with him traveling 8 miles per hour. How long will it take her to catch up with him? Rate Time Distance Mike Joy 2 8 Rate Time Distance Mike Joy 2 8 t + 6 t Rate Time Distance t + 6 2t + 12 = Mike Joy 2 8 t 8t − 2t + 12 = 8t 2t 2t − 12 = 6t 6 6 Basic table for Mike and Joy The given rates are ﬁlled in Joy, the faster person, we use t for time Mike′s time is t + 6 showing his 6 hour head start Distance column is found by multiplying the rate by time. Be sure to distribute the 2(t + 6) for Mike As they cover the same distance, = is put after Mike′s distance Now the distance column is the equation Subtract 2t from both sides The variable is multiplied by 6 Divide both sides by 6 82 2 = t Our solution for t, she catches him after 2 hours World View Note: The 10,000 race is the longest standard track event. 10,000 meters is approximately 6.2 miles. The current (at the time of printing) world record for this race is held by Ethiopian Kenenisa Bekele with a time of 26 minutes, 17.53 second. That is a rate of 12.7 miles per hour! As these example have shown, using the table can help keep all the given information organized, help ﬁll in the cells, and help ﬁnd the equation we will solve. The ﬁnal example clearly illustrates this. Example 118. On a 130 mile trip a car travled at an average speed of 55 mph and then reduced its speed to 40 mph for the remainder of the trip. The trip took 2.5 hours. For how long did the car travel 40 mph? Rate Time Distance 55 Fast Slow 40 Basic table for fast and slow speeds The given rates are ﬁlled in 2.5 Rate Time Distance Fast 55 Slow 40 t 2.5 − t 2.5 Rate Time Distance Fast 55 Slow 40 t 2.5 − 55t t 100 40t − 130 55t + 100 − 100 40t = 130 15t + 100 = 130 100 15t = 30 15 15 t = 2 − − Total time is put above the time column As we have total time, the ﬁrst time we have t The second time is the subtraction problem 2.5 t − t) for slow Distance column is found by multiplying rate by time. Be sure to distribute 40(2.5 − Total distance is put under distance The distance column gives our equation by adding Combine like terms 55t 40t Subtract 100 from both sides The variable is multiplied by 30 Divide both sides by 15 Our solution for t. − Fast Slow 2.5 Time 2 2 = 0.5 − To answer the question we plug 2 in for t The car traveled 40 mph for 0.5 hours (30 minutes) 83 1.10 Practice - Distance, Rate, and Time Problems 1. A is 60 miles from B. An automobile at A starts for B at the rate of 20 miles an hour at the same time that an automobile at B starts for A at the rate of 25 miles an hour. How long will it be before the automobiles meet? 2. Two automobiles are 276 miles apart and start at the same time to travel toward each other. They travel at rates diﬀering by 5 miles per hour. If they meet after 6 hours, ﬁnd the rate of each. 3. Two trains travel toward each other from points which are 195 miles apart. They travel at rate of 25 and 40 miles an hour respectively. If they start at the same time, how soon will they meet? 4. A and B start toward each other at the same time from points 150 miles apart. If A went at the rate of 20 miles an hour, at what rate must B travel if they meet in 5 hours? 5. A passenger and a freight train start toward each other at the same time from two points 300 miles apart. If the rate of the passenger train exceeds the rate of the freight train by 15 miles per hour, and they meet after 4 hours, what must the rate of each be? 6. Two automobiles started at the same time from a point, but traveled in opposite directions. Their rates were 25 and 35 miles per hour respectively. After how many hours were they 180 miles apart? 7. A man having ten hours at his disposal made an excursion, riding out at the rate of 10 miles an hour and returning on foot, at the rate of 3 miles an hour. 84 Find the distance he rode. 8. A man walks at the rate of 4 miles per hour. How far can he walk into the country and ride back on a trolley that travels at the rate of 20 miles per hour, if he must be back home 3 hours from the time he started? 9. A boy rides away from home in an automobile at the rate of 28 miles an hour and walks back at the rate of 4 miles an hour. The round trip requires 2 hours. How far does he ride? 10. A motorboat leaves a harbor and travels at an average speed of 15 mph toward an island. The average speed on the return trip was 10 mph. How far was the island from the harbor if the total trip took 5 hours? 11. A family drove to a resort at an average speed of 30 mph and later returned over the same road at an average speed of 50 mph. Find the distance to the resort if the total driving time was 8 hours. 12. As part of his ﬂight trainging, a student pilot was required to ﬂy to an airport and then return. The average speed to the airport was 90 mph, and the average speed returning was 120 mph. Find the distance between the two airports if the total ﬂying time was 7 hours. 13. A, who travels 4 miles an hour starts from a certain place 2 hours in advance of B, who travels 5 miles an hour in the same direction. How many hours must B travel to overtake A? 14. A man travels 5 miles an hour. After traveling for 6 hours another man starts at the same place, following at the rate of 8 miles an hour. When will the second man overtake the ﬁrst? 15. A motorboat leaves a harbor and travels at an average speed of 8 mph toward a small island. Two hours later a cabin cruiser leaves the same harbor and travels at an average speed of 16 mph toward the same island. In how many hours after the cabin cruiser leaves will the cabin cuiser be alongside the motorboat? 16. A long distance runner started on a course running at an average speed of 6 mph. One hour later, a second runner began the same course at an average speed of 8 mph. How long after the second runner started will the second runner overtake the ﬁrst runner? 17. A car traveling at 48 mph overtakes a cyclist who, riding at 12 mph, has had a 3 hour head start. How far from the starting point does the car overtake the cyclist? 18. A jet plane traveling at 600 mph overtakes a propeller-driven plane which has 85 had a 2 hour head start. The propeller-driven plane is traveling at 200 mph. How far from the starting point does the jet overtake the propeller-driven plane? 19. Two men are traveling in opposite directions at the rate of 20 and 30 miles an hour at the same time and from the same place. In how many hours will they be 300 miles apart? 20. Running at an average rate of 8 m/s, a sprinter ran to the end of a track and then jogged back to the starting point at an average rate of 3 m/s.
The sprinter took 55 s to run to the end of the track and jog back. Find the length of the track. 21. A motorboat leaves a harbor and travels at an average speed of 18 mph to an island. The average speed on the return trip was 12 mph. How far was the island from the harbor if the total trip took 5 h? 22. A motorboat leaves a harbor and travels at an average speed of 9 mph toward a small island. Two hours later a cabin cruiser leaves the same harbor and travels at an average speed of 18 mph toward the same island. In how many hours after the cabin cruiser leaves will the cabin cruiser be alongside the motorboat? 23. A jet plane traveling at 570 mph overtakes a propeller-driven plane that has had a 2 h head start. The propeller-driven plane is traveling at 190 mph. How far from the starting point does the jet overtake the propeller-driven plane? 24. Two trains start at the same time from the same place and travel in opposite directions. If the rate of one is 6 miles per hour more than the rate of the other and they are 168 miles apart at the end of 4 hours, what is the rate of each? 25. As part of ﬂight traning, a student pilot was required to ﬂy to an airport and then return. The average speed on the way to the airport was 100 mph, and the average speed returning was 150 mph. Find the distance between the two airports if the total ﬂight time was 5 h. 26. Two cyclists start from the same point and ride in opposite directions. One cyclist rides twice as fast as the other. In three hours they are 72 miles apart. Find the rate of each cyclist. 27. A car traveling at 56 mph overtakes a cyclist who, riding at 14 mph, has had a 3 h head start. How far from the starting point does the car overtake the cyclist? 28. Two small planes start from the same point and ﬂy in opposite directions. 86 The ﬁrst plan is ﬂying 25 mph slower than the second plane. In two hours the planes are 430 miles apart. Find the rate of each plane. 29. A bus traveling at a rate of 60 mph overtakes a car traveling at a rate of 45 mph. If the car had a 1 h head start, how far from the starting point does the bus overtake the car? 30. Two small planes start from the same point and ﬂy in opposite directions. The ﬁrst plane is ﬂying 25 mph slower than the second plane. In 2 h, the planes are 470 mi apart. Find the rate of each plane. 31. A truck leaves a depot at 11 A.M. and travels at a speed of 45 mph. At noon, a van leaves the same place and travels the same route at a speed of 65 mph. At what time does the van overtake the truck? 32. A family drove to a resort at an average speed of 25 mph and later returned over the same road at an average speed of 40 mph. Find the distance to the resort if the total driving time was 13 h. 33. Three campers left their campsite by canoe and paddled downstream at an average rate of 10 mph. They then turned around and paddled back upstream at an average rate of 5 mph to return to their campsite. How long did it take the campers to canoe downstream if the total trip took 1 hr? 34. A motorcycle breaks down and the rider has to walk the rest of the way to work. The motorcycle was being driven at 45 mph, and the rider walks at a speed of 6 mph. The distance from home to work is 25 miles, and the total time for the trip was 2 hours. How far did the motorcycle go before if broke down? 35. A student walks and jogs to college each day. The student averages 5 km/hr walking and 9 km/hr jogging. The distance from home to college is 8 km, and the student makes the trip in one hour. How far does the student jog? 36. On a 130 mi trip, a car traveled at an average speed of 55 mph and then reduced its speed to 40 mph for the remainder of the trip. The trip took a total of 2.5 h. For how long did the car travel at 40 mph? 37. On a 220 mi trip, a car traveled at an average speed of 50 mph and then reduced its average speed to 35 mph for the remainder of the trip. The trip took a total of 5 h. How long did the car travel at each speed? 38. An executive drove from home at an average speed of 40 mph to an airport where a helicopter was waiting. The executive boarded the helicopter and ﬂew to the corporate oﬃces at and average speed of 60 mph. The entire distance was 150 mi. The entire trip took 3 h. Find the distance from the airport to the corporate oﬃces. 87 Chapter 2 : Graphing 2.1 Points and Lines ......................................................................................89 2.2 Slope ........................................................................................................95 2.3 Slope-Intercept Form .............................................................................102 2.4 Point-Slope Form ...................................................................................107 2.5 Parallel and Perpendicular Lines ...........................................................112 88 2.1 Graphing - Points and Lines Objective: Graph points and lines using xy coordinates. Often, to get an idea of the behavior of an equation we will make a picture that represents the solutions to the equations. A graph is simply a picture of the solutions to an equation. Before we spend much time on making a visual representation of an equation, we ﬁrst have to understand the basis of graphing. Following is an example of what is called the coordinate plane. -4 -3 -2 -1 2 1 -1 -2 1 2 3 The plane is divided into four sections by a horizontal number line (x-axis) and a vertical number line (y-axis). Where the two lines meet in the center is called the origin. This center origin is where x = 0 and y = 0. As we move to the right the numbers count up from zero, representing x = 1, 2, 3 . To the left the numbers count down from zero, representing x = 3 . Similarly, as we move up the number count up from zero, y = 1, 2, 3 ., and as we move down count down from zero, y = 3. We can put dots on the − graph which we will call points. Each point has an “address” that deﬁnes its location. The ﬁrst number will be the value on the x axis or horizontal number line. This is the distance the point moves left/right from the origin. The second number will represent the value on the y axis or vertical number line. This is the distance the point moves up/down from the origin. The points are given as an ordered pair (x, y). 2, 2, 1, 1, − − − − − − − World View Note: Locations on the globe are given in the same manner, each number is a distance from a central point, the origin which is where the prime meridian and the equator. This “origin is just oﬀ the western coast of Africa. The following example ﬁnds the address or coordinate pair for each of several points on the coordinate plane. Example 119. Give the coordinates of each point. B A C 89 Tracing from the origin, point A is right 1, up 4. This becomes A(1, 4). Point B is left 5, up 3. Left is back5, wards or negative so we have B( − 3). C is straight down 2 units. There is no left or right. This means we go right zero so the point is C(0, 2). − A(1, 4), B( 5, 3), C(0, − − 2) Our Solution Just as we can give the coordinates for a set of points, we can take a set of points and plot them on the plane. Example 120. Graph the points A(3, 2), B ( G(0, 0) − 2, 1), C (3, 4), D ( 2, − − − 3), E ( − 3, 0), F (0, 2), A B Up 1 Left 2 Right 3 Up 2 The ﬁrst point, A is at (3, 2) this means x = 3 (right 3) and y = 2 (up 2). Following these instructions, starting from the origin, we get our point. The second point, B ( 2, 1), is left 2 (negative moves backwards), up 1. This is also illustrated on the graph. − A Right 3 B Left 2 Down 3 D Down 4 C The third point, C (3, down 4 (negative moves backwards). 4) is right 3, − The fourth point, D ( 3) is left 2, down 3 (both negative, both move backwards) − − 2, The last three points have zeros in them. We still treat these points just like the other points. If there is a zero there is just no movement. B F A Up 2 E Left 3 G D C − − axis. Next is E ( 3, 0). This is left 3 (negative is backwards), and up zero, right on the x Then is F (0, 2). This is right zero, and up two, right on the y Finally is G (0, 0). This point has no movement. Thus the point is right on the origin. axis. − 90 E B D F G A C Our Solution The main purpose of graphs is not to plot random points, but rather to give a picture of the solutions to an equation. We may have an equation such as y = 2x 3. We may be interested in what type of solution are possible in this equation. We can visualize the solution by making a graph of possible x and y combinations that make this equation a true statement. We will have to start by ﬁnding possible x and y combinations. We will do this using a table of values. − Example 121. Graph y = 2x − 3 We make a table of values − − − We will test three values for x. Any three can be used Evaluate each by replacing x with the given value x = 1; y = 2( 3 = x = 0; y = 2(0) x = 1; y = 2(1) 1, ( − − 5), (0, − 3), (1, − 1) These then become the points to graph on our equation 91 Plot each point. Once the point are on the graph, con- nect the dots to make a line. The graph is our solution − What this line tells us is that any point on the line will work in the equation y = 3. For example, notice the graph also goes through the point (2, 1). If we use 2x x = 2, we should get y = 1. Sure enough, y = 2(2) 3 = 1, just as the graph suggests. Thus we have the line is a picture of all the solutions for y = 2x 3. We can use this table of values method to draw a graph of any linear equation. 3 = 4 − − − Example 122. Graph 2x − 3y = 6 We will use a table of values y x 3 − 0 3 We will test three values for x. Any three can be used. 2( 3) − 6 − + 6 − − 3y = 6 3y = 6 3, multiply ﬁrst Substitute each value in for x and solve for y Start with x = + 6 Add 6 to both sides 3y = 12 Divide both sides by 3 y = Solution for y when x = 3, add this to table (0) 2(3) 6 6 − 3y = 6 Next x = 0 − 3y = 6 Multiplying clears the constant term − 3 − y = 3 − Solution for y when x = 0, add this to table 3 Divide
each side by 2 − − 3y = 6 Next x = 3 3y = 6 Multiply − − 6 − 3y = 0 3 3 − y = 0 − − Subtract 9 from both sides Divide each side by 3 − Solution for y when x = 3, add this to table − 92 Our completed table. 3, ( − − 4), (0, 2), (3, 0) Table becomes points to graph Graph points and connect dots Our Solution 93 2.1 Practice - Points and Lines State the coordinates of each point. K E 1) D J I H G C F B Plot each point. 2) L( − 5, 5) K(1, 0) J( − 3, 4) H( − 2) E (3, 4, 2) G(4, -2) 2) D(0, 3) − 3, 0) I( − F ( 2, − − C (0, 4) Sketch the graph of each line. 1 4 x 3 − 4) y = x 3) y = 5) y = − − 7) y = − 9 − 4x + 2 5 − 11) y = 4 5 x 3 − − 13) x + 5y = − 15) 4x + y = 5 15 17) 2x y = 2 − 19) x + y = 21) y = − 8) y = 5 3 x 10) y = − 12) y = 1 x 2 2 − 14) 8x y = 5 − 16) 3x + 4y = 16 18) 7x + 3y = − 20) 3x + 4y = 8 12 22) 9x y = 4 − − 94 2.2 Graphing - Slope Objective: Find the slope of a line given a graph or two points. As we graph lines, we will want to be able to identify diﬀerent properties of the lines we graph. One of the most important properties of a line is its slope. Slope is a measure of steepness. A line with a large slope, such as 25, is very steep. A line with a small slope, such as 1 10 is very ﬂat. We will also use slope to describe the direction of the line. A line that goes up from left to right will have a positive slope and a line that goes down from left to right will have a negative slope. As we measure steepness we are interested in how fast the line rises compared to how far the line runs. For this reason we will describe slope as the fraction rise run. Rise would be a vertical change, or a change in the y-values. Run would be a horizontal change, or a change in the x-values. So another way to describe slope would be the fraction change in y change in x . It turns out that if we have a graph we can draw vertical and horiztonal lines from one point to another to make what is called a slope triangle. The sides of the slope triangle give us our slope. The following examples show graphs that we ﬁnd the slope of using this idea. Example 123. Rise 4 − Run 6 To ﬁnd the slope of this line we will consider the rise, or verticle change and the run or horizontal change. Drawing these lines in makes a slope triangle that we can use to count from one point to the next the graph goes 4, run down 4, right 6. This is rise 4 6. As a fraction it would be, − 6 . Reduce the fraction to get − . 2 3 − 2 3 − Our Solution World View Note: When French mathematicians Rene Descartes and Pierre de Fermat ﬁrst developed the coordinate plane and the idea of graphing lines (and other functions) the y-axis was not a verticle line! Example 124. 95 Run 3 Rise 6 To ﬁnd the slope of this line, the rise is up 6, the run is right 3. Our slope is run or 6 3 . then written as a fraction, This fraction reduces to 2. This will be our slope. rise 2 Our Solution There are two special lines that have unique slopes that we need to be aware of. They are illustrated in the following example. Example 125. In this graph there is no rise, but the run is 3 units. This slope becomes 0 line, 3 and all horizontal lines have a zero slope. This 0. = slope This line has a rise of 5, but no run. The = undeﬁned. line, and all vertical lines, have no slope. becomes This 5 0 As you can see there is a big diﬀerence between having a zero slope and having no slope or undeﬁned slope. Remember, slope is a measure of steepness. The ﬁrst slope is not steep at all, in fact it is ﬂat. Therefore it has a zero slope. The second slope can’t get any steeper. It is so steep that there is no number large enough to express how steep it is. This is an undeﬁned slope. We can ﬁnd the slope of a line through two points without seeing the points on a graph. We can do this using a slope formula. If the rise is the change in y values, we can calculate this by subtracting the y values of a point. Similarly, if run is a change in the x values, we can calculate this by subtracting the x values of a point. In this way we get the following equation for slope. The slope of a line through (x1, y1) and (x2, y2) is y2 x2 − − y1 x1 96 When mathematicians began working with slope, it was called the modular slope. For this reason we often represent the slope with the variable m. Now we have the following for slope. Slope = m = rise run = change in y change in x = y2 x2 y1 x1 − − As we subtract the y values and the x values when calculating slope it is important we subtract them in the same order. This process is shown in the following examples. Example 126. Find the slope between ( 4, 3) and (2, 9) Identify x1, y1, x2, y2 − (x1, y1) and (x2, y2) Use slope formula) − − 12 m = − 6 m = − Simplify Reduce 2 Our Solution Example 127. Find the slope between (4, 6) and (2, 1) Identify x1, y1, x2, y2 − (x1, y1) and (x2, y2) Use slope formula, m = y2 − x2 − y1 x1 y2 − x2 − y1 x1 Simplify Reduce, dividing by 1 − Our Solution We may come up against a problem that has a zero slope (horiztonal line) or no slope (vertical line) just as with using the graphs. Example 128. Find the slope between ( 4, − − 1) and ( 4, − − 5) Identify x1, y1, x2, y2 97 (x1 , y1) and (x2, y2) Use slope formula) − − 4) − − 4 m = − 0 Simplify Can′t divide by zero, undeﬁned m = no slope Our Solution y2 − x2 − y1 x1 Example 129. Find the slope between (3, 1) and ( 2, 1) Identify x1, y1, x2, y2 − (x1, y1) and (x2, y2) Use slope formula − Simplify Reduce m = 0 Our Solution y2 − x2 − y1 x1 Again, there is a big diﬀerence between no slope and a zero slope. Zero is an integer and it has a value, the slope of a ﬂat horizontal line. No slope has no value, it is undeﬁned, the slope of a vertical line. Using the slope formula we can also ﬁnd missing points if we know what the slope is. This is shown in the following two examples. Example 130. Find the value of y between the points (2, y) and (5, 1) with slope − 3 − We will plug values into slope formula Simplify y1 x1 y m = y2 − x2 − (3 Multiply both sides by 3 (3) Simplify y Add 1 to both sides − y Divide both sides by 1 1 − 8 = y Our Solution 98 Example 131. Find the value of x between the points ( 3, 2) and (x, 6) with slope 2 5 − m = 2 5 = y2 − x2 − 2 6 − ( y1 x1 We will plug values into slope formula Simplify − = 3x + 3) = 4 Multiply by 5 to clear fraction Multiply both sides by (x + 3) 2 5 (5) 2 5 (x + 3) = 4(5) Simplify 2(x + 3) = 20 Distribute 2x + 6 = 20 6 6 Solve. Subtract 6 from both sides − − 2x = 14 Divide each side by 2 2 2 x = 7 Our Solution 99 2.2 Practice - Slope Find the slope of each line. 1) 3) 5) 7) 2) 4) 6) 8) 100 9) 10) Find the slope of the line through each pair of points. 11) ( − 2, 10), ( 2, − 13) ( 15, 10), (16, − 15) (10, 18), ( − − 15) 7) 10) 11, − − 14), (11, 16, − 4, 14), ( − 16, 8) − 19), (6, 14) 17) ( − 19) ( − 21) (12, 23) ( − 25) ( − 27) (7, 29) ( − 10), ( − − 17, 19), (10, 5, 20) 7) − 14), ( − 5, 7), ( 8, 9) − − 18, 14) − − 5, 14) 18) (13, 15), (2, 10) 12) (1, 2), ( 6, 14) − − 2), (7, 7) 14) (13, 16) ( − − 3, 6), ( 20, 13) − 20) (9, 6), ( 7, 7) − − − 16, 2), (15, 22) ( − 24) (8, 11), ( 10) − 13) 3, − − 2), (1, 17) 26) (11, − 18, 28) ( − 5), (14, 3) − − 30) (19, 15), (5, 11) Find the value of x or y so that the line through the points has the given slope. 31) (2, 6) and (x, 2); slope: 4 7 2) and (x, 6); slope: 3, − − 1, 1); slope: 6 7 8 5 34) ( − 36) (x, 33) ( 35) ( − − 37) (x, 8, y) and ( 7) and ( − − − 9, − 9); slope: 2 5 38) (2, − 32) (8, y) and ( 2, 4); slope: − − 2, y) and (2, 4); slope: 1 4 1) and ( 4, 6); slope: − − 5) and (3, y); slope: 6 1 5 7 10 − 39) (x, 5) and (8, 0); slope: 5 6 − 40) (6, 2) and (x, 6); slope: 4 5 − 101 2.3 Graphing - Slope-Intercept Form Objective: Give the equation of a line with a known slope and y-intercept. When graphing a line we found one method we could use is to make a table of values. However, if we can identify some properties of the line, we may be able to make a graph much quicker and easier. One such method is ﬁnding the slope and the y-intercept of the equation. The slope can be represented by m and the yintercept, where it crosses the axis and x = 0, can be represented by (0, b) where b is the value where the graph crosses the vertical y-axis. Any other point on the line can be represented by (x, y). Using this information we will look at the slope formula and solve the formula for y. Example 132. m, (0, b), (x, y) Using the slope formula gives Simplify − − − x b = mx Add b to both sides = m Multiply both sides by x − + b + b y = mx + b Our Solution This equation, y = mx + b can be thought of as the equation of any line that as a slope of m and a y-intercept of b. This formula is known as the slope-intercept equation. Slope − Intercept Equation: y = m x + b If we know the slope and the y-intercept we can easily ﬁnd the equation that represents the line. Example 133. Slope = 3 4 , y − intercept = 3 Use the slope intercept equation − − y = mx + b m is the slope, b is the y intercept − y = 3 4 x − 3 Our Solution We can also ﬁnd the equation by looking at a graph and ﬁnding the slope and yintercept. Example 134. 102 Identify the point where the graph crosses the y-axis (0,3). This means the y-intercept is 3. Idenﬁty one other point and draw a slope triangle to ﬁnd the slope. The slope is 2 3 − y = mx + b Slope-intercept equation y = 2 3 − x + 3 Our Solution We can also move the opposite direction, using the equation identify the slope and y-intercept and graph the equation from this information. However, it will be important for the equation to ﬁrst be in slope intercept form. If it is not, we will have to solve it for y so we can identify the slope and the y-intercept. Example 135. Write in slope − Solve for y Subtract 2x from both sides intercept form: 2x 2x − 4y = 4 − − 4y = 6 2x − 2x + 6 Put x term ﬁrst 4 4 Divide each term by 3 2 Our Solution − = Once we have an equation in slope-intercept form we can graph it by ﬁrst plotting the y-intercept, then using the slope, ﬁnd a second point and connecting the dots. Exam
ple 136. x 1 2 y = mx + b − Graph y = 4 Recall the slope intercept formula − Idenﬁty the slope, m, and the y intercept Make the graph Starting with a point at the y-intercept of 4, − Then use the slope rise run, so we will rise 1 unit and run 2 units to ﬁnd the next point. Once we have both points, connect the dots to get our graph. World View Note: Before our current system of graphing, French Mathematician Nicole Oresme, in 1323 sugggested graphing lines that would look more like a 103 bar graph with a constant slope! Example 137. Graph 3x + 4y = 12 Not in slope intercept form Subtract 3x from both sides 3x 3x − 4y = 4 y = m = − 4 − Divide each term by 4 3x + 12 Put the x term ﬁrst 4 3 4 y = mx + b 3 4 , b = 3 Make the graph x + 3 Recall slope Idenﬁty m and b − − − intercept equation Starting with a point at the y-intercept of 3, Then use the slope rise run, but its negative so it will go downhill, so we will drop 3 units and run 4 units to ﬁnd the next point. Once we have both points, connect the dots to get our graph. We want to be very careful not to confuse using slope to ﬁnd the next point with use a coordinate such as (4, 2) to ﬁnd an individule point. Coordinates such as (4, 2) start from the origin and move horizontally ﬁrst, and vertically second. Slope starts from a point on the line that could be anywhere on the graph. The numerator is the vertical change and the denominator is the horizontal change. − − Lines with zero slope or no slope can make a problem seem very diﬀerent. Zero slope, or horiztonal line, will simply have a slope of zero which when multiplied by x gives zero. So the equation simply becomes y = b or y is equal to the y-coordinate of the graph. If we have no slope, or a vertical line, the equation can’t be written in slope intercept at all because the slope is undeﬁned. There is no y in these equations. We will simply make x equal to the x-coordinate of the graph. Example 138. Give the equation of the line in the graph. Because we have a vertical line and no slope there is no slope-intercept equation we can use. Rather we make x equal to the x-coordinate of 4 − x = 4 − Our Solution 104 2.3 Practice - Slope-Intercept Write the slope-intercept form of the equation of each line given the slope and the y-intercept. 1) Slope = 2, y-intercept = 5 3) Slope = 1, y-intercept = − 3 4, y-intercept = 4 1 − 5) Slope = − 7) Slope = 1 3 , y-intercept = 1 2) Slope = 4) Slope = − − 6) Slope = − 8) Slope = 2 6, y-intercept = 4 1, y-intercept = − 1 4, y-intercept = 3 2 5 , y-intercept = 5 Write the slope-intercept form of the equation of each line. 9) 11) 13) 10) 12) 14) 105 15) x + 10y = 37 − 17) 2x + y = 1 − 3y = 24 19) 7x − 21) x = 8 − 23) y − 25) y − 27) y + 5 = 29) y + 1 = (x + 5) 4 = − 4 = 4(x − 4(x 1 2(x 1) 2) 4) − − − − Sketch the graph of each line. 31) y = 1 3 x + 4 33) y = 6 5 x 5 − 35) y = 3 2 x 37) x y + 3 = 0 − y − 3y = − − 39) 41) 4 + 3x = 0 5x + 9 − 10y = 3 16) x − 18) 6x 11y = 70 − − 20) 4x + 7y = 28 22) x 24) y 26) y − − − 7y = 5 = 5 − 2 (x 42 2) − 3 = − 2 3 (x + 3) 4 28) 0 = x − 30) y + 2 = 6 5 (x + 5) 32) y = 34 36) y = − 38) 4x + 5 = 5y 40) 42) 8 = 6x 3y = 3 2y 3 2 x − − − − 106 2.4 Graphing - Point-Slope Form Objective: Give the equation of a line with a known slope and point. The slope-intercept form has the advantage of being simple to remember and use, however, it has one major disadvantage: we must know the y-intercept in order to use it! Generally we do not know the y-intercept, we only know one or more points (that are not the y-intercept). In these cases we can’t use the slope intercept equation, so we will use a diﬀerent more ﬂexible formula. If we let the slope of an equation be m, and a speciﬁc point on the line be (x1, y1), and any other point on the line be (x, y). We can use the slope formula to make a second equation. Example 139. m, (x1, y1), (x, y) Recall slope formula y2 y1 − x2 − x1 y y1 − x1 x − y1 = m(x − y − = m Plug in values = m Multiply both sides by (x x1) − x1) Our Solution If we know the slope, m of an equation and any point on the line (x1, y1) we can easily plug these values into the equation above which will be called the pointslope formula. Point − Slope Formula: y y1 = m(x x1) − − Example 140. Write the equation of the line through the point (3, 4) with a slope of 3 5. − y1 = m(x x1) Plug values into point slope formula − − y y − − ( 4x (x 3) Simplify signs 3) Our Solution − − Often, we will prefer ﬁnal answers be written in slope intercept form. If the direc- 107 tions ask for the answer in slope-intercept form we will simply distribute the slope, then solve for y. Example 141. Write the equation of the line through the point ( slope-intercept form. − 6, 2) with a slope of 2 3 in − x1) Plug values into point slope formula − 6)) Simplify signs − − y − y1 = m(x (x + 6) Distribute slope 2 3 Solve for Our Solution An important thing to observe about the point slope formula is that the operation between the x’s and y’s is subtraction. This means when you simplify the signs you will have the opposite of the numbers in the point. We need to be very careful with signs as we use the point-slope formula. In order to ﬁnd the equation of a line we will always need to know the slope. If we don’t know the slope to begin with we will have to do some work to ﬁnd it ﬁrst before we can get an equation. Example 142. Find the equation of the line through the points ( 2, 5) and (4, 3). − − m = First we must ﬁnd the slope y2 − x2 − 8 = = − 6 − x1) With slope and either point, use point Plug values in slope formula and evaluate 5 2) y1 = m(x y1 x1 4 3 slope formula − x 4 3 − 5 = − 2)) ( − Simplify signs (x + 2) Our Solution − 4 3 Example 143. 108 Find the equation of the line through the points ( intercept form. − 3, 4) and ( 1, − − 2) in slope- m = y1 x1 First we must ﬁnd the slope y2 − x2 − 6 = = − 2 − x1) With slope and either point, point 3)) 3) y1 = m(x 3(x ( Simplify signs 3 Plug values in slope formula and evaluate − − slope formula − (x + 3) Distribute slope − y − 4 = − + 4 − 3x 9 Solve for y − + 4 Add 4 to both sides y = 3x − − 5 Our Solution Example 144. Find the equation of the line through the points (6, intercept form. − 2) and ( − 4, 1) in slope- m = y2 − x2 − = − First we must ﬁnd the slope y1 x1 3 10 x1) Use slope and either point, use point Plug values into slope formula and evaluate slope formula − 6) Simplify signs − 6) Distribute slope Solve for y. Subtract 2 from both sides Using 10 5 on right so we have a common denominator Our Solution ) − 6 − y = 3 10 y1 = m(x − ( − 2x 3 10 3 (x 10 3 10 − − x + 2 − y = 3 10 − − x − 9 5 10 5 1 5 World View Note: The city of Konigsberg (now Kaliningrad, Russia) had a river that ﬂowed through the city breaking it into several parts. There were 7 bridges that connected the parts of the city. In 1735 Leonhard Euler considered the question of whether it was possible to cross each bridge exactly once and only once. It turned out that this problem was impossible, but the work laid the foundation of what would become graph theory. 109 2.4 Practice - Point-Slope Form Write the point-slope form of the equation of the line through the given point with the given slope. 1) through (2, 3), slope = undeﬁned 2) through (1, 2), slope = undeﬁned 3) through (2, 2), slope = 1 2 4) through (2, 1), slope = 5) through ( 1, − − 5), slope = 9 6) through (2, − 4, 1), slope = 3 4 8) through (4, − 2), slope = 1 2 2 − − 2 3), slope = − 1, 1), slope = 4 3 10) through ( − 7) through ( − 9) through (0, − 11) through (0, 2), slope = − 5), slope = − 1 4 − 13) through ( 15) through ( 5, − 3), slope = 1 5 1, 4), slope = 5 4 − − − 12) through (0, 2), slope = 5 4 − 14) through ( 1, − − 4), slope = 16) through (1, 4), slope = − − 2 3 − 3 2 Write the slope-intercept form of the equation of the line through the given point with the given slope. 3 5 3 2 2 5 − 17) through: ( − 19) through: (5, 21) through: ( − 1, 5), slope = 2 − 1), slope = − − 4, 1), slope = 1 2 23) through: (4, 2), slope = 25) through: ( − 27) through: (2, 29) through:( − − 5, − 3), slope = − 2), slope = 1 − 3, 4), slope=undeﬁned 31) through: ( 4, 2), slope = − 1 2 − 18) through: (2, 2), slope = 20) through: ( − 22) through: (4, − 2, − − 2), slope = − 3), slope = − 2 2 3 − 7 4 24) through: ( 2, 0), slope = − 26) through: (3, 3), slope = 7 3 5 2 − 28) through: ( 4, − − 3), slope = 0 30) through: ( 2, 5), slope = 2 − 32) through: (5, 3), slope = 6 5 − 110 Write the point-slope form of the equation of the line through the given points. 33) through: ( 4, 3) and ( − 35) through: (5, 1) and ( − 3, 1) − 3, 0) 37) through: ( 4, − − 2) and (0, 4) 39) through: (3, 5) and ( 5, 3) − 41) through: (3, 3) and ( 4, 5) − − 34) through: (1, 3) and ( 3, 3) − 36) through: ( 38) through: ( 40) through: ( 42) through: ( − − − − 4, 5) and (4, 4) 4, 1) and (4, 4) 4) and ( 5) and ( 1, 1, − − − − 5, 0) 4) 5, − Write the slope-intercept form of the equation of the line through the given points. 43) through: ( 45) through: ( − − 5, 1) and ( − 5, 5) and (2, 1, 2) − 3) − 44) through: ( − 46) through: (1, 1) and (5, 5, − 1) and ( − 2) 4) − − 47) through: (4, 1) and (1, 4) 48) through: (0, 1) and ( − 49) through: (0, 2) and (5, 3) 50) through: (0, 2) and (2, 4) 5, − 3, 0) 51) through: (0, 3) and ( − 1, − 1) − 52) through: ( 2, 0) and (5, 3) − 111 2.5 Graphing - Parallel and Perpendicular Lines Objective: Identify the equation of a line given a parallel or perpendicular line. There is an interesting connection between the slope of lines that are parallel and the slope of lines that are perpendicular (meet at a right angle). This is shown in the following example. Example 145. The above graph has two parallel lines. The slope of the top line is 2 down 2, run 3, or 3. The slope of the bottom line is down 2, run 3 as well, or − 2 3. − The above graph has two perpendicular lines. The slope of the ﬂatter line or is 2 . The slope of the steeper line is down 3, 3 run 2 or run up 2, 3 3 2 . − World View Note: Greek Mathemati
cian Euclid lived around 300 BC and published a book titled, The Elements. In it is the famous parallel postulate which mathematicians have tried for years to drop from the list of postulates. The attempts have failed, yet all the work done has developed new types of geometries! As the above graphs illustrate, parallel lines have the same slope and perpendicular lines have opposite (one positive, one negative) reciprocal (ﬂipped fraction) slopes. We can use these properties to make conclusions about parallel and perpendicular lines. Example 146. Find the slope of a line parallel to 5y 2x = 7. − 5y 2x = 7 To ﬁnd the slope we will put equation in slope intercept form − − + 2x + 2x Add 2x to both sides 5y = 2x + 7 Put x term ﬁrst 5 Divide each term by The slope is the coeﬃcient of x 112 m = m = 2 5 2 5 Slope of ﬁrst line. Parallel lines have the same slope Our Solution Example 147. Find the slope of a line perpendicular to 3x 4y = 2 − 4y = 2 To ﬁnd slope we will put equation in slope − − 3x 3x − 4y = 4 − − 3x Subtract 3x from both sides 3x + 2 Put x term ﬁrst 4 4 Divide each term by − The slope is the coeﬃcient of x intercept form − m = m = − 3 4 4 3 Slope of ﬁrst lines. Perpendicular lines have opposite reciprocal slopes Our Solution Once we have a slope, it is possible to ﬁnd the complete equation of the second line if we know one point on the second line. Example 148. Find the equation of a line through (4, 5) and parallel to 2x 3y = 6. − − 2x 2x − 3y = 3y = 6 We ﬁrst need slope of parallel line 2x Subtract 2x from each side 2x + 6 Put x term ﬁrst 3 Divide each term by 3 − Identify the slope, the coeﬃcient of x Parallel lines have the same slope We will use this slope and our point (4, 5) − y y − − ( − y1 = m(x x1) Plug this information into point slope formula 5) = (x 4) Simplify signs x − 4) Our Solution 113 Example 149. Find the equation of the line through (6, slope-intercept form. − 9) perpendicular to y = 3 5 − x + 4 in y = 3 5 − x + 4 Identify the slope, coeﬃcient of Perpendicular lines have opposite reciprocal slopes We will use this slope and our point (6, 9) − y1 = m(x y − − x1) Plug this information into point slope formula − y ( − − 9x − (x − 6) Simplify signs 6) Distribute slope − − 10 Solve for y 9 Subtract 9 from both sides − 19 Our Solution Zero slopes and no slopes may seem like opposites (one is a horizontal line, one is a vertical line). Because a horizontal line is perpendicular to a vertical line we can say that no slope and zero slope are actually perpendicular slopes! Example 150. Find the equation of the line through (3, 4) perpendicular to x = 2 − x = 2 This equation has no slope, a vertical line no slope Perpendicular line then would have a zero slope − m = 0 Use this and our point (3, 4) y − y y1 = m(x 4 = 0(x y − x1) Plug this information into point 3) Distribute slope Solve for Add 4 to each side y = 4 Our Solution slope formula − Being aware that to be perpendicular to a vertical line means we have a horizontal line through a y value of 4, thus we could have jumped from this point right to the solution, y = 4. 114 2.5 Practice - Parallel and Perpendicular Lines Find the slope of a line parallel to each given line. 1) y = 2x + 4 2) y = 3) y = 4x − y = 4 5) x − 7) 7x + y = 2 − 5 10 3 x − − 5y = 20 6) 6x − 8) 3x + 4y = 8 − Find the slope of a line perpendicular to each given line. 9) x = 3 11) y = 1 3 x − 3y = 13) x − − 15) x + 2y = 8 1 2 x 1 − 10) y = − 12) y = 4 5 x 6 14) 3x 16) 8x − − 3 y = − 3y = 9 − Write the point-slope form of the equation of the line described. 17) through: (2, 5), parallel to x = 0 18) through: (5, 2), parallel to y = 7 5 x + 4 19) through: (3, 4), parallel to y = 9 2 x 5 − 20) through: (1, 1), parallel to y = − 21) through: (2, 3), parallel to − 22) through: ( − 1, 3), parallel to y = 3x 1 − − 23) through: (4, 2), parallel to x = 0 24) through: (1, 4), parallel to y = 7 5 x + 2 25) through: (1, 26) through: (1, − 5), perpendicular to − 2), perpendicular to x + y = 1 − x + 2y = 2 − 115 27) through: (5, 2), perpendicular to 5x + y = 3 − 28) through: (1, 3), perpendicular to x + y = 1 − 4x + y = 0 29) through: (4, 2), perpendicular to − 5), perpendicular to 3x + 7y = 0 30) through: ( − 31) through: (2, 3, 32) through: ( − − 2) perpendicular to 3y − 2, 5). perpendicular to y x = 0 2x = 0 − − Write the slope-intercept form of the equation of the line described. 33) through: (4, 34) through: ( 35) through: ( 36) through: ( 37) through: ( 3), parallel to y = − − 5, 2), parallel to y = 3 5 x 2x − 3, 1), parallel to y = 4, 0), parallel to ), parallel to y = 4, − 1 2 x + 1 − − − − 38) through: (2, 3), parallel to y = 5 2 x 1 − 39) through: ( 2, − − 1 2 x 1), parallel to y = − 4), parallel to y = 3 5 x 40) through: ( 5, − 41) through: (4, 3), perpendicular to − 42) through: ( 3, 5), perpendicular to x + 2y = − 43) through: (5, 2), perpendicular to x = 0 − 4 − 44) through: (5, 45) through: ( − 46) through: (2, 1), perpendicular to − 2, 5), perpendicular to 3), perpendicular to − 5x + 2y = 10 x + y = − 2x + 5y = 2 10 − − − − 47) through: (4, 3), perpendicular to x + 2y = − 4, 1), perpendicular to 4x + 3y = − 48) through: ( − 6 − 9 − 116 Chapter 3 : Inequalities 3.1 Solve and Graph Inequalities .................................................................118 3.2 Compound Inequalities ..........................................................................124 3.3 Absolute Value Inequalities ....................................................................128 117 3.1 Inequalities - Solve and Graph Inequalities Objective: Solve, graph, and give interval notation for the solution to linear inequalities. When we have an equation such as x = 4 we have a speciﬁc value for our variable. With inequalities we will give a range of values for our variable. To do this we will not use equals, but one of the following symbols: > Greater than > Greater than or equal to < Less than 6 Less than or equal to World View Note: English mathematician Thomas Harriot ﬁrst used the above symbols in 1631. However, they were not immediately accepted as symbols such as ⊏ and ⊐ were already coined by another English mathematician, William Oughtred. If we have an expression such as x < 4, this means our variable can be any number 2, 0, 3, 3.9 or even 3.999999999 as long as it is smaller smaller than 4 such as − 118 than 4. If we have an expression such as x > any number greater than or equal to − 2, such as 5, 0, 2, this means our variable can be 1, − − 1.9999, or even 2. − − Because we don’t have one set value for our variable, it is often useful to draw a picture of the solutions to the inequality on a number line. We will start from the value in the problem and bold the lower part of the number line if the variable is smaller than the number, and bold the upper part of the number line if the variable is larger. The value itself we will mark with brackets, either ) or ( for less than or greater than respectively, and ] or [ for less than or equal to or greater than or equal to respectively. Once the graph is drawn we can quickly convert the graph into what is called interval notation. Interval notation gives two numbers, the ﬁrst is the smallest value, the second is the largest value. If there is no largest value, we can use ∞ (inﬁnity). If there is no smallest value, we can use negative inﬁnity. If we use either positive or negative inﬁnity we will always use a curved bracket for that value. − ∞ Example 151. Graph the inequality and give the interval notation x < 2 Start at 2 and shade below Use ) for less than Our Graph Interval Notation ( − ∞ , 2) Example 152. Graph the inequality and give the interval notation y > 1 − Start at 1 and shade above Use [ for greater than or equal − Our Graph Interval Notation [ 1, ) ∞ − We can also take a graph and ﬁnd the inequality for it. 119 Example 153. Give the inequality for the graph: Graph starts at 3 and goes up or greater. Curved bracket means just greater than x > 3 Our Solution Example 154. Give the inequality for the graph: Graph starts at equal to − 4 and goes down or less. Square bracket means less than or x 6 − 4 Our Solution Generally when we are graphing and giving interval notation for an inequality we will have to ﬁrst solve the inequality for our variable. Solving inequalities is very similar to solving equations with one exception. Consider the following inequality and what happens when various operations are done to it. Notice what happens to the inequality sign as we add, subtract, multiply and divide by both positive and negative numbers to keep the statment a true statement. Subtract 2 from both sides 5 > 1 Add 3 to both sides 8 > 4 6 > 2 Multiply both sides by 3 12 > 6 Divide both sides by 2 6 > 3 Add 1 to both sides 5 > 2 9 > 6 Multiply both sides by 12 Divide both sides by − Subtract − − 6 2 4 from both sides − 18 < − − 3 > 2 Symbol ﬂipped when we multiply or divide by a negative! As the above problem illustrates, we can add, subtract, multiply, or divide on both sides of the inequality. But if we multiply or divide by a negative number, the symbol will need to ﬂip directions. We will keep that in mind as we solve inequalities. Example 155. Solve and give interval notation 2x > 11 5 − Subtract 5 from both sides 120 5 − 5 − 2x > 6 Divide both sides by − 2 Divide by a negative 2 − x 6 3 Graph, starting at 2 ﬂip symbol! − − 3, going down with ] for less than or equal to − − − ( 3] , − − ∞ Interval Notation The inequality we solve can get as complex as the linear equations we solved. We will use all the same patterns to solve these inequalities as we did for solving equations. Just remember that any time we multiply or divide by a negative the symbol switches directions (multiplying or dividing by a positive does not change the symbol!) Example 156. Solve and give interval notation 3(2x 6x − − 4) + 4x < 4(3x 12 + 4x < 12x 7) + 8 Distribute 28 + 8 Combine like terms − − 10x 10x − − 12 < 12x 10x − 12 < 2x − + 20 − 20 Move variable to
one side Subtract 10x from both sides 20 Add 20 to both sides − + 20 8 < 2x Divide both sides by 2 2 4 < x Be careful with graph, x is larger! 2 (4, ) ∞ Interval Notation It is important to be careful when the inequality is written backwards as in the previous example (4 < x rather than x > 4). Often students draw their graphs the wrong way when this is the case. The inequality symbol opens to the variable, this means the variable is greater than 4. So we must shade above the 4. 121 3.1 Practice - Solve and Graph Inequalities Draw a graph for each inequality and give interval notation. 5 − 2 > k 1) n > 3) − 5) 5 > x Write an inequality for each graph. 7) 8) 9) 10) 11) 12) 2) n > 4 4) 1 > k 6) − 5 < x 122 Solve each inequality, graph each solution, and give interval notation. 13) x 11 > 10 15) 2 + r < 3 17) 8 + n 3 > 6 19) 2 > a 2 − 5 14) − 16) m 5 2 6 n 13 6 6 5 − 18) 11 > 8 + x 2 20) v 9 − 4 − 6 2 21) 47 > 8 5x − − 22) 6 + x 12 6 1 − 10 > 60 24) 7n − 26) 5 > x − 5 + 1 2(3 + k) < 23) − 25) 18 < 27) 24 > 2( − 6(m − − 44 − 8 + p) 6) − 6) < 29) r 5(r − − − 31) 24 + 4b < 4(1 + 6b) − 18 28) 30) 32) − − − 8(n − 60 > 5) > 0 4( − − 2n) > 8(2 − 36 + 6x > 34) − − 36) 3(n + 3) + 7(8 6x 3) − 16 + n − 8(x + 2) + 4x 8n) < 5n + 5 + 2 − 5p) + 3 > 2(8 − − 5p) 33) 5v 5 < 5(4v + 1) − − − 35) 4 + 2(a + 5) < 2( a − − 4) 37) (k − − 2) > − − − k 20 38) (4 − − 123 3.2 Inequalities - Compound Inequalities Objective: Solve, graph and give interval notation to the solution of compound inequalities. Several inequalities can be combined together to form what are called compound inequalities. There are three types of compound inequalities which we will investigate in this lesson. The ﬁrst type of a compound inequality is an OR inequality. For this type of inequality we want a true statment from either one inequality OR the other inequality OR both. When we are graphing these type of inequalities we will graph each individual inequality above the number line, then move them both down together onto the actual number line for our graph that combines them together. When we give interval notation for our solution, if there are two diﬀerent parts to (union) symbol between two sets of interval notation, the graph we will put a one for each part. ∪ Example 157. Solve each inequality, graph the solution, and give interval notation of solution Solve each inequality 4 Add or subtract ﬁrst 2x − + or 4 4 − x > 2 Divide 2x > 8 or − 2 1 2 − x > 4 or x 6 − − − 1 Dividing by negative ﬂips sign 2 Graph the inequalities separatly above number line ( 2] , − ∪ (4, ∞ − ∞ ) Interval Notation World View Note: The symbol for inﬁnity was ﬁrst used by the Romans, although at the time the number was used for 1000. The greeks also used the symbol for 10,000. There are several diﬀerent results that could result from an OR statement. The graphs could be pointing diﬀerent directions, as in the graph above, or pointing in the same direction as in the graph below on the left, or pointing opposite directions, but overlapping as in the graph below on the right. Notice how interval notation works for each of these cases. 124 As the graphs overlap, we take the largest graph for our solution. When the graphs are combined they cover the entire number line. Interval Notation: ( , 1) − ∞ Interval Notation: ( , ∞ − ∞ ) or R The second type of compound inequality is an AND inequality. AND inequalities require both statements to be true. If one is false, they both are false. When we graph these inequalities we can follow a similar process, ﬁrst graph both inequalities above the number line, but this time only where they overlap will be drawn onto the number line for our ﬁnal graph. When our solution is given in interval notation it will be expressed in a manner very similar to single inequalities (there is a symbol that can be used for AND, the intersection , but we will not use it here). ∩ Example 158. Solve each inequality, graph the solution, and express it interval notation. 2x + 8 > 5x 2x − 2x 8 > 3x − 7 and 5x 3x − 7 and 2x − − − 3x 3 > 3x + 1 Move variables to one side 3 > 1 Add 7 or 3 to both sides + 7 − + 3 + 3 − + 7 15 > 3x and 2x > 4 Divide 3 2 2 3 5 > x and x > 2 Graph, x is smaller (or equal) than 5, greater than 2 (2, 5] Interval Notation Again, as we graph AND inequalities, only the overlapping parts of the individual graphs makes it to the ﬁnal number line. As we graph AND inequalities there are also three diﬀerent types of results we could get. The ﬁrst is shown in the above 125 example. The second is if the arrows both point the same way, this is shown below on the left. The third is if the arrows point opposite ways but don’t overlap, this is shown below on the right. Notice how interval notation is expressed in each case. In this graph, the overlap is only the smaller graph, so this is what makes it to the ﬁnal number line. Interval Notation: ( 2) , − − ∞ In this graph there is no overlap of the parts. Because their is no overlap, no values make it to the ﬁnal number line. Interval Notation: No Solution or ∅ The third type of compound inequality is a special type of AND inequality. When our variable (or expression containing the variable) is between two numbers, we can write it as a single math sentence with three parts, such as 5 < x 6 8, to show x is between 5 and 8 (or equal to 8). When solving these type of inequalities, because there are three parts to work with, to stay balanced we will do the same thing to all three parts (rather than just both sides) to isolate the variable in the middle. The graph then is simply the values between the numbers with appropriate brackets on the ends. Example 159. Solve the inequality, graph the solution, and give interval notation. 6 6 2 − 4x + 2 < 2 2 2 − − Subtract 2 from all three parts 8 6 4 − − − − 4x < 0 Divide all three parts by − Graph x between 0 and 2 − 4 − 4 Dividing by a negative ﬂips the symbols Flip entire statement so values get larger left to right (0, 2] Interval Notation 126 3.2 Practice - Compound Inequalities Solve each compound inequality, graph its solution, and give interval notation. 1) n 3 6 3 or − − 5n 6 10 − 2) 6m > − 24 or m 7 < 12 − 3) x + 7 > 12 or 9x < 45 − 13 or 6x 6 60 − 6 < − 5) x 7) v 8 − > 1 and v 2 < 1 − − 9) 8 + b < 3 and 4b < 20 − − 11) a + 10 > 3 and 8a 6 48 13) 3 6 9 + x 6 7 15) 11 < 8 + k 6 12 17) 19) 21) 23 − 16 6 2n 3m 6 11 10 6 22 − − 5b + 10 6 30 and 7b + 2 6 40 − 25) 3x − 9 < 2x + 10 and 5 + 7x 6 10x 10 − 8v and 7v + 9 6 6 + 10v − 27) 8 6v 6 8 − − 29) 1 + 5k 6 7k 3 or k − 31) 2x + 9 > 10x + 1 and 3x − 10 > 2k + 10 2 < 7x + 2 − − 5 < 4) 10r > 0 or r − 6) 9 + n < 2 or 5n > 40 − 12 8) − 9x < 63 and x 4 < 1 10) − 6n 6 12 and n 3 6 2 6 + v > 0 and 2v > 4 12) − 14) 0 > x 9 16) − 18) 1 6 p 8 > 1 − 11 6 n 6 0 9 6 5 − − 20) 3 + 7r > 59 or 6r 3 > 33 22) 6 8x > − − 24) n + 10 > 15 or 4n − − − 6 or 2 + 10x > 82 5 < 1 − − 6 or 10n 26) 4n + 8 < 3n − 2a > 2a + 1 or 10a 10r 6 8 + 4r or − 28) 5 30) 8 − − 8 > 9 + 9n − 10 > 9a + 9 − 6 + 8r < 2 + 8r 32) − 9m + 2 < 10 − − 6m or − m + 5 > 10 + 4m 127 3.3 Inequalities - Absolute Value Inequalities Objective: Solve, graph and give interval notation for the solution to inequalities with absolute values. When an inequality has an absolute value we will have to remove the absolute value in order to graph the solution or give interval notation. The way we remove the absolute value depends on the direction of the inequality symbol. Consider < 2. x \| \| Absolute value is deﬁned as distance from zero. Another way to read this inequality would be the distance from zero is less than 2. So on a number line we will shade all points that are less than 2 units away from zero. This graph looks just like the graphs of the three part compound inequalities! When the absolute value is less than a number we will remove the absolute value by changing the problem to a three part inequality, with the negative value on the 2 < x < 2, as the left and the positive value on the right. So graph above illustrates. < 2 becomes − x \| \| Consider > 2. x \| \| Absolute value is deﬁned as distance from zero. Another way to read this inequality would be the distance from zero is greater than 2. So on the number line we shade all points that are more than 2 units away from zero. This graph looks just like the graphs of the OR compound inequalities! When the absolute value is greater than a number we will remove the absolute value by changing the problem to an OR inequality, the ﬁrst inequality looking just like the problem with no absolute value, the second ﬂipping the inequality symbol and 2, as the graph changing the value to a negative. So above illustrates. > 2 becomes x > 2 or x < x \| − \| World View Note: The phrase “absolute value” comes from German mathematician Karl Weierstrass in 1876, though he used the absolute value symbol for complex numbers. The ﬁrst known use of the symbol for integers comes from a 1939 128 edition of a college algebra text! For all absolute value inequalities we can also express our answers in interval notation which is done the same way it is done for standard compound inequalities. We can solve absolute value inequalities much like we solved absolute value equations. Our ﬁrst step will be to isolate the absolute value. Next we will remove the absolute value by making a three part inequality if the absolute value is less than a number, or making an OR inequality if the absolute value is greater than a number. Then we will solve these inequalites. Remember, if we multiply or divide by a negative the inequality symbol will switch directions! Example 160. Solve, graph, and give interval notation for the solution 4x \| 5 > 6 OR 4x 5 − \| 5 6 4x − + 5 + 5 6 Solve − + 5 + 5 Add 5 to both sides − > 6 Absolute value is greater, use OR 4x > 11 OR 4x 6 4 4 x > 11 4 OR x 6 4 1 Divide both sides by 4 − 4 1 4 Graph − 1 4 , − 11 4 , ∞ ∪ − ∞ Interval notation Example 161. Solve, graph, and give interval notation for the solution \| − 16 Add 4 to both sides + 4 12 Divide bo
th sides by 3 Dividing by a negative switches the symbol − − > 4 Absolute value is greater, use OR − 3 \| 129 x > 4 OR x 6 4 Graph − ( 4] , − ∪ [4, ) ∞ − ∞ Interval Notation In the previous example, we cannot combine terms, the − adding 4, then divide by 3 has an absolute value attached. So we must ﬁrst clear the 3. The next example is similar. 3 because they are not like 4 by 4 and − − − − Example 162. Solve, graph, and give interval notation for the solution − 2 − 4x + 1 \| 2 \| − 9 9 \| > 4x + 1 \| 2 4x + 1 3 < 4x + 1 < 3 1 1 − \| 1 \| − − − − Subtract 9 from both sides > 3 9 6 Divide both sides by 2 Dividing by negative switches the symbol − − − < 3 Absolute value is less, use three part − 2 Solve Subtract 1 from all three parts − 4 < 4x < 2 Divide all three parts by 4 4 4 4 1 2 1 < x < − Graph 1, 1 2 − Interval Notation In the previous example, we cannot distribute the 2 into the absolute value. We can never distribute or combine things outside the absolute value with what is inside the absolute value. Our only way to solve is to ﬁrst isolate the absolute value by clearing the values around it, then either make a compound inequality (and OR or a three part) to solve. − 130 It is important to remember as we are solving these equations, the absolute value is always positive. If we end up with an absolute value is less than a negative number, then we will have no solution because absolute value will always be positive, greater than a negative. Similarly, if absolute value is greater than a negative, this will always happen. Here the answer will be all real numbers. Example 163. Solve, graph, and give interval notation for the solution 6x \| − 12 + 4 12 4 − 6x 4 6x 1 1 \| \| − − \| \| 1 Subtract 12 from both sides < 4 \| 12 − < 8 Divide both sides by 4 − 4 2 Absolute value can′t be less than a negative < − No Solution or ∅ Example 164. Solve, graph, and give interval notation for the solution \| \| x + 7 6 \| − 6 17 \| − − > \| Subtract 5 from both sides − 6 12 Divide both sides by 6 − 6 Dividing by a negative ﬂips the symbol 2 Absolute value always greater than negative All Real Numbers or R 131 3.3 Practice - Absolute Value Inequalities Solve each inequality, graph its solution, and give interval notation. 1) 3) 5) 7) 9) \| \| \| \| \| x \| 2x < − 3x 2 < 9 − 11) 1 + 2 \| x \| 2x 6 9 > = 3 1 \| 5 \| − − 13) 6 15) 17) 19) \| \| \| > 5 − \| 3x \| x = 3 > = 3 \| 3x 5 \| − > > 3 > = 10 6 15 − > 2x \| − − 7 < \| > 1 − 5 \| 21) 4 + 3 x \| − 23 \| 25) 2 3 \| − − 27) 4 29) 3 5 2 \| − 4x \| − − 31) 5 2 \| − − 4 − 2x − 3x − 2x + 6 6 < > 8 4 − − \| \| 33) 4 4 − \| − 10 + x 35) \| − > 8 \| 8 1 26) 28) 30) x 2) \| 6 8 \| x + 3 4) \| 6) x 8) \| 10) < 4 \| < 12 6 4 8 − \| x + 3 \| 2x + 5 \| 3 − \| > 5 12) 10 x \| x − 2x 14) 16) 18 \| − 20) 3 22) 3 24) 4 2 − \| 2 \| − − 3x − 4x 7 > \| 3x − − \| − − − \| − 3x 32) 6 1 3 \| − − 4x \| 2x 34) 4 3 − − \| − 132 Chapter 4 : Systems of Equations 4.1 Graphing ................................................................................................134 4.2 Substitution ............................................................................................139 4.3 Addition/Elimination .............................................................................146 4.4 Three Variables ......................................................................................151 4.5 Application: Value Problems .................................................................158 4.6 Application: Mixture Problems .............................................................167 133 4.1 Systems of Equations - Graphing Objective: Solve systems of equations by graphing and identifying the point of intersection. − We have solved problems like 3x 4 = 11 by adding 4 to both sides and then dividing by 3 (solution is x = 5). We also have methods to solve equations with more than one variable in them. It turns out that to solve for more than one variable we will need the same number of equations as variables. For example, to solve for two variables such as x and y we will need two equations. When we have several equations we are using to solve, we call the equations a system of equations. When solving a system of equations we are looking for a solution that works in both equations. This solution is usually given as an ordered pair (x, y). The following example illustrates a solution working in both equations Example 165. Show (2,1) is the solution to the system 3x 2, 1) Identify x and y from the orderd pair x = 2, y = 1 Plug these values into each equation 3(2) (1) = 5 First equation − 6 − 1 = 5 Evaluate 5 = 5 True (2) + (1) = 3 Second equation, evaluate 3 = 3 True As we found a true statement for both equations we know (2,1) is the solution to the system. It is in fact the only combination of numbers that works in both equations. In this lesson we will be working to ﬁnd this point given the equations. It seems to follow that if we use points to describe the solution, we can use graphs to ﬁnd the solutions. If the graph of a line is a picture of all the solutions, we can graph two lines on the same coordinate plane to see the solutions of both equations. We are inter- 134 ested in the point that is a solution for both lines, this would be where the lines intersect! If we can ﬁnd the intersection of the lines we have found the solution that works in both equations. Example 166 = First: m = − Second: m = 2 − (4,1) To graph we identify slopes and y intercepts − Now we can graph both lines on the same plane. To graph each equation, we start at the y-intercept and use the slope rise run to get the next point and connect the dots. Remember a negative slope is downhill! Find the intersection point, (4,1) (4,1) Our Solution Often our equations won’t be in slope-intercept form and we will have to solve both equations for y ﬁrst so we can idenﬁty the slope and y-intercept. Example 167. 6x 3y = − 2x + 2y = 9 6 − − Solve each equation for y − − − 6x 6x 3y = 3 3y = 9 − 6x − 6x 9 − 3 3 − − y = 2x + 3 − − − 6 − 2x 2x + 2y = 2x 2y = 2 − y = − − Subtract x terms 6 Put x terms ﬁrst − 2 3 Divide by coeﬃcient of y Identify slope and y − 2x 2 x intercepts − 135 First: m = 2 1 Second = (-2,-1) Now we can graph both lines on the same plane 3 − To graph each equation, we start at the y-intercept and use the slope rise run to get the next point and connect the dots. Remember a negative slope is downhill! Find the intersection point, ( 2, − − 1) 2, ( − − 1) Our Solution As we are graphing our lines, it is possible to have one of two unexpected results. These are shown and discussed in the next two example. Example 168 First: m = 3 2 Second Identify slope and y − intercept of each equation Now we can graph both equations on the same plane To graph each equation, we start at the y-intercept and use the slope rise run to get the next point and connect the dots. The two lines do not intersect! They are parallel! If the lines do not intersect we know that there is no point that works in both equations, there is no solution ∅ No Solution We also could have noticed that both lines had the same slope. Remembering 136 that parallel lines have the same slope we would have known there was no solution even without having to graph the lines. Example 169. 2x 3x − − 6y = 12 9y = 18 Solve each equation for y 9y = 18 3x Subtract x terms 3x + 18 Put x terms ﬁrst − 9 − 1 y = 3 x 9 Divide by coeﬃcient of y 2 Identify the slopes and y − − intercepts − 6y = 12 2x − − 2x 2x 2x + 12 6 6 − − 3x 3x − 9y = 9 − − − − − 6y = − First: m = 1 3 Second − Now we can graph both equations together To graph each equation, we start at the y-intercept and use the slope rise run to get the next point and connect the dots. Both equations are the same line! As one line is directly on top of the other line, we can say that the lines “intersect” at all the points! Here we say we have inﬁnite solutions Once we had both equations in slope-intercept form we could have noticed that the equations were the same. At this point we could have stated that there are inﬁnite solutions without having to go through the work of graphing the equations. World View Note: The Babylonians were the ﬁrst to work with systems of equations with two variables. However, their work with systems was quickly passed by the Greeks who would solve systems of equations with three or four variables and around 300 AD, developed methods for solving systems with any number of unknowns! 137 4.1 Practice - Graphing Solve each equation by graphing. 1) y = y = 3) y = y = 5) y = y = − 7) y = 1 3 y = − 9 5x − − 11) x + 3y = − 5x + 3y = 3 9 13) x y = 4 2x + y = − − 15) 2x + 3y = 2x + y = 2 − 1 17) 2x + y = 2 x y = 4 − 19) 2x + y = − x + 3y = 9 2) y = 2x + 2 4 y = x − 8) y = 2x 10 12) x + 4y = − 2x + y = 4 12 14) 6x + y = x + y = 2 − 3 16) 3x + 2y = 2 3x + 2y = − 6 18) x + 2y = 6 5x − 4y = 16 6 2 20) x y = 3 5x + 2y = 8 − 21) 0 = 6x − 12 = 6x 9y + 36 − 3y 23) 2x 0 = − − 25) 3 + y = − y = 2x − − x − 6x = 1 y 3 − 27) 29) 4 − − − − − y + 7x = 4 y 3 + 7x = 0 − 12 + x = 4y 5x = 4y − 12 − y − 22) 24) 2y + = − − − 2y = 2y = − − x − 2x = x 4 − 5x + 4 4y − 4 − − 4y 26) 16 = = − 5x + − 28) 30) 138 4.2 Systems of Equations - Substitution Objective: Solve systems of equations using substitution. When solving a system by graphing has several limitations. First, it requires the graph to be perfectly drawn, if the lines are not straight we may arrive at the wrong answer. Second, graphing is not a great method to use if the answer is really large, over 100 for example, or if the answer is a decimal the that graph will not help us ﬁnd, 3.2134 for example. For these reasons we will rarely use graphing to solve our systems. Instead, an algebraic approach will be used. The ﬁrst algebraic approach is called substitution. We will build the concepts of substitution through several example, then end with a ﬁve-step process to solve problems using this method. Example 170. We already know x = 5, substitute this into the other equation x = 5 y = 2x y = 2(5) y = 10 − − − 3 3 Evaluate, multiply ﬁrst
3 Subtract y = 7 We now also have y (5, 7) Our Solution When we know what one variable equals we can plug that value (or expression) in for the variable in the other equation. It is very important that when we substitute, the substituted value goes in parenthesis. The reason for this is shown in the next example. Example 171. 2x − y = 3x 3y = 7 7 − 7) = 7 2x − 3(3x − We know y = 3x − 7, substitute this into the other equation Solve this equation, distributing 3 ﬁrst − 139 2x − − Subtract 21 9x + 21 = 7 Combine like terms 2x 7x + 21 = 7 21 14 Divide by 7 21 7x = 7 − − 7 9x − − − − − x = 2 We now have our x, plug into the y = equation to ﬁnd y − y = 3(2) y = 6 y = (2, − − − − Subtract 7 Evaluate, multiply ﬁrst 7 1 We now also have y 1) Our Solution By using the entire expression 3x 7 to replace y in the other equation we were able to reduce the system to a single linear equation which we can easily solve for our ﬁrst variable. However, the lone variable (a variable without a coeﬃcient) is not always alone on one side of the equation. If this happens we can isolate it by solving for the lone variable. − Example 172. − 3x + 2y = 1 x 5y = 6 + 5y + 5y x = 6 + 5y 3(6 + 5y) + 2y = 1 Lone variable is x, isolate by adding 5y to both sides. Substitute this into the untouched equation Solve this equation, distributing 3 ﬁrst 18 + 15y + 2y = 1 Combine like terms 15y + 2y Subtract 18 from both sides 18 + 17y = 1 18 18 17 Divide both sides by 17 17 17y = 17 − − − x = 6 + 5( y = − x = 6 1 We have our y, plug this into the x = equation to ﬁnd x − 1) Evaluate, multiply ﬁrst 5 Subtract x = 1 We now also have x − (1, − 1) Our Solution The process in the previous example is how we will solve problems using substitu- 140 tion. This process is described and illustrated in the following table which lists the ﬁve steps to solving by substitution. Problem 1. Find the lone variable 2. Solve for the lone variable 3. Substitute into the untouched equation 4. Solve 5. Plug into lone variable equation and evaluate Solution 5 − 2y = 2 5 4x − 2x + y = − Second Equation, y 2x + y = 2x − y = 4x − 4x + 10 + 4x = 2 8x + 10 = 2 10 10 2x − 2x − 2( − 5 − − 5 2x) = 2 − 8x = , ( − 8 − 8 1 − 2) − − 1) − Sometimes we have several lone variables in a problem. In this case we will have the choice on which lone variable we wish to solve for, either will give the same ﬁnal result. Example 173. − Find the lone variable: x or y in ﬁrst, or x in second. We will chose x in the ﬁrst Solve for the lone variable, subtract y from both sides Plug into the untouched equation, the second equation x = 5 y = 1 − 2y = 1 − 5 6 Divide both sides by 2 Solve, parenthesis are not needed here, combine like terms Subtract 5 from both sides 2y = 2 y We have our y! (5 − − x = 5 (3) Plug into lone variable equation, evaluate − x = 2 Now we have our x 141 (2, 3) Our Solution Just as with graphing it is possible to have no solution ∅ (parallel lines) or inﬁnite solutions (same line) with the substitution method. While we won’t have a parallel line or the same line to look at and conclude if it is one or the other, the process takes an interesting turn as shown in the following example. Example 174. Solve for the lone variable, subtract 4 from both sides Find the lone variable, y in the ﬁrst equation − − y + 4 = 3x 6x = 2y 8 y + 4 = 3x 4 4 − y = 3x 4 Plug into untouched equation 6x = 8 8 Combine like terms 6x 6x = 8 Variables are gone!A true statement. 8 = Solve, distribute through parenthesis 4) 8 − − 6x − − − − − − − − Inﬁnite solutions Our Solution 2(3x − 6x Because we had a true statement, and no variables, we know that anything that works in the ﬁrst equation, will also work in the second equation. However, we do not always end up with a true statement. Example 175. 6x 3y = 9 − − 2x + y = 5 − 2x + y = 5 − + 2x + 2x Find the lone variable, y in the second equation Solve for the lone variable, add 2x to both sides y = 5 + 2x Plug into untouched equation 6x − 6x 15 3(5 + 2x) = − 6x = − 15 − − − − Solve, distribute through parenthesis 9 9 Combine like terms 6x 9 Variables are gone!A false statement. 6x − No Solution ∅ Our Solution Because we had a false statement, and no variables, we know that nothing will work in both equations. 142 World View Note: French mathematician Rene Descartes wrote a book which included an appendix on geometry. It was in this book that he suggested using letters from the end of the alphabet for unknown values. This is why often we are solving for the variables x, y, and z. One more question needs to be considered, what if there is no lone variable? If there is no lone variable substitution can still work to solve, we will just have to select one variable to solve for and use fractions as we solve. Example 176. − − − 5x 5x = 5 6y = 14 − 2x + 4y = 12 − 6y = 5x 14 − + 6y + 6y 14 + 6y 5 5 6y 14 x = − 5 5 6y 5 12y 5 + 4y = 12 + 4y = 12 + No lone variable, we will solve for x in the ﬁrst equation Solve for our variable, add 6y to both sides Divide each term by 5 Plug into untouched equation Solve, distribute through parenthesis Clear fractions by multiplying by 5 + 4y(5) = 12(5) Reduce fractions and multiply − 12y + 20y = 60 28 + 8y = 60 28 28 − 8y = 32 8 8 − Combine like terms Subtract 28 from both sides − 12y + 20y Divide both sides by 8 − 2 + 14 − 5 28 5 − 28(5) 5 − 12y(5) 5 28 Plug into lone variable equation, multiply + 14 x = − 5 14 x = − 5 y = 4 We have our y 6(4) 5 24 5 10 5 Add fractions Reduce fraction + x = x = 2 Now we have our x (2, 4) Our Solution Using the fractions does make the problem a bit more tricky. This is why we have another method for solving systems of equations that will be discussed in another lesson. 143 4.2 Practice - Substitution Solve each system by substitution. 1) y = − y = 6x 3) y = − y = 2x 3x − 2x − 9 9 − 1 2) y = x + 5 2x y = 4 − − 6x + 3 4) y = − y = 6x + 3 5) y = 6x + 4 6) y = 3x + 13 y = 8) y = y = − − − 10) y = 7x y = 2x 22 − 2x 5x 9 21 − − 24 − 3x + 16 − x + 3y = 12 − y = 6x + 21 24 − 12) y = 3x − − 7) y = 3x + 2 5 y = 3x + 8 − 9) y = 2x y = − 11) y = 6x 3x − 13) y = 3x − 15) y = 3 − 2x + 9 6 − 3y = − 6 − 6y = 30 14) 6x y = 4y = − 6x + 2 − − 8 7 5 − 3x + 4y = 17 − 16) 7x + 2y = y = 5x + 5 − 17) 2x + 2y = 18 − y = 7x + 15 19) y = 8x + 19 x + 6y = 16 − − 21) 7x − y = 7 2y = 7 − 19 18) y = x + 4 4y = 3x − 20) y = − 2x + 8 6y = − 7x − 22) x − 2y = 4x + 2y = 18 − − − 13 23) x 5y = 7 2x + 7y = − 20 24) 3x 4y = 15 − 7x + y = 4 5 26) 6x + 4y = 16 8 20 2x + y = − 3 − 13 28) 7x + 5y = 4y = x − 16 − − 5x 5y = 2x + y = 7 − − − − − y = − 23 25) 2x − 8y = − x − − 6x + y = 20 3y = 3x 27) − − 29) 3x + y = 9 2x + 8y = − − 18 − 16 30) 144 31) 2x + y = 2 3x + 7y = 14 33) x + 5y = 15 3x + 2y = 6 − 32) 2x + y = − 5x + 3y = − 7 21 34) 2x + 3y = 7x + y = 3 − 10 35) 37) 2x + 4y = 2 − y = − 16 − 6x + 6y = 3y = 16 − 8x − 12 − 39) 2x + 3y = 16 7x y = 20 − − − − 2x + 2y = 7y = 5x 22 19 − − − 8x + 2y = 6 − 2x + 3y = 11 − − 4y = x 14 − 6x + 8y = 12 − − − 36) 38) 40) 145 4.3 Systems of Equations - Addition/Elimination Objective: Solve systems of equations using the addition/elimination method. When solving systems we have found that graphing is very limited when solving equations. We then considered a second method known as substituion. This is probably the most used idea in solving systems in various areas of algebra. However, substitution can get ugly if we don’t have a lone variable. This leads us to our second method for solving systems of equations. This method is known as either Elimination or Addition. We will set up the process in the following examples, then deﬁne the ﬁve step process we can use to solve by elimination. Example 177. 4y = 8 3x − 5x + 4y = 8x = 8 − − 24 16 8 Notice opposites in front of y ′s. Add columns. Solve for x, divide by 8 x = 2) + 4y = 10 + 4y = 2 We have our x! − 24 Plug into either original equation, simplify − 24 Add 10 to both sides − + 10 5( − − + 10 4y = 4 − 14 Divide by 4 4 7 y = − 2 7 2, − 2 Our Solution − Now we have our y! 4y and 4y. In the previous example one variable had opposites in front of it, Adding these together eliminated the y completely. This allowed us to solve for the x. This is the idea behind the addition method. However, generally we won’t have opposites in front of one of the variables. In this case we will manipulate the equations to get the opposites we want by multiplying one or both equations (on both sides!). This is shown in the next example. − Example 178. − 6x + 5y = 22 2x + 3y = 2 We can get opposites in front of x, by multiplying the second equation by 3, to get 6x and + 6x − 3(2x + 3y) = (2)3 Distribute to get new second equation. 146 − 6x + 9y = 6 New second equation 6x + 5y = 22 First equation still the same, add 14y = 28 Divide both sides by 14 14 14 y = 2 We have our y! 2x + 3(2) = 2 Plug into one of the original equations, simplify Subtract 6 from both sides 2x + 6 = 2 6 6 − − 2x = 4 Divide both sides by 2 − 2 2 x = 2 We also have our x! ( − 2, 2) Our Solution − − When we looked at the x terms, 6x and 2x we decided to multiply the 2x by 3 to get the opposites we were looking for. What we are looking for with our opposites is the least common multiple (LCM) of the coeﬃcients. We also could have solved the above problem by looking at the terms with y, 5y and 3y. The LCM of 3 and 5 is 15. So we would want to multiply both equations, the 5y by 3, and the 3y by 15y. This illustrates an important point, some problems we will have to multiply both equations by a constant (on both sides) to get the opposites we want. 5 to get opposites, 15y and − − Example 179. 3x + 6y = 2x + 9y = 9 26 − − We can get opposites in front of x, ﬁnd LCM of 6 and 9, 18y The LCM is 18. We will multiply to get 18y and − 3(3x + 6y) = ( 9x + 18y = 9)3 Multiply the ﬁrst equation by 3, both sides! 27 − − − 2(2x + 9y) = ( 4x − − − 2) Multiply the second equation by 26)( − 18y = 52 2, both sides! − 9x + 18y = 27 Add two new equations together − 18y = 52 − − 4x 5x 5 = 25 Divide both sides by 5 5 x = 5 We have ou
r solution for x 3(5) + 6y = 15 + 6y = 15 9 Plug into either original equation, simplify − 9 − 15 − Subtract 15 from both sides − 147 6y = 6 − y = (5, − − 24 Divide both sides by 6 6 4 Now we have our solution for y 4) Our Solution It is important for each problem as we get started that all variables and constants are lined up before we start multiplying and adding equations. This is illustrated in the next example which includes the ﬁve steps we will go through to solve a problem using elimination. Problem 2x 5y = − − 3y + 4 = 13 5x − Second Equation: − 1. Line up the variables and constants 2. Multiply to get opposites (use LCD) 3. Add 4. Solve 5. Plug into either original and solve Solution 4 − 3y + 4 = − 3y = 5y = 3y = 5x − − 4 + 5x + 5x 5x 4 2x 13 5x 4 First Equation: multiply by 5(2x 5y) = ( 5) 10x + 25y = 65 − − − − − − 13)( − − − − − 5 − Second Equation: multiply by 2 2(5x 10x 3y) = ( − 6y = 8 4)2 − − − − 10x + 25y = 65 6y = 10x 8 19y = 57 − − 19y = 57 19 19 y = 3 2x 2x 13 5(3) = − − 13 15 = − + 15 + 15 − 2x 2 = 2 2 x = 1 (1, 3) World View Note: The famous mathematical text, The Nine Chapters on the Mathematical Art, which was printed around 179 AD in China describes a formula very similar to Gaussian elimination which is very similar to the addition method. 148 Just as with graphing and substution, it is possible to have no solution or inﬁnite solutions with elimination. Just as with substitution, if the variables all disappear from our problem, a true statment will indicate inﬁnite solutions and a false statment will indicate no solution. Example 180. 2x 5y = 3 6x + 15y = − − To get opposites in front of x, multiply ﬁrst equation by 3 9 − 3(2x 5y) = (3)3 Distribute − 6x 15y = 9 − 6x 15y = 9 Add equations together − 6x + 15y = 9 − − 0 = 0 True statement Inﬁnite solutions Our Solution Example 181. 4x 6x − − 6y = 8 9y = 15 LCM for x′s is 12. 3(4x 6y) = (8)3 Multiply ﬁrst equation by 3 − 12x − 18y = 24 2(6x − 9y) = (15)( − 12x + 18y = − 2) Multiply second equation by 30 2 − − − 12x 18y = 24 Add both new equations together − − 12x + 18y = − 0 = − 30 6 False statement No Solution Our Solution We have covered three diﬀerent methods that can be used to solve a system of two equations with two variables. While all three can be used to solve any system, graphing works great for small integer solutions. Substitution works great when we have a lone variable, and addition works great when the other two methods fail. As each method has its own strengths, it is important you are familiar with all three methods. 149 4.3 Practice - Addition/Elimination Solve each system by elimination. 28 − 22 − 1) 4x + 2y = 0 9y = 4x − − 3) 5) 9x + 5y = 5y = 13 − 6x + 9y = 3 9y = 9 − 9x − 6x − − 10 14 6y = 6y = − − 5y = 28 − − x x + 4y = − 7) 4x 4x 9) − − 11) 2x y = 5 − 5x + 2y = 17 − 28 − 13) 10x + 6y = 24 6x + y = 4 − 15) 2x + 4y = 24 12y = 8 4x 17) − 10x − 7x + 4y = 8y = 4 − 8 − − 19) 5x + 10y = 20 6x 5y = 3 − − − 3y = 12 5y = 20 18 10 − − 7x 6x 21) − − 23) 9x 5x − − 2y = 7y = − − 25) 9x + 6y = 21 9y = 28 − 10x − 7x + 5y = 3x 8 − 3y = 12 − − − − 8x 8y = − − 10x + 9y = 1 − 8 27) 29) 31) 9y = 7 x − 18y + 4x = − 33) 0 = 9x + 5y y = 2 7 x 26 − 2) 4) − − 7x + y = y = 9x x − 2y = − − x + 2y = 7 10 22 7 − − − 6) 5x 5x − − 8) 10) 12) 15 15 5y = 5y = − − − − 3x + 3y = 3x + 9y = 12 24 − − 5y = 0 10y = 10x 10x − − 5x + 6y = 2y = 5 − − − x 30 − 17 − − 14) x + 3y = 1 − − 10x + 6y = 10 6x + 4y = 12 − 12x + 6y = 18 16) 18) 20) 22) − − − 3x 6x + 4y = 4 y = 26 3x − 9x 5y = 19 − 7y = − 11 − − 5x + 4y = 4 10y = 7x 10 − − − 24) 3x + 7y = 4x + 6y = − 8 4 − − 5y = 12 4x 10x + 6y = 30 26) − − − 28) 8x + 7y = 6x + 3y = 24 18 − − 30) 7x + 10y = 13 − 4x + 9y = 22 9x 21 + 12y 32=0 3 3 42y = 7 2 − y = 0 12x 34) − x 6 − 1 2 − − 150 4.4 Systems of Equations - Three Variables Objective: Solve systems of equations with three variables using addition/elimination. Solving systems of equations with 3 variables is very similar to how we solve systems with two variables. When we had two variables we reduced the system down to one with only one variable (by substitution or addition). With three variables we will reduce the system down to one with two variables (usually by addition), which we can then solve by either addition or substitution. To reduce from three variables down to two it is very important to keep the work organized. We will use addition with two equations to eliminate one variable. This new equation we will call (A). Then we will use a diﬀerent pair of equations and use addition to eliminate the same variable. This second new equation we will call (B). Once we have done this we will have two equations (A) and (B) with the same two variables that we can solve using either method. This is shown in the following examples. Example 182. z = 1 − − 3x + 2y 2x − 5x + 2y − 2y + 3z = 5 We will eliminate y using two diﬀerent pairs of equations z = 3 − 151 z = 1 Using the ﬁrst two equations, − − 2y + 3z = 5 Add the ﬁrst two equations 3x + 2y 2x x − − (A) + 2z = 4 This is equation (A), our ﬁrst equation − (B) 2x − 5x + 2y 3x 2y + 3z = 5 Using the second two equations z = 3 Add the second two equations − + 2z = 8 This is equation (B), our second equation (A) x + 2z = 4 Using (A) and (B) we will solve this system. (B) 3x + 2z = 8 We will solve by addition − 1(x + 2z) = (4)( 2z = x 1) Multiply (A) by Add to the second equation, unchanged x 2z = − 3x + 2z = 8 2x = 4 2 2 x = 2 We now have x! Plug this into either (A) or (B) Solve, divide by 2 (2) + 2z = 4 We plug it into (A), solve this equation, subtract 2 2 − 2 − 2z = 2 Divide by 2 2 z = 1 We now have z! Plug this and x into any original equation 2 3(2) + 2y (1) = − 2y + 5 = 5 − 2y = 2 y = Solve, subtract 5 1 We use the ﬁrst, multiply 3(2) = 6 and combine with − 1 − 5 − 6 Divide by 2 − 2 3 We now have y! − 1 − (2, − 3, 1) Our Solution As we are solving for x, y, and z we will have an ordered triplet (x, y, z) instead of 152 just the ordered pair (x, y). In this above problem, y was easily eliminated using the addition method. However, sometimes we may have to do a bit of work to get a variable to eliminate. Just as with addition of two equations, we may have to multiply equations by something on both sides to get the opposites we want so a variable eliminates. As we do this remmeber it is improtant to eliminate the same variable both times using two diﬀerent pairs of equations. Example 183. 3y + 2z = z = 4x − 6x + 2y 8x − − − − 29 No variable will easily eliminate. 16 We could choose any variable, so we chose x − y + 3z = 23 We will eliminate x twice. 4x − 6x + 2y 3y + 2z = z = − 3(4x 3y + 2z) = ( − 12x − 9y + 6z = − 29 16 Make the ﬁrst equation have 12x, the second Start with ﬁrst two equations. LCM of 4 and 6 is 12. 12x − − − 29)3 Multiply the ﬁrst equation by 3 87 − 2(6x + 2y − z) = ( − 12x − − 16)( 4y + 2z = 32 − − 2) Multiply the second equation by 2 − 12x 9y + 6z = 87 Add these two equations together − 4y + 2z = 32 − (A) − 13y + 8z = − 12x − 55 This is our (A) equation − 6x + 2y 8x − − z = 16 Now use the second two equations (a diﬀerent pair) − y + 3z = 23 The LCM of 6 and − 8 is 24. − 4(6x + 2y z) = ( − 24x + 8y − 4 = − 16)4 Multiply the ﬁrst equation by 4 64 − 8x 3( − − 24x y + 3z) = (23)3 Multiply the second equation by 3 3y + 9z = 69 − − 24x + 8y 24x − − 64 Add these two equations together − − 4 = 3y + 9z = 69 5y + 5z = 5 This is our (B) equation (B) 153 (A) (B) − 13y + 8z = − 5y + 5z = 5 55 Using (A) and (B) we will solve this system The second equation is solved for z to use substitution − − − Solving for z, subtract 5y 5y + 5z = 5 5y 5y 5y Divide each term by 5 5z = 5 5 5 5 z = 1 y Plug into untouched equation − y) = 55 Distribute − − 8y = 55 Combine like terms − − 21y + 8 = 55 − 8 − 63 Divide by − 21 − y = 3 We have our y! Plug this into z = equations 8 − 21y = 21 Subtract 8 13y 8y 21 − − − − − − 13y + 8(1 13y + 8 − − z = 1 (3) Evaluate − z = − 2 We have z, now ﬁnd x from original equation. 4x − 3(3) + 2( 4x 29 Multiply and combine like terms 29 Add 13 2) = 13 = − − + 13 + 13 − − 4x = 4 − x = − 16 Divide by 4 4 4 We have our x! 4, 3, ( − − 2) Our Solution! World View Note: Around 250, The Nine Chapters on the Mathematical Art were published in China. This book had 246 problems, and chapter 8 was about solving systems of equations. One problem had four equations with ﬁve variables! Just as with two variables and two equations, we can have special cases come up with three variables and three equations. The way we interpret the result is identical. Example 184. 4y + 3z = 5x 10x + 8y − 4 − − − 6z = 8 We will eliminate x, start with ﬁrst two equations 154 15x − 12y + 9z = 12 − 4y + 3z = 5x 10x + 8y − 4 − 6z = 8 − − LCM of 5 and 10 is 10. − 2(5x 4y + 3z) = 4(2) Multiply the ﬁrst equation by 2 − 10x − 8y + 6z = − 8 − 10x 8y + 6z = 8 Add this to the second equation, unchanged − 10x + 8y − − 6 = 8 0 = 0 A true statment − Inﬁnite Solutions Our Solution Example 185. 3x − 9x + 12y 4y + z = 2 3z = 4x − 2y − z = 3 − − 5 We will eliminate z, starting with the ﬁrst two equations 3x 4y + z = 2 The LCM of 1 and − 9x + 12y 3z = 5 − − 3 is 3 − − − 3(3x 4y + z) = (2)3 Multiply the ﬁrst equation by 3 − 9x − 12y + 3z = 6 12y + 3z = 6 Add this to the second equation, unchanged 9x 9x + 12y − − − 3z = 5 − 0 = 1 A false statement No Solution ∅ Our Solution Equations with three (or more) variables are not any more diﬃcult than two variables if we are careful to keep our information organized and eliminate the same variable twice using two diﬀerent pairs of equations. It is possible to solve each system several diﬀerent ways. We can use diﬀerent pairs of equations or eliminate variables in diﬀerent orders, but as long as our information is organized and our algebra is correct, we will arrive at the same ﬁnal solution. 155 4.4 Practice - Three Variables Solve each of the following systems of equation. 1) a 1 2b + c = 5 c = − 2c = − 2a + b − 3a + 3b − − z = 11 − x + 3y = z + 13 3z = 11 x + y 4 1 − 2) 2x + 3y = z 3x = 8z 1 − 5y + 7z
= − 4) x + y + z = 2 1 4y + 5z = 31 6x 5x + 2y + 2z = 13 − 3) 3x + y − 5) x + 6y + 3z = 4 2x + y + 2z = 3 2y + z = 0 3x − 7 − − − 2y + 3z = 6 − 2x x 3 z = 0 9 + 2z = 0 − 11) 2x + y 3z = 1 − − x 4y + z = 6 4x + 16y + 4z = 24 − 3z = 0 13) 2x + y − x 4y + z = 0 4x + 16y + 4z = 0 − 15) 3x + 2y + 2z = 3 x + 2y 2x z = 5 4y + z = 0 − − 2y + 3z = 4 y + z = 1 17) x 2x − 4x + y + z = 1 − − y + 2z = 0 2y + 3z = 2y + z = 1 3 − − 19) x − x − 2x − 21) 4x − 5x + 9y 9x + 8y 3y + 2z = 40 7z = 47 3z = 97 − − 6) x − y + 2z = 3 − x + 2y + 3z = 4 2x + y + z = 3 − 8) x + y z = 0 − x + 2y − 2x + y + z = 0 4z = 0 10) x + 2y 4x 5x z = 4 3y + z = 8 y = 12 − − − 12) 4x + 12y + 16z = 4 3x + 4y + 5z = 3 x + 8y + 11z = 1 14) 4x + 12y + 16z = 0 3x + 4y + 5z = 0 x + 8y + 11z = 0 16) p + q + r = 1 p + 2q + 3r = 4 4p + 5q + 6r = 7 18) x + 2y 2x 3x − − − − 3z = 9 y + 2z = y − 4z = 3 8 20) 4x 7y + 3z = 1 2z = 4 7y + 3z = 6 − − 3x + y 4x − 22) 3x + y y 8x 2y 5x − − − − − z = 10 6z = − 5z = 1 3 156 23) 3x + 3y 6x + 2y 2y 5x 2z = 13 5z = 13 5z = 1 − − − − − 4y + 2z = 1 25) 3x − 2x + 3y x + 10y 1 3z = − 8z = 7 − − 27) m + 6n + 3p = 8 3m + 4n = − 5m + 7n = 1 3 10 − 30) 29) 2z = 2w + 2x + 2y − w + x + y + z = 5 3w + 2x + 2y + 4z = − − w + 3x − − 2y + 2z = 11 − 6 31 + 2x + 2y + 4z = 1 w + x y w + 3x + y − z = z = 6 2 − − − − − − 24) 2x 3y + 5z = 1 − 3x + 2y 4x + 7y − − 26) 2x + y = z z = 4 7z = 7 4x + z = 4y y = x + 1 28) 3x + 2y = z + 2 y = 1 3z = 2x 2y − − w + 2x − w + x + y 3y + = 22 14 − − 32 + 2x + 2y + z = 5 w + 3x + y 4 2w + x + y z = − 3z = − − − − − 7 − 157 4.5 Systems of Equations - Value Problems Objective: Solve value problems by setting up a system of equations. One application of system of equations are known as value problems. Value problems are ones where each variable has a value attached to it. For example, if our variable is the number of nickles in a person’s pocket, those nickles would have a value of ﬁve cents each. We will use a table to help us set up and solve value problems. The basic structure of the table is shown below. Number Value Total Item 1 Item 2 Total The ﬁrst column in the table is used for the number of things we have. Quite often, this will be our variables. The second column is used for the that value each item has. The third column is used for the total value which we calculate by multiplying the number by the value. For example, if we have 7 dimes, each with a value of 10 cents, the total value is 7 10 = 70 cents. The last row of the table is for totals. We only will use the third row (also marked total) for the totals that · 158 are given to use. This means sometimes this row may have some blanks in it. Once the table is ﬁlled in we can easily make equations by adding each column, setting it equal to the total at the bottom of the column. This is shown in the following example. Example 186. In a child’s bank are 11 coins that have a value of S1.85. The coins are either quarters or dimes. How many coins each does child have? Quarter Dime Total Quarter Dime Total Quarter Dime Total Number Value Total q d 25 10 Using value table, use q for quarters, d for dimes Each quarter′s value is 25 cents, dime′s is 10 cents Number Value Total 25q 10d 25 10 q d Number Value Total 25q 10d 185 q d 11 25 10 Multiply number by value to get totals We have 11 coins total. This is the number total. We have 1.85 for the ﬁnal total, Write ﬁnal total in cents (185) Because 25 and 10 are cents q + d = 11 25q + 10d = 185 First and last columns are our equations by adding Solve by either addition or substitution. − 10(q + d) = (11)( 10d = 10q − − Using addition, multiply ﬁrst equation by 10 − 10) 110 − − − 10q − − 10d = 110 25q + 10d = 185 15q = 75 15 15 Add together equations Divide both sides by 15 q = 5 We have our q, number of quarters is 5 (5) + d = 11 5 5 Plug into one of original equations Subtract 5 from both sides d = 6 We have our d, number of dimes is 6 − − 159 5 quarters and 6 dimes Our Solution World View Note: American coins are the only coins that do not state the value of the coin. On the back of the dime it says “one dime” (not 10 cents). On the back of the quarter it says “one quarter” (not 25 cents). On the penny it says “one cent” (not 1 cent). The rest of the world (Euros, Yen, Pesos, etc) all write the value as a number so people who don’t speak the language can easily use the coins. Ticket sales also have a value. Often diﬀerent types of tickets sell for diﬀerent prices (values). These problems can be solve in much the same way. Example 187. There were 41 tickets sold for an event. Tickets for children cost S1.50 and tickets for adults cost S2.00. Total receipts for the event were S73.50. How many of each type of ticket were sold? Child Adult Total Child Adult Total Child Adult Total Number Value Total c a 1.5 2 Using our value table, c for child, a for adult Child tickets have value 1.50, adult value is 2.00 (we can drop the zeros after the decimal point) Number Value Total 1.5c 2a 1.5 2 c a Number Value Total 1.5c 2a 73.5 c a 41 1.5 2 Multiply number by value to get totals We have 41 tickets sold. This is our number total The ﬁnal total was 73.50 Write in dollars as 1.5 and 2 are also dollars c + a = 41 First and last columns are our equations by adding 1.5c + 2a = 73.5 We can solve by either addition or substitution − − − Solve for a by subtracting c c + a = 41 We will solve by substitution. c c a = 41 c c) = 73.5 2c = 73.5 0.5c + 82 = 73.5 82 − 0.5c = Substitute into untouched equation Distribute Combine like terms Subtract 82 from both sides Divide both sides by 82 8.5 − − 0.5 − − − − 1.5c + 2(41 1.5c + 82 − 160 0.5 − a = 41 0.5 − c = 17 We have c, number of child tickets is 17 (17) Plug into a = equation to ﬁnd a − a = 24 We have our a, number of adult tickets is 24 17 child tickets and 24 adult tickets Our Solution Some problems will not give us the total number of items we have. Instead they will give a relationship between the items. Here we will have statements such as “There are twice as many dimes as nickles”. While it is clear that we need to multiply one variable by 2, it may not be clear which variable gets multiplied by 2. Generally the equations are backwards from the English sentence. If there are twice as many dimes, than we multiply the other variable (nickels) by two. So the equation would be d = 2n. This type of problem is in the next example. Example 188. A man has a collection of stamps made up of 5 cent stamps and 8 cent stamps. There are three times as many 8 cent stamps as 5 cent stamps. The total value of all the stamps is S3.48. How many of each stamp does he have? Number Value Total f 3f 5 8 5f 24f 348 Number Value Total f e 5 8 5f 8e Number Value Total f e 5 8 5f 8e 348 Five Eight Total Five Eight Total Five Eight Total Use value table, f for ﬁve cent stamp, and e for eight Also list value of each stamp under value column Multiply number by value to get total The ﬁnal total was 338(written in cents) We do not know the total number, this is left blank. e = 3f 5f + 8e = 348 3 times as many 8 cent stamples as 5 cent stamps Total column gives second equation 5f + 8(3f ) = 348 Substitution, substitute ﬁrst equation in second 5f + 24f = 348 Multiply ﬁrst 29f = 348 29 29 Combine like terms Divide both sides by 39 f = 12 We have f . There are 12 ﬁve cent stamps e = 3(12) Plug into ﬁrst equation 161 12 ﬁve cent, 36 eight cent stamps Our Solution e = 36 We have e, There are 36 eight cent stamps The same process for solving value problems can be applied to solving interest problems. Our table titles will be adjusted slightly as we do so. Invest Rate Interest Account 1 Account 2 Total Our ﬁrst column is for the amount invested in each account. The second column is the interest rate earned (written as a decimal - move decimal point twice left), and the last column is for the amount of interset earned. Just as before, we multiply the investment amount by the rate to ﬁnd the ﬁnal column, the interest earned. This is shown in the following example. Example 189. A woman invests S4000 in two accounts, one at 6% interest, the other at 9% interest for one year. At the end of the year she had earned S270 in interest. How much did she have invested in each account? Invest Rate Interest 0.06 0.09 x y Invest Rate Interest 0.06x 0.06 0.09y 0.09 x y Account 1 Account 2 Total Account 1 Account 2 Total Account 1 Account 2 Total Invest Rate Interest 0.06x 0.06 0.09y 0.09 270 x y 4000 Use our investment table, x and y for accounts Fill in interest rates as decimals Multiply across to ﬁnd interest earned. Total investment is 4000, Total interest was 276 x + y = 4000 0.06x + 0.09y = 270 First and last column give our two equations Solve by either substitution or addition − 0.06(x + y) = (4000)( 0.06x − 0.06y = − − − 0.06) Use Addition, multiply ﬁrst equation by 0.06 − 240 162 − 0.06x − − Add equations together 0.06y = 240 0.06x + 0.09y = 270 0.03y = 30 0.03 0.03 y = 1000 We have y, S1000 invested at 9% Divide both sides by 0.03 x + 1000 = 4000 1000 1000 − − Plug into original equation Subtract 1000 from both sides x = 3000 We have x, S3000 invested at 6% S1000 at 9% and S3000 at 6% Our Solution The same process can be used to ﬁnd an unknown interest rate. Example 190. John invests S5000 in one account and S8000 in an account paying 4% more in interest. He earned S1230 in interest after one year. At what rates did he invest? Interest Invest Rate 5000 8000 x + 0.04 x Account 1 Account 2 Total Our investment table. Use x for ﬁrst rate The second rate is 4% higher, or x + 0.04 Be sure to write this rate as a decimal! Account 1 Account 2 Total Account 2 Account 2 Total Invest Rate 5000 8000 x + 0.04 8000x + 320 Interest 5000x x Invest Rate 5000 8000 x + 0.04 8000x + 320 Interest 5000x x 1230 5000x + 8000x + 320 = 1230 13000x + 320 = 1230 320 − 13000x = 910 13000 13000 320 − Multiply to ﬁll in interest column. Be sure to distribute 8000(x + 0.04) Total interest was 1230. Last column gives our equation Combine like terms
Subtract 320 from both sides Divide both sides by 13000 x = 0.07 We have our x, 7% interest (0.07) + 0.04 0.11 S5000 at 7% and S8000 at 11% Second account is 4% higher The account with S8000 is at 11% Our Solution 163 4.5 Practice - Value Problems Solve. 1) A collection of dimes and quaters is worth S15.25. There are 103 coins in all. How many of each is there? 2) A collection of half dollars and nickels is worth S13.40. There are 34 coins in all. How many are there? 3) The attendance at a school concert was 578. Admission was S2.00 for adults and S1.50 for children. The total receipts were S985.00. How many adults and how many children attended? 4) A purse contains S3.90 made up of dimes and quarters. If there are 21 coins in all, how many dimes and how many quarters were there? 5) A boy has S2.25 in nickels and dimes. If there are twice as many dimes as nickels, how many of each kind has he? 6) S3.75 is made up of quarters and half dollars. If the number of quarters exceeds the number of half dollars by 3, how many coins of each denomination are there? 7) A collection of 27 coins consisting of nickels and dimes amounts to S2.25. How many coins of each kind are there? 8) S3.25 in dimes and nickels, were distributed among 45 boys. If each received one coin, how many received dimes and how many received nickels? 9) There were 429 people at a play. Admission was S1 each for adults and 75 cents each for children. The receipts were S372.50. How many children and how many adults attended? 10) There were 200 tickets sold for a women’s basketball game. Tickets for students were 50 cents each and for adults 75 cents each. The total amount of money collected was S132.50. How many of each type of ticket was sold? 11) There were 203 tickets sold for a volleyball game. For activity-card holders, the price was S1.25 each and for noncard holders the price was S2 each. The total amount of money collected was S310. How many of each type of ticket was sold? 12) At a local ball game the hotdogs sold for S2.50 each and the hamburgers sold for S2.75 each. There were 131 total sandwiches sold for a total value of S342. How many of each sandwich was sold? 13) At a recent Vikings game S445 in admission tickets was taken in. The cost of a student ticket was S1.50 and the cost of a non-student ticket was S2.50. A total of 232 tickets were sold. How many students and how many nonstudents attented the game? 164 14) A bank contains 27 coins in dimes and quarters. The coins have a total value of S4.95. Find the number of dimes and quarters in the bank. 15) A coin purse contains 18 coins in nickels and dimes. The coins have a total value of S1.15. Find the number of nickels and dimes in the coin purse. 16) A business executive bought 40 stamps for S9.60. The purchase included 25c stamps and 20c stamps. How many of each type of stamp were bought? 17) A postal clerk sold some 15c stamps and some 25c stamps. Altogether, 15 stamps were sold for a total cost of S3.15. How many of each type of stamps were sold? 18) A drawer contains 15c stamps and 18c stamps. The number of 15c stamps is four less than three times the number of 18c stamps. The total value of all the stamps is S1.29. How many 15c stamps are in the drawer? 19) The total value of dimes and quarters in a bank is S6.05. There are six more quarters than dimes. Find the number of each type of coin in the bank. 20) A child’s piggy bank contains 44 coins in quarters and dimes. The coins have a total value of S8.60. Find the number of quaters in the bank. 21) A coin bank contains nickels and dimes. The number of dimes is 10 less than twice the number of nickels. The total value of all the coins is S2.75. Find the number of each type of coin in the bank. 22) A total of 26 bills are in a cash box. Some of the bills are one dollar bills, and the rest are ﬁve dollar bills. The total amount of cash in the box is S50. Find the number of each type of bill in the cash box. 23) A bank teller cashed a check for S200 using twenty dollar bills and ten dollar bills. In all, twelve bills were handed to the customer. Find the number of twenty dollar bills and the number of ten dollar bills. 24) A collection of stamps consists of 22c stamps and 40c stamps. The number of 22c stamps is three more than four times the number of 40c stamps. The total value of the stamps is S8.34. Find the number of 22c stamps in the collection. 25) A total of S27000 is invested, part of it at 12% and the rest at 13%. The total interest after one year is S3385. How much was invested at each rate? 26) A total of S50000 is invested, part of it at 5% and the rest at 7.5%. The total interest after one year is S3250. How much was invested at each rate? 27) A total of S9000 is invested, part of it at 10% and the rest at 12%. The total interest after one year is S1030. How much was invested at each rate? 28) A total of S18000 is invested, part of it at 6% and the rest at 9%. The total interest after one year is S1248. How much was invested at each rate? 29) An inheritance of S10000 is invested in 2 ways, part at 9.5% and the remainder at 11%. The combined annual interest was S1038.50. How much was invested at each rate? 165 30) Kerry earned a total of S900 last year on his investments. If S7000 was invested at a certain rate of return and S9000 was invested in a fund with a rate that was 2% higher, ﬁnd the two rates of interest. 31) Jason earned S256 interest last year on his investments. If S1600 was invested at a certain rate of return and S2400 was invested in a fund with a rate that was double the rate of the ﬁrst fund, ﬁnd the two rates of interest. 32) Millicent earned S435 last year in interest. If S3000 was invested at a certain rate of return and S4500 was invested in a fund with a rate that was 2% lower, ﬁnd the two rates of interest. 33) A total of S8500 is invested, part of it at 6% and the rest at 3.5%. The total interest after one year is S385. How much was invested at each rate? 34) A total of S12000 was invested, part of it at 9% and the rest at 7.5%. The total interest after one year is S1005. How much was invested at each rate? 35) A total of S15000 is invested, part of it at 8% and the rest at 11%. The total interest after one year is S1455. How much was invested at each rate? 36) A total of S17500 is invested, part of it at 7.25% and the rest at 6.5%. The total interest after one year is S1227.50. How much was invested at each rate? 37) A total of S6000 is invested, part of it at 4.25% and the rest at 5.75%. The total interest after one year is S300. How much was invested at each rate? 38) A total of S14000 is invested, part of it at 5.5% and the rest at 9%. The total interest after one year is S910. How much was invested at each rate? 39) A total of S11000 is invested, part of it at 6.8% and the rest at 8.2%. The total interest after one year is S797. How much was invested at each rate? 40) An investment portfolio earned S2010 in interest last year. If S3000 was invested at a certain rate of return and S24000 was invested in a fund with a rate that was 4% lower, ﬁnd the two rates of interest. 41) Samantha earned S1480 in interest last year on her investments. If S5000 was invested at a certain rate of return and S11000 was invested in a fund with a rate that was two-thirds the rate of the ﬁrst fund, ﬁnd the two rates of interest. 42) A man has S5.10 in nickels, dimes, and quarters. There are twice as many nickels as dimes and 3 more dimes than quarters. How many coins of each kind were there? 43) 30 coins having a value of S3.30 consists of nickels, dimes and quarters. If there are twice as many quarters as dimes, how many coins of each kind were there? 44) A bag contains nickels, dimes and quarters having a value of S3.75. If there are 40 coins in all and 3 times as many dimes as quarters, how many coins of each kind were there? 166 4.6 Systems of Equations - Mixture Problems Objective: Solve mixture problems by setting up a system of equations. One application of systems of equations are mixture problems. Mixture problems are ones where two diﬀerent solutions are mixed together resulting in a new ﬁnal solution. We will use the following table to help us solve mixture problems: Amount Part Total Item 1 Item 2 Final The ﬁrst column is for the amount of each item we have. The second column is labeled “part”. If we mix percentages we will put the rate (written as a decimal) in this column. If we mix prices we will put prices in this column. Then we can multiply the amount by the part to ﬁnd the total. Then we can get an equation by adding the amount and/or total columns that will help us solve the problem and answer the questions. These problems can have either one or two variables. We will start with one variable problems. Example 191. A chemist has 70 mL of a 50% methane solution. How much of a 80% solution must she add so the ﬁnal solution is 60% methane? Amount Part Total Start Add Final 70 x 0.5 0.8 Set up the mixture table. We start with 70, but don′t know how much we add, that is x. The part is the percentages, 0.5 for start, 0.8 for add. 167 Amount Part Total Start Add Final 70 x 70 + x 0.5 0.8 0.6 Amount Part 0.5 0.8 0.6 70 x 70 + x Total 35 0.8x 42 + 0.6x Start Add Final Add amount column to get ﬁnal amount. The part for this amount is 0.6 because we want the ﬁnal solution to be 60% methane. Multiply amount by part to get total. be sure to distribute on the last row: (70 + x)0.6 35 + 0.8x = 42 + 0.6x The last column is our equation by adding 0.6x Move variables to one side, subtract 0.6x 0.6x − − 35 + 0.2x = 42 35 35 − 0.2x = 7 0.2 0.2 − Subtract 35 from both sides Divide both sides by 0.2 x = 35 We have our x! 35 mL must be added Our Solution The same process can be used if the starting and ﬁnal amount have a price attached to them, rather than a percentage. Example 192. A coﬀee mix is to be made that sells for S2.50 by mixing two types of coﬀee. The cafe has 40 mL of coﬀee that costs S3.00. How much of
another coﬀee that costs S1.50 should the cafe mix with the ﬁrst? Amount Part Total Start Add Final 40 x 3 1.5 Amount Part Total Start Add Final 40 x 40 + x 3 1.5 2.5 Set up mixture table. We know the starting amount and its cost, S3. The added amount we do not know but we do know its cost is S1.50. Add the amounts to get the ﬁnal amount. We want this ﬁnal amount to sell for S2.50. 168 Amount Part Start Add Final 40 x 40 + x 3 1.5 2.5 Total 120 1.5x 100 + 2.5x Multiply amount by part to get the total. Be sure to distribute on the last row (40 + x)2.5 120 + 1.5x = 100 + 2.5x Adding down the total column gives our equation 1.5x Move variables to one side by subtracting 1.5x 1.5x − − 120 = 100 + x 100 100 20 = x We have our x. − − Subtract 100 from both sides 20mL must be added. Our Solution World View Note: Brazil is the world’s largest coﬀee producer, producing 2.59 million metric tons of coﬀee a year! That is over three times as much coﬀee as second place Vietnam! The above problems illustrate how we can put the mixture table together and get an equation to solve. However, here we are interested in systems of equations, with two unknown values. The following example is one such problem. Example 193. A farmer has two types of milk, one that is 24% butterfat and another which is 18% butterfat. How much of each should he use to end up with 42 gallons of 20% butterfat? Amount Part Total Milk 1 Milk 2 Final x y 42 0.24 0.18 0.2 Amount Part Total 0.24 0.24x 0.18 0.18y 8.4 0.2 x y 42 Milk 1 Milk 2 Final We don′t know either start value, but we do know ﬁnal is 42. Also ﬁll in part column with percentage of each type of milk including the ﬁnal solution Multiply amount by part to get totals. x + y = 42 0.24x + 0.18y = 8.4 The amount column gives one equation The total column gives a second equation. 169 − 0.18(x + y) = (42)( 0.18y = 0.18x − − 0.18) 7.56 − − − 0.18x − − 0.18y = 7.56 0.24x + 0.18y = 8.4 0.06x = 0.84 0.06 0.06 Use addition. Multiply ﬁrst equation by 0.18 − Add the equations together Divide both sides by 0.06 x = 14 We have our x, 14 gal of 24% butterfat (14) + y = 42 14 14 Plug into original equation to ﬁnd y Subtract 14 from both sides y = 28 We have our y, 28 gal of 18% butterfat − − 14 gal of 24% and 28 gal of 18% Our Solution The same process can be used to solve mixtures of prices with two unknowns. Example 194. In a candy shop, chocolate which sells for S4 a pound is mixed with nuts which are sold for S2.50 a pound are mixed to form a chocolate-nut candy which sells for S3.50 a pound. How much of each are used to make 30 pounds of the mixture? Amount Part Total Chocolate Nut Final c n 30 4 2.5 3.5 Amount Part Total Chocolate Nut Final c n 30 4 2.5 3.5 4c 2.5n 105 Using our mixture table, use c and n for variables We do know the ﬁnal amount (30) and price, include this in the table Multiply amount by part to get totals c + n = 30 4c + 2.5n = 105 First equation comes from the ﬁrst column Second equation comes from the total column c + n = 30 We will solve this problem with substitution Solve for c by subtracting n from the ﬁrst equation n − c = 30 n n − − 170 4(30 120 − − n) + 2.5n = 105 4n + 2.5n = 105 1.5n = 105 120 120 120 15 1.5 − − Substitute into untouched equation Distribute Combine like terms Subtract 120 from both sides Divide both sides by 1.5 − 1.5n = − 1.5 − n = 10 We have our n, 10 lbs of nuts − − − c = 30 (10) Plug into c = equation to ﬁnd c − c = 20 We have our c, 20 lbs of chocolate 10 lbs of nuts and 20 lbs of chocolate Our Solution With mixture problems we often are mixing with a pure solution or using water which contains none of our chemical we are interested in. For pure solutions, the percentage is 100% (or 1 in the table). For water, the percentage is 0%. This is shown in the following example. Example 195. A solution of pure antifreeze is mixed with water to make a 65% antifreeze solution. How much of each should be used to make 70 L? Amount Part Final Antifreeze Water Final a w 70 1 0 0.65 We use a and w for our variables. Antifreeze is pure, 100% or 1 in our table, written as a decimal. Water has no antifreeze, its percentage is 0. We also ﬁll in the ﬁnal percent Amount Part Final Antifreeze Water Final a w 70 1 0 0.65 a 0 45.5 Multiply to ﬁnd ﬁnal amounts a + w = 70 a = 45.5 First equation comes from ﬁrst column Second equation comes from second column (45.5) + w = 70 We have a, plug into to other equation 45.5 Subtract 45.5 from both sides 45.5 − − 45.5L of antifreeze and 24.5L of water w = 24.5 We have our w Our Solution 171 4.6 Practice - Mixture Problems Solve. 1) A tank contains 8000 liters of a solution that is 40% acid. How much water should be added to make a solution that is 30% acid? 2) How much antifreeze should be added to 5 quarts of a 30% mixture of antifreeze to make a solution that is 50% antifreeze? 3) Of 12 pounds of salt water 10% is salt; of another mixture 3% is salt. How many pounds of the second should be added to the ﬁrst in order to get a mixture of 5% salt? 4) How much alcohol must be added to 24 gallons of a 14% solution of alcohol in order to produce a 20% solution? 5) How many pounds of a 4% solution of borax must be added to 24 pounds of a 12% solution of borax to obtain a 10% solution of borax? 6) How many grams of pure acid must be added to 40 grams of a 20% acid solution to make a solution which is 36% acid? 7) A 100 LB bag of animal feed is 40% oats. How many pounds of oats must be added to this feed to produce a mixture which is 50% oats? 8) A 20 oz alloy of platinum that costs S220 per ounce is mixed with an alloy that costs S400 per ounce. How many ounces of the S400 alloy should be used to make an alloy that costs S300 per ounce? 9) How many pounds of tea that cost S4.20 per pound must be mixed with 12 lb of tea that cost S2.25 per pound to make a mixture that costs S3.40 per pound? 10) How many liters of a solvent that costs S80 per liter must be mixed with 6 L of a solvent that costs S25 per liter to make a solvent that costs S36 per liter? 11) How many kilograms of hard candy that cost S7.50 per kilogram must be mixed with 24 kg of jelly beans that cost S3.25 per kilogram to make a mixture that sells for S4.50 per kilogram? 12) How many kilograms of soil supplement that costs S7.00 per kilogram must be mixed with 20 kg of aluminum nitrate that costs S3.50 per kilogram to make a fertilizer that costs S4.50 per kilogram? 13) How many pounds of lima beans that cost 90c per pound must be mixed with 16 lb of corn that cost 50c per pound to make a mixture of vegetables that costs 65c per pound? 14) How many liters of a blue dye that costs S1.60 per liter must be mixed with 18 L of anil that costs S2.50 per liter to make a mixture that costs S1.90 per liter? 15) Solution A is 50% acid and solution B is 80% acid. How much of each should be used to make 100cc. of a solution that is 68% acid? 172 16) A certain grade of milk contains 10% butter fat and a certain grade of cream 60% butter fat. How many quarts of each must be taken so as to obtain a mixture of 100 quarts that will be 45% butter fat? 17) A farmer has some cream which is 21% butterfat and some which is 15% butter fat. How many gallons of each must be mixed to produce 60 gallons of cream which is 19% butterfat? 18) A syrup manufacturer has some pure maple syrup and some which is 85% maple syrup. How many liters of each should be mixed to make 150L which is 96% maple syrup? 19) A chemist wants to make 50ml of a 16% acid solution by mixing a 13% acid solution and an 18% acid solution. How many milliliters of each solution should the chemist use? 20) A hair dye is made by blending 7% hydrogen peroxide solution and a 4% hydrogen peroxide solution. How many mililiters of each are used to make a 300 ml solution that is 5% hydrogen peroxide? 21) A paint that contains 21% green dye is mixed with a paint that contains 15% green dye. How many gallons of each must be used to make 60 gal of paint that is 19% green dye? 22) A candy mix sells for S2.20 per kilogram. It contains chocolates worth S1.80 per kilogram and other candy worth S3.00 per kilogram. How much of each are in 15 kilograms of the mixture? 23) To make a weed and feed mixture, the Green Thumb Garden Shop mixes fertilizer worth S4.00/lb. with a weed killer worth S8.00/lb. The mixture will cost S6.00/lb. How much of each should be used to prepare 500 lb. of the mixture? 24) A grocer is mixing 40 cent per lb. coﬀee with 60 cent per lb. coﬀee to make a mixture worth 54c per lb. How much of each kind of coﬀee should be used to make 70 lb. of the mixture? 25) A grocer wishes to mix sugar at 9 cents per pound with sugar at 6 cents per pound to make 60 pounds at 7 cents per pound. What quantity of each must he take? 26) A high-protein diet supplement that costs S6.75 per pound is mixed with a vitamin supplement that costs S3.25 per pound. How many pounds of each should be used to make 5 lb of a mixture that costs S4.65 per pound? 27) A goldsmith combined an alloy that costs S4.30 per ounce with an alloy that costs S1.80 per ounce. How many ounces of each were used to make a mixture of 200 oz costing S2.50 per ounce? 28) A grocery store oﬀers a cheese and fruit sampler that combines cheddar cheese that costs S8 per kilogram with kiwis that cost S3 per kilogram. How many kilograms of each were used to make a 5 kg mixture that costs S4.50 per kilogram? 173 29) The manager of a garden shop mixes grass seed that is 60% rye grass with 70 lb of grass seed that is 80% rye grass to make a mixture that is 74% rye grass. How much of the 60% mixture is used? 30) How many ounces of water evaporated from 50 oz of a 12% salt solution to produce a 15% salt solution? 31) A caterer made an ice cream punch by combining fruit juice that cost S2.25 per gallon with ice cream that costs S3.25 per gallon. How many gallons of each were used to make 100 gal of punch costing S2.50 per pound? 32) A clothing manufacturer has some pure silk thread and some thread that
is 85% silk. How many kilograms of each must be woven together to make 75 kg of cloth that is 96% silk? 33) A carpet manufacturer blends two ﬁbers, one 20% wool and the second 50% wool. How many pounds of each ﬁber should be woven together to produce 600 lb of a fabric that is 28% wool? 34) How many pounds of coﬀee that is 40% java beans must be mixed with 80 lb of coﬀee that is 30% java beans to make a coﬀee blend that is 32% java beans? 35) The manager of a specialty food store combined almonds that cost S4.50 per pound with walnuts that cost S2.50 per pound. How many pounds of each were used to make a 100 lb mixture that cost S3.24 per pound? 36) A tea that is 20% jasmine is blended with a tea that is 15% jasmine. How many pounds of each tea are used to make 5 lb of tea that is 18% jasmine? 37) How many ounces of dried apricots must be added to 18 oz of a snack mix that contains 20% dried apricots to make a mixture that is 25% dried apricots? 38) How many mililiters of pure chocolate must be added to 150 ml of chocolate topping that is 50% chocolate to make a topping that is 75% chocolate? 39) How many ounces of pure bran ﬂakes must be added to 50 oz of cereal that is 40% bran ﬂakes to produce a mixture that is 50% bran ﬂakes? 40) A ground meat mixture is formed by combining meat that costs S2.20 per pound with meat that costs S4.20 per pound. How many pounds of each were used to make a 50 lb mixture tha costs S3.00 per pound? 41) How many grams of pure water must be added to 50 g of pure acid to make a solution that is 40% acid? 42) A lumber company combined oak wood chips that cost S3.10 per pound with pine wood chips that cost S2.50 per pound. How many pounds of each were used to make an 80 lb mixture costing S2.65 per pound? 43) How many ounces of pure water must be added to 50 oz of a 15% saline solution to make a saline solution that is 10% salt? 174 175 Chapter 5 : Polynomials 5.1 Exponent Properties ..............................................................................177 5.2 Negative Exponents ...............................................................................183 5.3 Scientiﬁc Notation .................................................................................188 5.4 Introduction to Polynomials ..................................................................192 5.5 Multiply Polynomials .............................................................................196 5.6 Multiply Special Products .....................................................................201 5.7 Divide Polynomials ................................................................................205 176 5.1 Polynomials - Exponent Properties Objective: Simplify expressions using the properties of exponents. Problems with expoenents can often be simpliﬁed using a few basic exponent properties. Exponents represent repeated multiplication. We will use this fact to discover the important properties. World View Note: The word exponent comes from the Latin “expo” meaning out of and “ponere” meaning place. While there is some debate, it seems that the Babylonians living in Iraq were the ﬁrst to do work with exponents (dating back to the 23rd century BC or earlier!) Example 196. a3a2 Expand exponents to multiplication problem (aaa)(aa) Now we have 5a′s being multiplied together a5 Our Solution A quicker method to arrive at our answer would have been to just add the exponents: a3a2 = a3+2 = a5 This is known as the product rule of exponents Product Rule of Exponents: aman = am+n The product rule of exponents can be used to simplify many problems. We will add the exponent on like variables. This is shown in the following examples Example 197. Example 198. 32 36 · Same base, add the exponents 2 + 6 + 1 3 · 39 Our Solution 2x3y5z 5xy2z3 Multiply 2 · 10x4y7z4 Our Solution · 5, add exponents on x, y and z Rather than multiplying, we will now try to divide with exponents Example 199. a5 a2 aaaaa aa aaa Expand exponents Divide out two of the a′s Convert to exponents a3 Our Solution 177 A quicker method to arrive at the solution would have been to just subtract the exponents, a5 2 = a3. This is known as the quotient rule of exponents. a2 = a5 − Quotient Rule of Exponents: am an = am−n The quotient rule of exponents can similarly be used to simplify exponent problems by subtracting exponents on like variables. This is shown in the following examples. Example 200. Example 201. Same base, subtract the exponents 713 75 78 Our Solution 5a3b5c2 2ab3c Subtract exponents on a, b and c 5 2 a2b2c Our Solution A third property we will look at will have an exponent problem raised to a second exponent. This is investigated in the following example. Example 202. 3 a2 This means we have a2 three times a2 a2 · · a2 Add exponents a6 Our solution A quicker method to arrive at the solution would have been to just multiply the 3 = a6. This is known as the power of a power rule of expoexponents, (a2)3 = a2 nents. · Power of a Power Rule of Exponents: (am)n = amn This property is often combined with two other properties which we will investigate now. Example 203. (ab)3 This means we have (ab) three times (ab)(ab)(ab) Three a′s and three b′s can be written with exponents a3b3 Our Solution 178 A quicker method to arrive at the solution would have been to take the exponent of three and put it on each factor in parenthesis, (ab)3 = a3b3. This is known as the power of a product rule or exponents. Power of a Product Rule of Exponents: (ab)m = ambm It is important to be careful to only use the power of a product rule with multiplication inside parenthesis. This property does NOT work if there is addition or subtraction. Warning 204. (a + b)m am + bm These are NOT equal, beware of this error! Another property that is very similar to the power of a product rule is considered next. Example 205 This means we have the fraction three timse Multiply fractions across the top and bottom, using exponents a3 b3 Our Solution A quicker method to arrive at the solution would have been to put the exponent on every factor in both the numerator and denominator, b3 . This is known as the power of a quotient rule of exponents. = a3 a b 3 Power of a Quotient Rule of Exponents: a b m = am bm The power of a power, product and quotient rules are often used together to simplify expressions. This is shown in the following examples. Example 206. (x3yz2)4 Put the exponent of 4 on each factor, multiplying powers x12y4z8 Our solution 179 Example 207. a3b c8d5 2 Put the exponent of 2 on each factor, multiplying powers a6b2 c8d10 Our Solution As we multiply exponents its important to remember these properties apply to exponents, not bases. An expressions such as 53 does not mean we multipy 5 by 3, 5 = 125. This is shown in the next 5 rather we multiply 5 three times, 5 example. × × Example 208. (4x2y5)3 Put the exponent of 3 on each factor, multiplying powers 43x6y15 Evaluate 43 64x6y15 Our Solution In the previous example we did not put the 3 on the 4 and multipy to get 12, this would have been incorrect. Never multipy a base by the exponent. These properties pertain to exponents only, not bases. In this lesson we have discussed 5 diﬀerent exponent properties. These rules are summarized in the following table. Rules of Exponents Product Rule of Exponents Quotient Rule of Exponents aman = am+n am an = am−n (am)n = amn Power of a Power Rule of Exponents Power of a Product Rule of Exponents (ab)m = ambm Power of a Quotient Rule of Exponents a b m = am bm These ﬁve properties are often mixed up in the same problem. Often there is a bit of ﬂexibility as to which property is used ﬁrst. However, order of operations still applies to a problem. For this reason it is the suggestion of the auther to simplify inside any parenthesis ﬁrst, then simplify any exponents (using power rules), and ﬁnally simplify any multiplication or division (using product and quotient rules). This is illustrated in the next few examples. Example 209. (4x3y 5x4y2)3 In parenthesis simplify using product rule, adding exponents · (20x7y3)3 With power rules, put three on each factor, multiplying exponents 203x21y9 Evaluate 203 8000x21y9 Our Solution 180 Example 210. 7a3(2a4)3 Parenthesis are already simpliﬁed, next use power rules 7a3(8a12) Using product rule, add exponents and multiply numbers 56a15 Our Solution Example 211. 3a3b 10a4b3 · 2a4b2 Simplify numerator with product rule, adding exponents 30a7b4 2a4b2 Now use the quotient rule to subtract exponents 15a3b2 Our Solution Example 212. 3m8n12 (m2n3)3 Use power rule in denominator 3m8n12 m6n9 Use quotient rule 3m2n3 Our solution Example 213. 3ab2(2a4b2)3 6a5b7 3ab2(8a12b6) 6a5b7 24a13b8 6a5b7 2 2 2 Simplify inside parenthesis ﬁrst, using power rule in numerator Simplify numerator using product rule Simplify using the quotient rule 4a8b)2 Now that the parenthesis are simpliﬁed, use the power rules 16a16b2 Our Solution Clearly these problems can quickly become quite involved. Remember to follow order of operations as a guide, simplify inside parenthesis ﬁrst, then power rules, then product and quotient rules. 181 5.1 Practice - Exponent Properties Simplify. 44 · 2) 4 42 44 · · 44 22 1) 4 · 3) 4 · 5) 3m 4mn · 7) 2m4n2 4nm2 · 9) (33)4 11) (44)2 13) (2u3v2)2 15) (2a4)4 17) 45 43 19) 32 3 21) 3nm2 3n 23) 4x3y4 3xy3 25) (x3y4 2x2y3)2 · 27) 2x(x4y4)4 29) 31) 33) 35) 2x 7y5 3x3y 4x2y3 · (2x)3 x3 2 2y17 (2x2y4)4 3 2m4n4 2m n4 · mn4 37) 2xy5 · 2xy4 2x2y3 y3 · 3 39) q3r2 · (2p2q2r3)2 2p3 41) zy3 z3 x4y4 4 · x3y3z3 43) 2x2y2z6 · (x2z3)2 2zx2y2 4) 3 · 6) 3x 33 32 · 4x2 · 8) x2 y4 · 10) (43)4 xy2 12) (32)3 14) (xy)3 16) (2xy)4 18) 37 33 20) 34 3 22) x2y4 4xy 24) xy3 4xy 26) (u2v2 2u4)3 · 2v3 · 2u3v 2b4 · 3a3b4 · 28) 3vu5 uv2 30) 2ba7 ba2 32) 2a2b2a7 (ba4)2 · 34) yx2 (y4)2 · 2y4 36) n3(n4)2 2mn 38) (2y 3x2)2 x2 2x2y4 · 40) 2x4y5 · 2z10 x2y7 (xy2z2)4 2q3 p3r4 2p3 · (qrp3)2 4 42) 182 5.2 Polynomials - Negative Exponents Objective: Simplify expressions with negative exponents
using the properties of exponents. There are a few special exponent properties that deal with exponents that are not positive. The ﬁrst is considered in the following example, which is worded out 2 diﬀerent ways: Example 214. a3 a3 Use the quotient rule to subtract exponents a0 Our Solution, but now we consider the problem a the second way: a3 a3 Rewrite exponents as repeated multiplication aaa aaa Reduce out all the a′s 1 1 = 1 Our Solution, when we combine the two solutions we get: a0 = 1 Our ﬁnal result. This ﬁnal result is an imporant property known as the zero power rule of exponents Zero Power Rule of Exponents: a0 = 1 Any number or expression raised to the zero power will always be 1. This is illustrated in the following example. Example 215. (3x2)0 Zero power rule 1 Our Solution Another property we will consider here deals with negative exponents. Again we will solve the following example two ways. 183 Example 216. a3 a5 Using the quotient rule, subtract exponents a− 2 Our Solution, but we will also solve this problem another way. a3 a5 Rewrite exponents as repeated multiplication aaa aaaaa 1 aa Reduce three a′s out of top and bottom Simplify to exponents 1 a2 Our Solution, putting these solutions together gives: a− 2 = 1 a2 Our Final Solution This example illustrates an important property of exponents. Negative exponents yield the reciprocal of the base. Once we take the reciprical the exponent is now positive. Also, it is important to note a negative exponent does not mean the expression is negative, only that we need the reciprocal of the base. Following are the rules of negative exponents Rules of Negative Exponets: a−m = 1 m 1 a−m = am a b −m = bm am Negative exponents can be combined in several diﬀerent ways. As a general rule if we think of our expression as a fraction, negative exponents in the numerator must be moved to the denominator, likewise, negative exponents in the denominator need to be moved to the numerator. When the base with exponent moves, the exponent is now positive. This is illustrated in the following example. Example 217. 2c a3b− 4f 2 Negative exponents on b, d, and e need to ﬂip 1 e− 2d− a 3cde4 2b2f 2 Our Solution 184 As we simpliﬁed our fraction we took special care to move the bases that had a negative exponent, but the expression itself did not become negative because of those exponents. Also, it is important to remember that exponents only eﬀect what they are attached to. The 2 in the denominator of the above example does not have an exponent on it, so it does not move with the d. We now have the following nine properties of exponents. It is important that we are very familiar with all of them. Properties of Exponents aman = am+n (ab)m = ambm am an = am−n (am)n = amn a b m = am bm a0 = 1 a−m = 1 am 1 a−m = am a b −m = bm am World View Note: Nicolas Chuquet, the French mathematician of the 15th century wrote 121m¯ to indicate 12x− 1. This was the ﬁrst known use of the negative exponent. Simplifying with negative exponents is much the same as simplifying with positive exponents. It is the advice of the author to keep the negative exponents until the end of the problem and then move them around to their correct location (numerator or denominator). As we do this it is important to be very careful of rules for adding, subtracting, and multiplying with negatives. This is illustrated in the following examples Example 218. 4x− 2 3 5y− 6x− 3x3y− · 5y3 5 12x− 6x− 2y− 5y3 Simplify numerator with product rule, adding exponents Quotient rule to subtract exponets, be careful with negatives! ( ( 2) 5) 5) = ( 2) + 5 = 3 3) = 8 Negative exponent needs to move down to denominator ( − 3 = ( − 5 2x3y− 2x3 y8 Our Solution 185 Example 219. 3 (3ab3)− 2a− 2ab− 4b0 In numerator, use power rule with In denominator, b0 = 1 − 2, multiplying exponents 3− 3 2a− 2b− 6ab− 2a− 4 3− 9 2a− 2a− 1b− 4 3− 9 2a3b− 2 a3 322b9 In numerator, use product rule to add exponents Use quotient rule to subtract exponents, be careful with negatives 1) ( − − ( − 4) = ( − 1) + 4 = 3 Move 3 and b to denominator because of negative exponents Evaluate 322 a3 18b9 Our Solution In the previous example it is important to point out that when we simpliﬁed 3− we moved the three to the denominator and the exponent became positive. We did not make the number negative! Negative exponents never make the bases negative, they simply mean we have to take the reciprocal of the base. One ﬁnal example with negative exponents is given here. 2 Example 220. 3x− 2y5z3 6x− · 9(x2y− 2)− 6y− 3 2z − 3 3 − 3 − 18x− 9x− 8y3z0 6y6 In numerator, use product rule, adding exponents In denominator, use power rule, multiplying exponets Use quotient rule to subtract exponents, be careful with negatives: ( 3 ( 8) − − 6 = 3 + ( 6) = ( 8) + 6 = 2 − − − 6) = 3 Parenthesis are done, use power rule with 3z0)− 3x6y9z0 Move 2 with negative exponent down and z0 = 1 − − − − 3 3 2y− 2− (2x− x6y9 23 x6y9 8 Evaluate 23 Our Solution 186 5.2 Practice - Negative Exponents Simplify. Your answer should contain only positive expontents. 1) 2x4y− 2 3) (a4b− · 3)3 (2xy3)4 2a3b− 2 · 5) (2x2y2)4x− 4 · 2x−3y2 3x0 9) 3x−3y3 11) 4xy−3 · x−4y0 · 4y−1 2) 2a− 3 2b− (2a0b4)4 · (2x3)0 4) 2x3y2 · 6) (m0n3 2m− 3n− 3)0 · 1n− 3y3 2x4y−3 · 3x3y2 10) 3yx3 12) 4y −2 · 3x−2y−4 (2m− 1n− 3)4 3 · 7) (x3y4)3 x− 4y4 8) 2m− 13) 15) u2v−1 2u0v4 · u2 4u0v3 · 2uv 3v2 17) 2y (x0y2)4 19) ( 2a2b3 a−1 )4 2nm4 (2m2n2)4 21) 23) (2mn)4 m0n−2 25) y3 x−3y2 (x4y2)3 · 27) 2u−2v3 · 2u−4v0 (2uv4)−1 29) ( 2x0 · y4 y4 )3 31) y(2x4y2)2 2x4y0 33) 2x4y4z−2 (zy2)4 2yzx2 · 2h−3k0 35) 2kh0 · (2kj3)2 37) (cb3)2 · (a3b−2c3)3 2a−3b2 4x3y−4 4x · 14) 2xy2 · 4x−4y−4 16) 2x−2y2 4yx2 18) (a4)4 2b 20) ( 2y −4 x2 )− 2 22) 2y2 (x4y0)−4 24) 2x−3 (x4y−3)−1 26) 2x−2y0 · (xy0)−1 2xy4 28) 2yx2 x−2 · (2x0y4)−1 30) u−3v−4 2v(2u−3v4)0 b−1 32) (2a4b0)0 34) 2b4c−2 · a−2b4 2a−3b2 · (2b3c2)−4 36) ( (2x−3y0z−1)3 2x3 · x−3y2 2 )− 38) 2q4 m2p2q4 (2m−4p2)3 · 39) (yx−4z2)−1 x2y3z−1 z3 · 40) 2mpn−3 (m0n−4p2)3 2n2p0 · 187 5.3 Polynomials - Scientiﬁc Notation Objective: Multiply and divide expressions using scientiﬁc notation and exponent properties. One application of exponent properties comes from scientiﬁc notation. Scientiﬁc notation is used to represent really large or really small numbers. An example of really large numbers would be the distance that light travels in a year in miles. An example of really small numbers would be the mass of a single hydrogen atom in grams. Doing basic operations such as multiplication and division with these numbers would normally be very combersome. However, our exponent properties make this process much simpler. First we will take a look at what scientiﬁc notation is. Scientiﬁc notation has two parts, a number between one and ten (it can be equal to one, but not ten), and that number multiplied by ten to some exponent. Scientiﬁc Notation: a 10b where 1 6 a < 10 × The exponent, b, is very important to how we convert between scientiﬁc notation and normal numbers, or standard notation. The exponent tells us how many times we will multiply by 10. Multiplying by 10 in aﬀect moves the decimal point one place. So the exponent will tell us how many times the exponent moves between scientiﬁc notation and standard notation. To decide which direction to move the decimal (left or right) we simply need to remember that positive exponents mean in standard notation we have a big number (bigger than ten) and negative exponents mean in standard notation we have a small number (less than one). Keeping this in mind, we can easily make conversions between standard notation and scientiﬁc notation. Example 221. Convert 14, 200 to scientiﬁc notation Put decimal after ﬁrst nonzero number 1.42 Exponent is how many times decimal moved, 4 104 Positive exponent, standard notation is big 104 Our Solution × × 1.42 Example 222. Convert 0.0042 to scientiﬁc notation Put decimal after ﬁrst nonzero number 4.2 Exponent is how many times decimal moved, 3 3 Negative exponent, standard notation is small 3 Our Solution 10− 10− 4.2 × × 188 Example 223. Convert 3.21 × 105 to standard notation Positive exponent means standard notation big number. Move decimal right 5 places 321, 000 Our Solution Example 224. Conver 7.4 10− × 3 to standard notation Negative exponent means standard notation is a small number. Move decimal left 3 places 0.0074 Our Solution Converting between standard notation and scientiﬁc notation is important to understand how scientiﬁc notation works and what it does. Here our main interest is to be able to multiply and divide numbers in scientiﬁc notation using exponent properties. The way we do this is ﬁrst do the operation with the front number (multiply or divide) then use exponent properties to simplify the 10’s. Scientiﬁc notation is the only time where it will be allowed to have negative exponents in our ﬁnal solution. The negative exponent simply informs us that we are dealing with small numbers. Consider the following examples. Example 225. (2.1 × 7)(3.7 10− (2.1)(3.7) = 7.77 Multiply numbers × 105) Deal with numbers and 10′s separately 10− 7105 = 10− 10− 7.77 2 Use product rule on 10′s and add exponents 2 Our Solution × Example 226. 3 Deal with numbers and 10′s separately = 1.6 Divide Numbers 4.96 3.1 104 × 10− × 4.96 3.1 104 10− 3 = 107 Use quotient rule to subtract exponents, be careful with negatives! Be careful with negatives, 4 ( − − 3) = 4 + 3 = 7 107 Our Solution 1.6 × 189 Example 227. 4)3 Use power rule to deal with numbers and 10′s separately × 10− (1.8 1.83 = 5.832 Evaluate 1.83 4)3 = 10− 10− 12 Multiply exponents 12 Our Solution 5.832 (10− × Often when we multiply or divide in scientiﬁc notation the end result is not in scientiﬁc notation. We will then have to convert the front number into scientiﬁc notation and then combine the 10’s using the product property of exponents and adding the exponents. This is shown in the following examples. Example 228. (4.7 10− 3)(6.1 109) Deal with numbers and 10′s separately × (4.7)(6
.1) = 28.67 Multiply numbers × 10110− × 101 Convert this number into scientiﬁc notation 2.867 3109 = 107 Use product rule, add exponents, using 101 from conversion 2.867 107 Our Solution × World View Note: Archimedes (287 BC - 212 BC), the Greek mathematician, developed a system for representing large numbers using a system very similar to scientiﬁc notation. He used his system to calculate the number of grains of sand it would take to ﬁll the universe. His conclusion was 1063 grains of sand because he ﬁgured the universe to have a diameter of 1014 stadia or about 2 light years. Example 229. 3 Deal with numbers and 10′s separately = 0.53 Divide numbers 2.014 3.8 10− × 7 10− × 2.014 3.8 0.53 = 5.3 10−110− 10− × 3 10− 1 Change this number into scientiﬁc notation 7 = 103 Use product and quotient rule, using 10− 1 from the conversion 5.3 × Be careful with signs: ( 1) + ( ( 3) − 103 Our Solution − − − 7) = ( 1) + ( − − 3) + 7 = 3 190 5.3 Practice - Scientiﬁc Notation Write each number in scientiﬁc notiation 1) 885 3) 0.081 5) 0.039 Write each number in standard notation 7) 8.7 x 105 9) 9 x 10− 4 11) 2 x 100 2) 0.000744 4) 1.09 6) 15000 8) 2.56 x 102 10) 5 x 104 12) 6 x 10− 5 Simplify. Write each answer in scientiﬁc notation. 13) (7 x 10− 1)(2 x 10− 3) 15) (5.26 x 10− 5)(3.16 x 10− 2) 17) (2.6 x 10− 2)(6 x 10− 2) 101 19) 4.9 × 10−3 2.7 × 21) 5.33 9.62 × × 23) (5.5 10−6 10−2 25) (7.8 10− 5)2 10− 2)5 × × 27) (8.03 104)− 4 × 10−6 10−4 29) 6.1 5.1 × × 31) (3.6 33) (1.8 × 104 10−2 10−3 × 10−6 35) 9 7.83 × × 37) 3.22 7 × 39) 2.4 6.5 41) 6 5.8 10−6 100 × × 103 10−3 × × 100)(6.1 10− 3) × × 10− 3 5)− 14) (2 10− 6)(8.8 5) 10− × × 106)(9.84 10− 1) × 16) (5.1 × 104 18) 7.4 × 10−4 1.7 × 20) 7.2 7.32 10−1 10−1 × × 10−3 100 22) 3.2 × 5.02 × 24) (9.6 26) (5.4 × 28) (6.88 × 105 × 10−2 30) 8.4 7 × 32) (3.15 × 103)− 4 106)− 3 34) 9.58 1.14 × × 36) (8.3 38) 5 × 6.69 101)5 × 106 102 × 10− × 3 2)− 104)(6 101) × × 40) (9 42) (2 191 10− 4)(4.23 101) × 10− 1) × 103)(8 × 10−2 10−3 5.4 Polynomials - Introduction to Polynomials Objective: Evaluate, add, and subtract polynomials. Many applications in mathematics have to do with what are called polynomials. Polynomials are made up of terms. Terms are a product of numbers and/or variables. For example, 5x, 2y2, 5, ab3c, and x are all terms. Terms are connected to each other by addition or subtraction. Expressions are often named based on the number of terms in them. A monomial has one term, such as 3x2. A binomial has two terms, such as a2 b2. A Trinomial has three terms, such as ax2 + bx + c. The term polynomial means many terms. Monomials, binomials, trinomials, and expressions with more terms all fall under the umbrella of “polynomials”. − − If we know what the variable in a polynomial represents we can replace the variable with the number and evaluate the polynomial as shown in the following example. Example 230. 2x2 4x + 6 when x = − 2( − 4)2 − 2(16) − − 4( 4( 32 + 16 + 6 Add − − 4 Replace variable x with 4 − 4) + 6 Exponents ﬁrst 4) + 6 Multiplication (we can do all terms at once) 54 Our Solution − 9 because the exponent is only attached to the 3. Also, ( It is important to be careful with negative variables and exponents. Remember the exponent only eﬀects the number it is physically attached to. This means − 32 = 3)2 = 9 because the exponent is attached to the parenthesis and eﬀects everything inside. When we replace a variable with parenthesis like in the previous example, the substi4)2 = 16 in the example. However, contuted value is in parenthesis. So the ( sider the next example. − − Example 231. − x2 + 2x + 6 when x = 3 Replace variable x with 3 − (3)2 + 2(3) + 6 Exponent only on the 3, not negative − 9 + 2(3) + 6 Multiply 9 + 6 + 6 Add − 3 Our Solution 192 World View Note: Ada Lovelace in 1842 described a Diﬀerence Engine that would be used to caluclate values of polynomials. Her work became the foundation for what would become the modern computer (the programming language Ada was named in her honor), more than 100 years after her death from cancer. Generally when working with polynomials we do not know the value of the variable, so we will try and simplify instead. The simplest operation with polynomials is addition. When adding polynomials we are mearly combining like terms. Consider the following example Example 232. (4x3 − 2x + 8) + (3x3 9x2 7x3 − − 9x2 − 2x − 11) Combine like terms 4x3 + 3x3 and 8 3 Our Solution − 11 − Generally ﬁnal answers for polynomials are written so the exponent on the variable counts down. Example 3 demonstrates this with the exponent counting down 3, 2, 1, 0 (recall x0 = 1). Subtracting polynomials is almost as fast. One extra step comes from the minus in front of the parenthesis. When we have a negative in front of parenthesis we distribute it through, changing the signs of everything inside. The same is done for the subtraction sign. Example 233. (5x2 2x + 7) − 5x2 − 2x + 7 − (3x2 + 6x 3x2 − 2x2 − 6x + 4 Combine like terms 5x2 − 8x + 11 Our Solution − 4) Distribute negative through second part 3x3, 2x − − − 6x, and 7 + 4 Addition and subtraction can also be combined into the same problem as shown in this ﬁnal example. Example 234. (2x2 4x + 3) + (5x2 6x + 1) (x2 − 2x2 − 4x + 3 + 5x2 − 6x + 1 − − 9x + 8) Distribute negative through − − x2 + 9x 6x2 x − 8 Combine like terms 4 Our Solution − − 193 5.4 Practice - Introduction to Polynomials Simplify each expression. 1) a3 a2 + 6a 21 when a = − − 2) n2 + 3n − 11 when n = 6 4 − − 7n2 + 15n − 20 when n = 2 3) n3 − 4) n3 9n2 + 23n 21 when n = 5 − − 9n2 − 5n4 5) − 6) x4 11n3 n 5 when n = − 5x3 − − x + 13 when x = 5 − 1 − − − 7) x2 + 9x + 23 when x = 3 − 8) − 9) x4 6x3 + 41x2 6x3 + x2 − − − 32x + 11 when x = 6 24 when x = 6 10) m4 + 8m3 + 14m2 + 13m + 5 when m = 6 − 11) (5p 5p4) (8p 8p4) − − 12) (7m2 + 5m3) − (6m3 5m2) 13) (3n2 + n3) − − (2n3 − 7n2) − 14) (x2 + 5x3) + (7x2 + 3x3) 15) (8n + n4) (3n − 16) (3v4 + 1) + (5 − 4n4) − v4) 17) (1 + 5p3) − 18) (6x3 + 5x) (1 8p3) − (8x + 6x3) − 19) (5n4 + 6n3) + (8 20) (8x2 + 1) (6 − − − x2 3n3 − x4) − 5n4) 194 21) (3 + b4) + (7 + 2b + b4) 22) (1 + 6r2) + (6r2 − 2 23) (8x3 + 1) (5x4 − − 3r4) − 6x3 + 2) 24) (4n4 + 6) − 25) (2a + 2a4) (4n 1 − (3a2 n4) − 5a4 + 4a) − − 26) (6v + 8v3) + (3 + 4v3 3v) 6p + 3) − − 27) (4p2 3 2p) (3p2 − − 28) (7 + 4m + 8m4) − (5m4 + 1 + 6m) − 29) (4b3 + 7b2 3) + (8 + 5b2 + b3) − 8n4) (3n + 7n4 + 7) 30) (7n + 1 − 31) (3 + 2n2 + 4n4) + (n3 − − 32) (7x2 + 2x4 + 7x3) + (6x3 33) (n − 5n4 + 7) + (n2 − 34) (8x2 + 2x4 + 7x3) + (7x4 7n2 4n4) − 8x4 n) − 7n4 7x2) − − 7x3 + 2x2) − − 8 + 2x + 6x3) 35) (8r4 − 36) (4x3 + x − 37) (2n2 + 7n4 5r3 + 5r2) + (2r2 + 2r3 7r4 + 1) 7x2) + (x2 − 2) + (2 + 2n3 + 4n2 + 2n4) 38) (7b3 39) (8 − − 4b + 4b4) − b + 7b3) (8b3 − − (3b4 + 7b − − 8n3) + (7n4 + 2 40) (1 3n4 − − 4b2 + 2b4 8b) − 8 + 7b2) + (3 3b + 6b3) − 6n2 + 3n3) + (4n3 + 8n4 + 7) − 41) (8x4 + 2x3 + 2x) + (2x + 2 2x3 42) (6x 5x4 4x2) (2x − − − − − 7x2 − 4x4 − x4) − (x3 + 5x4 + 8x) 8) − − (8 − 6x2 − 4x4) 195 5.5 Polynomials - Multiplying Polynomials Objective: Multiply polynomials. Multiplying polynomials can take several diﬀerent forms based on what we are multiplying. We will ﬁrst look at multiplying monomials, then monomials by polynomials and ﬁnish with polynomials by polynomials. Multiplying monomials is done by multiplying the numbers or coeﬃcients and then adding the exponents on like factors. This is shown in the next example. Example 235. (4x3y4z)(2x2y6z3) Multiply numbers and add exponents for x, y, and z 8x5y10z4 Our Solution In the previous example it is important to remember that the z has an exponent of 1 when no exponent is written. Thus for our answer the z has an exponent of 1 + 3 = 4. Be very careful with exponents in polynomials. If we are adding or subtracting the exponnets will stay the same, but when we multiply (or divide) the exponents will be changing. Next we consider multiplying a monomial by a polynomial. We have seen this operation before with distributing through parenthesis. Here we will see the exact same process. Example 236. 2x + 5) Distribute the 4x3, multiplying numbers, adding exponents 4x3(5x2 20x5 − − 8x4 + 20x3 Our Solution Following is another example with more variables. When distributing the exponents on a are added and the exponents on b are added. Example 237. 2a3b(3ab2 6a4b3 − − 4a) Distribute, multiplying numbers and adding exponents 8a4b Our Solution There are several diﬀerent methods for multiplying polynomials. All of which work, often students prefer the method they are ﬁrst taught. Here three methods will be discussed. All three methods will be used to solve the same two multiplication problems. Multiply by Distributing 196 Just as we distribute a monomial through parenthesis we can distribute an entire polynomial. As we do this we take each term of the second polynomial and put it in front of the ﬁrst polynomial. Example 238. (4x + 7y)(3x 2y) Distribute (4x + 7y) through parenthesis 3x(4x + 7y) 2y(4x + 7y) Distribute the 3x and 2y − 12x2 + 21xy 8xy − 12x2 + 13xy − 14y2 Combine like terms 21xy 14y2 Our Solution 8xy − − − − This example illustrates an important point, the negative/subtraction sign stays with the 2y. Which means on the second step the negative is also distributed through the last set of parenthesis. Multiplying by distributing can easily be extended to problems with more terms. First distribute the front parenthesis onto each term, then distribute again! Example 239. 4x2(2x − 8x3 (2x 7x(2x − − − 5)(4x2 5) + 3(2x 14x2 + 35x + 6x 34x2 + 41x 8x3 − 5) − − 20x2 − − − − 7x + 3) Distribute (2x 5) through parenthesis − 5) Distribute again through each parenthesis 15 Combine like terms 15 Our Solution This process of multiplying by distributing can easily be reversed to do an important procedure known as factoring. Factoring will be addressed in a future lesson. Multiply by FOIL Another form of multiplying is known as FOIL. Using the FOIL method we multiply each term in the ﬁrst binomial by each term in the second
binomial. The letters of FOIL help us remember every combination. F stands for First, we multiply the ﬁrst term of each binomial. O stand for Outside, we multiply the outside two terms. I stands for Inside, we multiply the inside two terms. L stands for Last, we multiply the last term of each binomial. This is shown in the next example: Example 240. (4x + 7y)(3x (4x)(3x) = 12x2 − (4x)( 2y) = (7y)(3x) = 21xy − − 2y) Use FOIL to multiply − − − − First terms (4x)(3x) Outside terms (4x)( Inside terms (7y)(3x) Last terms (7y)( F 8xy O I 14y2 L 14y2 Combine like terms 14y2 Our Solution 2y) − 2y) − 8xy + 21xy − 12x2 (7y)( 2y) = − 8xy + 21xy 12x2 + 13xy − − − − 197 Some students like to think of the FOIL method as distributing the ﬁrst term 4x through the (3x 2y). Thinking about FOIL in this way makes it possible to extend this method to problems with more terms. 2y) and distributing the second term 7y through the (3x − − Example 241. (2x)(4x2) + (2x)( − (2x 5)(4x2 7x + 3) Distribute 2x and 5 − 7x) + (2x)(3) 8x3 − 14x2 + 6x 8x3 − − 5(4x2) 5( − 7x) − − − 20x2 + 35x 34x2 + 41x − − − − 5(3) Multiply out each term 15 Combine like terms 15 Our Solution The second step of the FOIL method is often not written, for example, consider the previous example, a student will often go from the problem (4x + 7y)(3x 2y) − and do the multiplication mentally to come up with 12x2 14y2 and − then combine like terms to come up with the ﬁnal solution. 8xy + 21xy − Multiplying in rows A third method for multiplying polynomials looks very similar to multiplying numbers. Consider the problem: 35 27 × 245 Multiply 7 by 5 then 3 700 Use 0 for placeholder, multiply 2 by 5 then 3 945 Add to get Our Solution World View Note: The ﬁrst known system that used place values comes from Chinese mathematics, dating back to 190 AD or earlier. The same process can be done with polynomials. Multiply each term on the bottom with each term on the top. Example 242. (4x + 7y)(3x 2y) Rewrite as vertical problem − 4x + 7y 2y 3x × 8xy − 12x2 + 21xy 12x2 + 13xy − − − 14y2 Multiply 2y by 7y then 4x − Multiply 3x by 7y then 4x. Line up like terms 14y2 Add like terms to get Our Solution This same process is easily expanded to a problem with more terms. 198 Example 243. (2x − 8x3 8x3 − − − − − × 5)(4x2 4x3 7x + 3) Rewrite as vertical problem 7x + 3 Put polynomial with most terms on top 2x 20x2 + 35x 14x2 + 6x 34x2 + 41x 5 − 15 Multiply − 15 Add like terms to get our solution 5 by each term Multiply 2x by each term. Line up like terms − − This method of multiplying in rows also works with multiplying a monomial by a polynomial! Any of the three described methods work to multiply polynomials. It is suggested that you are very comfortable with at least one of these methods as you work through the practice problems. All three methods are shown side by side in the example. Example 244. (2x − y)(4x 5y) − 4x(2x 8x2 Distribute 5y(2x 10xy y) − 4xy − − 14xy − 5y2 y) − 5y2 − − 8x2 − 2x(4x) + 2x( FOIL 5y) − 10xy − − y( y(4x) 4xy + 5y2 − 14xy + 5y2 8x2 − 8x2 − 5y) − Rows y 2x 5y 4x − × − 10xy + 5y2 4xy 14xy + 5y2 − 8x2 − 8x2 − When we are multiplying a monomial by a polynomial by a polynomial we can solve by ﬁrst multiplying the polynomials then distributing the coeﬃcient last. This is shown in the last example. Example 245. 3(2x 3(2x2 + 10x − − 4)(x + 5) Multiply the binomials, we will use FOIL 4x 3(2x2 + 6x 6x2 + 18x 20) Combine like terms 20) Distribute the 3 60 Our Solution − − − A common error students do is distribute the three at the start into both parenthesis. While we can distribute the 3 into the (2x 4) factor, distributing into both would be wrong. Be careful of this error. This is why it is suggested to multiply the binomials ﬁrst, then distribute the coeﬃenct last. − 199 5.5 Practice - Multiply Polynomials Find each product. 1) 6(p 7) − 3) 2(6x + 3) 5) 5m4(4m + 4) 7) (4n + 6)(8n + 8) 9) (8b + 3)(7b 5) − 11) (4x + 5)(2x + 3) 13) (3v 4)(5v 2) − − 15) (6x 7)(4x + 1) − 17) (5x + y)(6x 4y) − 2) 4k(8k + 4) 4) 3n2(6n + 7) 6) 3(4r 7) − 8) (2x + 1)(x 4) − 10) (r + 8)(4r + 8) 12) (7n − 6)(n + 7) 14) (6a + 4)(a 8) − 6)(4x 16) (5x − − 18) (2u + 3v)(8u 1) 7v) − 19) (x + 3y)(3x + 4y) 20) (8u + 6v)(5u 8v) 21) (7x + 5y)(8x + 3y) 22) (5a + 8b)(a − 3b) − 7)(6r2 r + 5) 24) (4x + 8)(4x2 + 3x + 5) 23) (r − 25) (6n − 4)(2n2 − − 27) (6x + 3y)(6x2 2n + 5) 7xy + 4y2) − 29) (8n2 + 4n + 6)(6n2 5n + 6) − 31) (5k2 + 3k + 3)(3k2 + 3k + 6) 33) 3(3x − 4)(2x + 1) 35) 3(2x + 1)(4x 5) − 37) 7(x − 39) 6(4x 5)(x 2) − 1)(4x + 1) − 26) (2b − 3)(4b2 + 4b + 4) 28) (3m − 2n)(7m2 + 6mn + 4n2) 30) (2a2 + 6a + 3)(7a2 6a + 1) − 32) (7u2 + 8uv − 6v2)(6u2 + 4uv + 3v2) 34) 5(x 4)(2x − − 36) 2(4x + 1)(2x 3) 6) − 38) 5(2x 1)(4x + 1) − 40) 3(2x + 3)(6x + 9) 200 5.6 Polynomials - Multiply Special Products Objective: Recognize and use special product rules of a sum and diﬀerence and perfect squares to multiply polynomials. There are a few shortcuts that we can take when multiplying polynomials. If we can recognize them the shortcuts can help us arrive at the solution much quicker. These shortcuts will also be useful to us as our study of algebra continues. The ﬁrst shortcut is often called a sum and a diﬀerence. A sum and a diﬀerence is easily recognized as the numbers and variables are exactly the same, but the sign in the middle is diﬀerent (one sum, one diﬀerence). To illustrate the shortcut consider the following example, multiplied by the distributing method. Example 246. (a + b)(a b) Distribute (a + b) a(a + b) a2 + ab − − b(a + b) Distribute a and ab a2 b2 Combine like terms ab b2 Our Solution − b ab − − − − The important part of this example is the middle terms subtracted to zero. Rather than going through all this work, when we have a sum and a diﬀerence we will jump right to our solution by squaring the ﬁrst term and squaring the last term, putting a subtraction between them. This is illustrated in the following example Example 247. (x − 5)(x + 5) Recognize sum and diﬀerence x2 − 25 Square both, put subtraction between. Our Solution This is much quicker than going through the work of multiplying and combining like terms. Often students ask if they can just multiply out using another method and not learn the shortcut. These shortcuts are going to be very useful when we get to factoring polynomials, or reversing the multiplication process. For this reason it is very important to be able to recognize these shortcuts. More examples are shown here. 201 Example 248. (3x + 7)(3x 9x2 − − 7) Recognize sum and diﬀerence 49 Square both, put subtraction between. Our Solution Example 249. (2x − 6y)(2x + 6y) Recognize sum and diﬀerence 4x2 − 36y2 Square both, put subtraction between. Our Solution b2 It is interesting to note that while we can multiply and get an answer like a2 (with subtraction), it is impossible to multiply real numbers and end up with a product such as a2 + b2 (with addition). − Another shortcut used to multiply is known as a perfect square. These are easy to recognize as we will have a binomial with a 2 in the exponent. The following example illustrates multiplying a perfect square Example 250. (a + b)2 Squared is same as multiplying by itself (a + b)(a + b) Distribute (a + b) a(a + b) + b(a + b) Distribute again through ﬁnal parenthesis a2 + ab + ab + b2 Combine like terms ab + ab a2 + 2ab + b2 Our Solution This problem also helps us ﬁnd our shortcut for multiplying. The ﬁrst term in the answer is the square of the ﬁrst term in the problem. The middle term is 2 times the ﬁrst term times the second term. The last term is the square of the last term. This can be shortened to square the ﬁrst, twice the product, square the last. If we can remember this shortcut we can square any binomial. This is illustrated in the following example Example 251. (x − 5)2 Recognize perfect square x2 Square the ﬁrst 2(x)( 10x Twice the product − 5)2 = 25 10x + 25 Our Solution Square the last 5) = − ( x2 − − 202 Be very careful when we are squaring a binomial to NOT distribute the square 5)2 = x2 through the parenthesis. A common error is to do the following: (x 25 (or x2 + 25). Notice both of these are missing the middle term, 10x. This is why it is important to use the shortcut to help us ﬁnd the correct solution. Another important observation is that the middle term in the solution always has the same sign as the middle term in the problem. This is illustrated in the next examples. − − − Example 252. Example 253. (2x + 5)2 Recognize perfect square Square the ﬁrst (2x)2 = 4x2 2(2x)(5) = 20x Twice the product 52 = 25 Square the last 4x2 + 20x + 25 Our Solution (3x 7y)2 Recognize perfect square − 42xy + 49y2 9x2 − Square the ﬁrst, twice the product, square the last. Our Solution Example 254. (5a + 9b)2 Recognize perfect square 25a2 + 90ab + 81b2 Square the ﬁrst, twice the product, square the last. Our Solution These two formulas will be important to commit to memory. The more familiar we are with them, the easier factoring, or multiplying in reverse, will be. The ﬁnal example covers both types of problems (two perfect squares, one positive, one negative), be sure to notice the diﬀerence between the examples and how each formula is used Example 255. (4x 7)(4x + 7) 49 − 16x2 − (4x + 7)2 16x2 + 56x + 49 (4x 7)2 − 56x + 49 16x2 − World View Note: There are also formulas for higher powers of binomials as well, such as (a + b)3 = a3 + 3a2b + 3ab2 + b3. While French mathematician Blaise Pascal often gets credit for working with these expansions of binomials in the 17th century, Chinese mathematicians had been working with them almost 400 years earlier! 203 5.6 Practice - Multiply Special Products Find each product. 1) (x + 8)(x 8) − 3) (1 + 3p)(1 3p) − 7n)(1 + 7n) 5) (1 − 7) (5n 8)(5n + 8) − 9) (4x + 8)(4x 8) − x)(4y + x) 11) (4y − 13) (4m − 8n)(4m + 8n) 15) (6x − 2y)(6x + 2y) 17) (a + 5)2 19) (x 8)2 − 21) (p + 7)2 23) (7 − 25) (5m 5n)2 8)2 − 27) (5x + 7y)2 29) (2x + 2y)2 31) (5 + 2r)2 33) (2 + 5x)2 35) (4v − 7) (4v + 7) 37) (n 5)(n +
5) − 39) (4k + 2)2 2) (a 4) (x − − 4)(a + 4) 3)(x + 3) 6) (8m + 5)(8m 5) − 3) 8) (2r + 3)(2r − 7)(b + 7) 10) (b − 12) (7a + 7b)(7a 7b) − 3x)(3y + 3x) 14) (3y − 16) (1 + 5n)2 18) (v + 4)2 20) (1 − 22) (7k 6n)2 − 7)2 5)2 24) (4x − 26) (3a + 3b)2 28) (4m n)2 − 30) (8x + 5y)2 32) (m 7)2 − 34) (8n + 7)(8n 36) (b + 4)(b − 38) (7x + 7)2 7) − 4) 40) (3a − 8)(3a + 8) 204 5.7 Polynomials - Divide Polynomials Objective: Divide polynomials using long division. Dividing polynomials is a process very similar to long division of whole numbers. But before we look at that, we will ﬁrst want to be able to master dividing a polynomial by a monomial. The way we do this is very similar to distributing, but the operation we distribute is the division, dividing each term by the monomial and reducing the resulting expression. This is shown in the following examples Example 256. 9x5 + 6x4 18x3 − 3x2 − 24x2 9x5 3x2 + 6x4 3x2 − 18x3 3x2 − 24x2 3x2 Divide each term in the numerator by 3x2 Reduce each fraction, subtracting exponents 3x3 + 2x2 6x − − 8 Our Solution Example 257. 8x3 + 4x2 − 4x2 2x + 6 Divide each term in the numerator by 4x2 8x3 4x2 + 4x2 4x2 − 2x 4x2 + 6 4x2 Reduce each fraction, subtracting exponents Remember negative exponents are moved to denominator 2x + 1 1 2x + 3 2x2 Our Solution − The previous example illustrates that sometimes we will have fractions in our solution, as long as they are reduced this will be correct for our solution. Also interesting in this problem is the second term 4x2 4x2 divided out completely. Remember that this means the reduced answer is 1 not 0. Long division is required when we divide by more than just a monomial. Long division with polynomials works very similar to long division with whole numbers. 205 An example is given to review the (general) steps that are used with whole numbers that we will also use with polynomials Example 258. 4 631 Divide front numbers: = 1 \| 6 4 4 1 631 Multiply this number by divisor: 1 \| 4 − 23 Change the sign of this number (make it subtract) and combine Bring down next number 4 = 4 · 15 Repeat, divide front numbers: 23 4 = 5 4 − − 631 \| 4 23 20 31 Bring down next number Multiply this number by divisor: 5 Change the sign of this number (make it subtract) and combine 4 = 20 · 157 Repeat, divide front numbers: = 7 31 4 4 \| − − 631 4 23 20 31 Multiply this number by divisor: 7 28 Change the sign of this number (make it subtract) and combine 4 = 28 · − 3 We will write our remainder as a fraction, over the divisor, added to the end 157 3 4 Our Solution This same process will be used to multiply polynomials. The only diﬀerence is we will replace the word “number” with the word “term” Dividing Polynomials 1. Divide front terms 2. Multiply this term by the divisor 206 3. Change the sign of the terms and combine 4. Bring down the next term 5. Repeat Step number 3 tends to be the one that students skip, not changing the signs of the terms would be equivalent to adding instead of subtracting on long division with whole numbers. Be sure not to miss this step! This process is illustrated in the following two examples. Example 259. 3x3 − 5x2 x 32x + 7 − 4 − Rewrite problem as long division x 4 \| − 3x3 − 5x2 − 32x + 7 Divide front terms: 3x3 x = 3x2 x 4 − − 5x2 3x2 3x3 − − \| 3x3 + 12x2 7x2 − 32x + 7 Multiply this term by divisor: 3x2(x 32x Change the signs and combine Bring down the next term 4) = 3x3 12x2 − − 3x2 + 7x Repeat, divide front terms: 7x2 x = 7x Multiply this term by divisor: 7x(x Change the signs and combine − 4) = 7x2 28x − 4x + 7 Bring down the next term − 4x Repeat, divide front terms: − x = 4 − x − 4 \| − 32x + 7 5x2 3x3 − − 3x3 + 12x2 7x2 32x − 7x2 + 28x − x − 4 \| − 32x + 7 3x2 + 7x 4 − 5x2 3x3 − − 3x3 + 12x2 7x2 32x − 7x2 + 28x − 4x + 7 Multiply this term by divisor: 16 Change the signs and combine − + 4x 4(x − 4) = − 4x + 16 − − − 9 Remainder put over divisor and subtracted (due to negative) 207 3x2 + 7x 4 − − x 9 − 4 Example 260. Our Solution 6x3 − 8x2 + 10x + 103 2x + 4 Rewrite problem as long division 2x + 4 \| 6x3 − 8x2 + 10x + 103 Divide front terms: 6x3 2x = 3x2 2x + 4 \| − 3x2 6x3 − 6x3 − − 8x2 + 10x + 103 Multiply term by divisor: 3x2(2x + 4) = 6x3 + 12x2 12x2 Change the signs and combine 20x2 + 10x Bring down the next term 3x2 − 10x 6x3 2x + 4 \| 6x3 − − 8x2 + 10x + 103 Repeat, divide front terms: − 12x2 Multiply this term by divisor: 20x2 + 10x 40x − + 20x2 + 40x Change the signs and combine 10x(2x + 4) = − − 20x2 − − 20x2 2x 10x = − 3x2 − 2x + 4 \| − 6x3 6x3 50x + 103 Bring down the next term 10x + 25 − 8x2 + 10x + 103 Repeat, divide front terms: 12x2 − 20x2 + 10x − + 20x2 + 40x 50x 2x = 25 50x + 103 Multiply this term by divisor: 25(2x + 4) = 50x + 100 50x − − 100 Change the signs and combine 3 Remainder is put over divsor and added (due to positive) 3x2 − 10x + 25 + 3 2x + 4 Our Solution In both of the previous example the dividends had the exponents on our variable counting down, no exponent skipped, third power, second power, ﬁrst power, zero power (remember x0 = 1 so there is no variable on zero power). This is very important in long division, the variables must count down and no exponent can be skipped. If they don’t count down we must put them in order. If an exponent is skipped we will have to add a term to the problem, with zero for its coeﬃcient. This is demonstrated in the following example. 208 Example 261. 2x3 + 42 x + 3 − 4x Reorder dividend, need x2 term, add 0x2 for this 2x3 + 0x2 x + 3 \| − 4x + 42 Divide front terms: 2x3 x = 2x2 x + 3 − 2x2 2x3 + 0x2 − \| 6x2 2x3 6x2 − − − 4x + 42 Multiply this term by divisor: 2x2(x + 3) = 2x3 + 6x2 4x Change the signs and combine Bring down the next term 4x + 42 Repeat, divide front terms: − 6x2 x 6x = − Multiply this term by divisor: Change the signs and combine − 6x(x + 3) = 6x2 − − 18x 2x2 6x − − 2x3 + 0x2 6x2 2x3 6x2 4x + 6x2 + 18x − − − x + 3 \| − x + 3 \| − 14x + 42 Bring down the next term 4x + 42 Repeat, divide front terms: − 14x x = 14 2x2 6x + 14 − 2x3 + 0x2 6x2 2x3 6x2 − 4x − − + 6x2 + 18x 14x + 42 Multiply this term by divisor: 14(x + 3) = 14x + 42 14x − − 42 Change the signs and combine 0 No remainder 2x2 − 6x + 14 Our Solution It is important to take a moment to check each problem to verify that the exponents count down and no exponent is skipped. If so we will have to adjust the problem. Also, this ﬁnal example illustrates, just as in regular long division, sometimes we have no remainder in a problem. World View Note: Paolo Ruﬃni was an Italian Mathematician of the early 19th century. In 1809 he was the ﬁrst to describe a process called synthetic division which could also be used to divide polynomials. 209 5.7 Practice - Divide Polynomials 10n Divide. 1) 20x4 + x3 + 2x2 4x3 3) 20n4 + n3 + 40n2 5) 12x4 + 24x3 + 3x2 7) 10n4 + 50n3 + 2n2 10n2 9) x2 2x − − x + 8 6x 71 11) n2 + 13n + 32 n + 5 13) v2 − v 15) a2 − 10 89 38 2v − 4a − − a − 8 17) 45p2 + 56p + 19 9p + 4 19) 10x2 − 10x 32x + 9 2 − 21) 4r2 1 r − − 4r + 3 23) n2 n 4 2 − − 25) 27b2 + 87b + 35 3b + 8 27) 4x2 33x + 28 5 − 4x − 29) a3 + 15a2 + 49a a + 7 35) x3 − 46x + 22 x + 7 37) 9p3 + 45p2 + 27p 9p + 9 5 − 39) r3 − r2 r 16r + 8 4 − − 41) 12n3 + 12n2 − 2n + 3 15n 4 − 43) 4v3 − 21v2 + 6v + 19 4v + 3 2) 5x4 + 45x3 + 4x2 9x 4) 3k3 + 4k2 + 2k 8k 6) 5p4 + 16p3 + 16p2 4p 8) 3m4 + 18m3 + 27m2 9m2 10) r2 12) b2 3r 53 − r − 9 − 10b + 16 − b 7 − 14) x2 + 4x − x + 7 26 16) x2 10x + 22 − x 4 − 18) 48k2 − 6k 70k + 16 2 − 20) n2 + 7n + 15 n + 4 22) 3m2 + 9m 9 − 24) 2x2 3 3m − 5x − 2x + 3 8 − 26) 3v2 3v 32 9 − − 28) 4n2 38 23n − 4n + 5 − 36) 2n3 + 21n2 + 25n 2n + 3 38) 8m3 57m2 + 42 − 8m + 7 40) 2x3 + 12x2 + 4x 2x + 6 37 − 38b2 + 29b 4b 7 − 60 − 42) 24b3 − 210 55 − 30) 8k3 − 31) x3 41 − 26x x + 4 − 33) 3n3 + 9n2 64n − n + 6 68 − 32) x3 − 34) k3 − 66k2 + 12k + 37 8 k − 16x2 + 71x 56 − x 8 − 4k2 k 6k + 4 − 1 − Chapter 6 : Factoring 6.1 Greatest Common Factor .......................................................................212 6.2 Grouping ................................................................................................216 6.3 Trinomials where a =1 ...........................................................................221 6.4 Trinomials where a 1 .........................................................................226 6.5 Factoring Special Products ....................................................................229 6.6 Factoring Strategy .................................................................................234 6.7 Solve by Factoring .................................................................................237 211 6.1 Factoring - Greatest Common Factor Objective: Find the greatest common factor of a polynomial and factor it out of the expression. The opposite of multiplying polynomials together is factoring polynomials. There are many beniﬁts of a polynomial being factored. We use factored polynomials to help us solve equations, learn behaviors of graphs, work with fractions and more. Because so many concepts in algebra depend on us being able to factor polynomials it is very important to have very strong factoring skills. In this lesson we will focus on factoring using the greatest common factor or GCF of a polynomial. When we multiplied polynomials, we multiplied monomials by polynomials by distributing, solving problems such as 4x2(2x2 − 12x3 + 32x. In this lesson we will work the same problem backwards. We will start with 8x2 12x3 + 32x and try and work backwards to the 4x2(2x 3x + 8) = 8x4 3x + 8). − − − To do this we have to be able to ﬁrst identify what is the GCF of a polynomial. We will ﬁrst introduce this by looking at ﬁnding the GCF of several numbers. To ﬁnd a GCF of sevearal numbers we are looking for the largest number that can be divided by each of the numbers. This can often be done with quick mental math and it is shown in the following example Example 262. Find the GCF of 15, 24, and 27 15 3 = 5, 24 3 = 6, 27 3 GCF = 3 Our Solution = 9 Each of the numbers can be divided by 3 When ther
e are variables in our problem we can ﬁrst ﬁnd the GCF of the num- 212 bers using mental math, then we take any variables that are in common with each term, using the lowest exponent. This is shown in the next example Example 263. GCF of 24x4y2z, 18x2y4, and 12x3yz5 24 6 = 4, 18 6 = 3, 12 6 = 2 Each number can be divided by 6 x2y x and y are in all 3, using lowest exponets GCF = 6x2y Our Solution To factor out a GCF from a polynomial we ﬁrst need to identify the GCF of all the terms, this is the part that goes in front of the parenthesis, then we divide each term by the GCF, the answer is what is left inside the parenthesis. This is shown in the following examples Example 264. 4x2 4 = x2, − 20x 4 4x2 20x + 16 GCF is 4, divide each term by 4 = − 4(x2 − 5x, − 16 4 = 4 This is what is left inside the parenthesis 5x + 4) Our Solution With factoring we can always check our solutions by multiplying (distributing in this case) out the answer and the solution should be the original equation. Example 265. 25x4 25x4 5x2 = 5x2, − 15x3 5x2 = − 3x, − 5x2(5x2 15x3 + 20x2 GCF is 5x2, divide each term by this 20x2 5x2 = 4 This is what is left inside the parenthesis 3x + 4) Our Solution − Example 266. 3x3y2z + 5x4y3z5 − 4xy4 GCF is xy2, divide each term by this 213 3x3y2z xy2 = 3x2z, 5x4y3z5 xy2 = 5x3yz5, − 4xy4 xy2 = xy2(3x2z + 5x3yz5 − 4y2 This is what is left in parenthesis − 4y2) Our Solution World View Note: The ﬁrst recorded algorithm for ﬁnding the greatest common factor comes from Greek mathematician Euclid around the year 300 BC! Example 267. 21x3 7x = 3x2, 21x3 + 14x2 + 7x GCF is 7x, divide each term by this = 2x, 14x2 7x 7x(3x2 + 2x + 1) Our Solution 7x 7x = 1 This is what is left inside the parenthesis It is important to note in the previous example, that when the GCF was 7x and 7x was one of the terms, dividing gave an answer of 1. Students often try to factor out the 7x and get zero which is incorrect, factoring will never make terms dissapear. Anything divided by itself is 1, be sure to not forget to put the 1 into the solution. Often the second line is not shown in the work of factoring the GCF. We can simply identify the GCF and put it in front of the parenthesis as shown in the following two examples. Example 268. 12x5y2 − 2x3y2(6x2 6x4y4 + 8x3y5 GCF is 2x3y2, put this in front of parenthesis and divide 3xy2 + 4y3) Our Solution − Example 269. 18a4 b3 27a3b3 + 9a2b3 GCF is 9a2b3, divide each term by this − 9a2b3(2a2 3a + 1) Our Solution − Again, in the previous problem, when dividing 9a2b3 by itself, the answer is 1, not zero. Be very careful that each term is accounted for in your ﬁnal solution. 214 6.1 Practice - Greatest Common Factor Factor the common factor out of each expression. 1) 9 + 8b2 3) 45x2 5) 56 − 25 − 35p 7) 7ab 35a2b − 3a2b + 6a3b2 9) − 5x2 11) − 13) 20x4 5x3 15x4 − − 30x + 30 − 15) 28m4 + 40m3 + 8 2) x 5 − 4) 1 + 2n2 80y 6) 50x − 8) 27x2y5 72x3y2 − 10) 8x3y2 + 4x3 12) − 32n9 + 32n6 + 40n5 14) 21p6 + 30p2 + 27 16) − 10x4 + 20x2 + 12x 18) 27y7 + 12y2x + 9y2 17) 30b9 + 5ab 15a2 − 56a3b 19) − 48a2b2 − 56a5b − 21) 20x8y2z2 + 15x5y2z + 35x3y3z 23) 50x2y + 10y2 + 70xz2 25) 30qpr 5qp + 5q − 18n5 + 3n3 21n + 3 − 27) 29) 31) − − − 40x11 − 20x12 + 50x13 50x14 − 32mn8 + 4m6n + 12mn4 + 16mn 20) 30m6 + 15mn2 25 22) 3p + 12q − − 15q2r2 24) 30y4z3x5 + 50y4z5 10y4z3x − 26) 28b + 14b2 + 35b3 + 7b5 28) 30a8 + 6a5 + 27a3 + 21a2 24x6 4x4 + 12x3 + 4x2 − 10y7 + 6y10 4y10x 8y8x − − 30) 32) − − 215 6.2 Factoring - Grouping Objective: Factor polynomials with four terms using grouping. The ﬁrst thing we will always do when factoring is try to factor out a GCF. This GCF is often a monomial like in the problem 5x y + 10xz the GCF is the monomial 5x, so we would have 5x(y + 2z). However, a GCF does not have to be a monomial, it could be a binomial. To see this, consider the following two example. Example 270. 3ax x(3a − − 7bx Both have x in common, factor it out 7b) Our Solution Now the same problem, but instead of x we have (2a + 5b). Example 271. 3a(2a + 5b) 7b(2a + 5b) Both have (2a + 5b) in common, factor it out − (2a + 5b)(3a 7b) Our Solution − In the same way we factored out a GCF of x we can factor out a GCF which is a binomial, (2a + 5b). This process can be extended to factor problems where there is no GCF to factor out, or after the GCF is factored out, there is more factoring that can be done. Here we will have to use another strategy to factor. We will use a process known as grouping. Grouping is how we will factor if there are four terms in the problem. Remember, factoring is like multiplying in reverse, so ﬁrst we will look at a multiplication problem and then try to reverse the process. Example 272. (2a + 3)(5b + 2) Distribute (2a + 3) into second parenthesis 5b(2a + 3) + 2(2a + 3) Distribute each monomial 10ab + 15b + 4a + 6 Our Solution The solution has four terms in it. We arrived at the solution by looking at the two parts, 5b(2a + 3) and 2(2a + 3). When we are factoring by grouping we will always divide the problem into two parts, the ﬁrst two terms and the last two terms. Then we can factor the GCF out of both the left and right sides. When we do this our hope is what is left in the parenthesis will match on both the left and right. If they match we can pull this matching GCF out front, putting the rest in parenthesis and we will be factored. The next example is the same problem worked backwards, factoring instead of multiplying. 216 Example 273. 10ab + 15b + 4a + 6 10ab + 15b + 4a + 6 5b(2a + 3) + 2(2a + 3) Split problem into two groups GCF on left is 5b, on the right is 2 (2a + 3) is the same! Factor out this GCF (2a + 3)(5b + 2) Our Solution The key for grouping to work is after the GCF is factored out of the left and right, the two binomials must match exactly. If there is any diﬀerence between the two we either have to do some adjusting or it can’t be factored using the grouping method. Consider the following example. Example 274. 6x2 + 9xy 6x2 + 9xy − 3x(2x + 3y) + 7( − − 14x 14x 2x − − − 21y 21y 3y) Split problem into two groups GCF on left is 3x, on right is 7 The signs in the parenthesis don′t match! when the signs don’t match on both terms we can easily make them match by factoring the opposite of the GCF on the right side. Instead of 7 we will use 7. This will change the signs inside the second parenthesis. − 3x(2x + 3y) 7(2x + 3y) (2x + 3y) is the same! Factor out this GCF − (2x + 3y)(3x 7) Our Solution − Often we can recognize early that we need to use the opposite of the GCF when factoring. If the ﬁrst term of the ﬁrst binomial is positive in the problem, we will also want the ﬁrst term of the second binomial to be positive. If it is negative then we will use the opposite of the GCF to be sure they match. Example 275. 8x − 8x − − 10y + 16 10y + 16 5xy 5xy x(5y − − Split the problem into two groups GCF on left is x, on right we need a negative, so we use (5y 8) is the same! Factor out this GCF − 2 8) (5y − − 2(5y 8)(x − − 8) − 2) Our Solution 217 Sometimes when factoring the GCF out of the left or right side there is no GCF to factor out. In this case we will use either the GCF of 1 or 1. Often this is all we need to be sure the two binomials match. − Example 276. 12ab 12ab 2a(6b 14a − 14a Split the problem into two groups GCF on left is 2a, on right, no GCF, use (6b − 7) is the same! Factor out this GCF 6b + 7 6b + 7 7) − 1) Our Solution − − 1(6b 7)(2a 1 − 7) (6b − − − − − Example 277. 6x3 3x2(2x 5 6x3 15x2 + 2x − 15x2 + 2x 5) + 1(2x (2x − − − − 5 5) − − 5)(3x2 + 1) Our Solution − Split problem into two groups GCF on left is 3x2, on right, no GCF, use 1 (2x 5) is the same! Factor out this GCF Another problem that may come up with grouping is after factoring out the GCF on the left and right, the binomials don’t match, more than just the signs are different. In this case we may have to adjust the problem slightly. One way to do this is to change the order of the terms and try again. To do this we will move the second term to the end of the problem and see if that helps us use grouping. Example 278. 4a2 4a2 1(4a2 − − 14ab2 21b3 + 6ab − 21b3 + 6ab 14ab2 − 21b3) + 2ab(3 7b) 21b3 14ab2 14ab2 21b3 − 7b2(2a + 3b) − − − − − (2a + 3b)(2a − 4a2 + 6ab 4a2 + 6ab 2a(2a + 3b) Split the problem into two groups GCF on left is 1, on right is 2ab Binomials don′t match! Move second term to end Start over, split the problem into two groups GCF on left is 2a, on right is (2a + 3b) is the same! Factor out this GCF 7b2 − 7b2) Our Solution − When rearranging terms the problem can still be out of order. Sometimes after factoring out the GCF the terms are backwards. There are two ways that this can happen, one with addition, one with subtraction. If it happens with addition, for 218 example the binomials are (a + b) and (b + a), we don’t have to do any extra work. This is because addition is the same in either order (5 + 3 = 3 + 5 = 8). Example 279. 7 + y 7 + y 1(7 + y) 21x 3xy − 3xy 21x − 3x(y + 7) − − − (y + 7)(1 3x) Our Solution − Split the problem into two groups GCF on left is 1, on the right is y + 7 and 7 + y are the same, use either one 3x − However, if the binomial has subtraction, then we need to be a bit more careful. For example, if the binomials are (a a), we will factor out the opposite of the GCF on one part, usually the second. Notice what happens when we factor out b) and (b − − 1. − Example 280. − 1( − − 1(a a) Factor out (b b + a) Addition can be in either order, switch order − − 1 b) The order of the subtraction has been switched! − Generally we won’t show all the above steps, we will simply factor out the opposite of the GCF and switch the order of the subtraction to make it match the other binomial. Example 281. 8xy 8xy 12y + 15 − 12y 15 − 3) + 5(3 3) (2x − − − 5(2x 3)(4y 10x Split the problem into two groups GCF on left is 4y, on right, 5 10x Need to switch subtraction order, use 2x) Now 2x 3) 5) Our Solution − − − 4y(2x 4y(2y − − − − 3 match on both! Factor out this GCF 5 in middle − Worl
d View Note: Soﬁa Kovalevskaya of Russia was the ﬁrst woman on the editorial staﬀ of a mathematical journal in the late 19th century. She also did research on how the rings of Saturn rotated. 219 6.2 Practice - Grouping Factor each completely. 1) 40r3 8r2 − − 25r + 5 9n + 6 3) 3n3 2n2 − − 5) 15b3 + 21b2 35b 49 − 7) 3x3 + 15x2 + 2x + 10 − 2) 35x3 10x2 − − 56x + 16 4) 14v3 + 10v2 7v 5 − − 6) 6x3 − 48x2 + 5x 40 − 8) 28p3 + 21p2 + 20p + 15 35 − 12) 42r3 − 9) 35x3 − 11) 7xy − 28x2 20x + 16 − 49x + 5y 13) 32xy + 40x2 + 12y + 15x 15) 16xy 17) 2xy − 56x + 2y − 8x2 + 7y3 7 − 28y2x 19) 40xy + 35x 21) 32uv − − 8y2 − 20u + 24v 7y 15 − − 10) 7n3 + 21n2 5n 15 − − 49r2 + 18r 14) 15ab 16) 3mn − − 6a + 5b3 − 8m + 15n 18) 5mn + 2m 20) 8xy + 56x 25n − − y 7 − − 21 − 2b2 − 40 10 22) 4uv + 14u2 + 12v + 42u 20x − 49a − 30y3 − 16b 23) 10xy + 30 + 25x + 12y 25) 3uv + 14u 27) 16xy 3x − − − 6u2 7v − 6x2 + 8y 24) 24xy + 25y2 26) 56ab + 14 − 220 6.3 Factoring - Trinomials where a = 1 Objective: Factor trinomials where the coeﬃcient of x2 is one. Factoring with three terms, or trinomials, is the most important type of factoring to be able to master. As factoring is multiplication backwards we will start with a multipication problem and look at how we can reverse the process. Example 282. x(x + 6) x2 + 6x (x + 6)(x − 4) Distribute (x + 6) through second parenthesis 4(x + 6) Distribute each monomial through parenthesis − 4x − x2 + 2x 24 Combine like terms 24 Our Solution − − You may notice that if you reverse the last three steps the process looks like grouping. This is because it is grouping! The GCF of the left two terms is x and the GCF of the second two terms is 4. The way we will factor trinomials is to make them into a polynomial with four terms and then factor by grouping. This is shown in the following example, the same problem worked backwards − Example 283. x2 + 6x x(x + 6) x2 + 2x − 4x − 4(x + 6) − − (x + 6)(x Split middle term into + 6x 24 24 Grouping: GCF on left is x, on right is 4x − − (x + 6) is the same, factor out this GCF 4 4) Our Solution − − 4x and not + 5x 3x? The reason is because 6x The trick to make these problems work is how we split the middle term. Why did we pick + 6x 4x is the only combination that works! So how do we know what is the one combination that works? To ﬁnd the correct way to split the middle term we will use what is called the ac method. In the next lesson we will discuss why it is called the ac method. The way the ac method works is we ﬁnd a pair of numers that multiply to a certain number and add to another number. Here we will try to multiply to get the last term and add to get the coeﬃcient of the middle term. In the previous − − 221 example that would mean we wanted to multiply to only numbers that can do this are 6 and This process is shown in the next few examples 4 (6 · − − − 4 = − 24 and add to + 2. The 4) = 2). 24 and 6 + ( − Example 284. x2 + 9x + 18 Want to multiply to 18, add to 9 x2 + 6x + 3x + 18 x(x + 6) + 3(x + 6) 6 and 3, split the middle term Factor by grouping (x + 6)(x + 3) Our Solution Example 285. x(x x2 − 3x − 1(x 3)(x 4 1, split the middle term 4x + 3 Want to multiply to 3, add to x + 3 3) 1) Our Solution − − Factor by grouping 3 and − x2 − 3) − (x − − − − Example 286. x2 x2 8x − 10x + 2x − 10) + 2(x (x − − − − x(x − 20, add to 10 and 2, split the middle term 20 Want to multiply to 20 10) − Factor by grouping − 8 − 10)(x + 2) Our Solution Often when factoring we have two variables. These problems solve just like problems with one variable, using the coeﬃcients to decide how to split the middle term Example 287. a2 a(a − − a2 7ab 7b) (a − − − − 9ab + 14b2 Want to multiply to 14, add to 2ab + 14b2 7b) 2b(a 2b) Our Solution 7b)(a 9 2, split the middle term − − Factor by grouping 7 and − − − 222 As the past few examples illustrate, it is very important to be aware of negatives as we ﬁnd the pair of numbers we will use to split the middle term. Consier the following example, done incorrectly, ignoring negative signs Warning 288. x2 + 5x x2 + 2x + 3x x(x + 2) + 3(x 6 Want to multiply to 6, add 5 − 2 and 3, split the middle term 6 − Factor by grouping 2) − ??? Binomials do not match! Because we did not use the negative sign with the six to ﬁnd our pair of numbers, the binomials did not match and grouping was not able to work at the end. Now the problem will be done correctly. Example 289. x2 + 6x x2 + 5x 6 Want to multiply to x 6 − 1(x + 6) 6 and Factor by grouping − − − − x(x + 6) 1, split the middle term 6, add to 5 − (x + 6)(x 1) Our Solution − You may have noticed a shortcut for factoring these problems. Once we identify the two numbers that are used to split the middle term, these are the two numbers in our factors! In the previous example, the numbers used to split the middle 1, our factors turned out to be (x + 6)(x term were 6 and 1). This pattern does not always work, so be careful getting in the habit of using it. We can use it however, when we have no number (technically we have a 1) in front of x2. In all the problems we have factored in this lesson there is no number in front of x2. If this is the case then we can use this shortcut. This is shown in the next few examples. − − Example 290. x2 − 7x − 18 Want to multiply to 18, add to 9 and 2, write the factors − 7 − − 9)(x + 2) Our Solution (x − 223 Example 291. m2 − mn − 30n2 Want to multiply to 30, add to − 1 − 6, write the factors, don′t forget second variable 5 and 6n) Our Solution − (m + 5n)(m − It is possible to have a problem that does not factor. If there is no combination of numbers that multiplies and adds to the correct numbers, then we say we cannot factor the polynomial, or we say the polynomial is prime. This is shown in the following example. Example 292. x2 + 2x + 6 Want to multiply to 6, add to 2 1 3 Only possibilities to multiply to six, none add to 2 6 and 2 · Prime, can′t factor Our Solution · When factoring it is important not to forget about the GCF. If all the terms in a problem have a common factor we will want to ﬁrst factor out the GCF before we factor using any other method. Example 293. 3x2 3(x2 − − 24x + 45 GCF of all terms is 3, factor this out 8x + 15) Want to multiply to 15, add to 3, write the factors − 8 5 and 3) Our Solution − − 3(x − 5)(x − Again it is important to comment on the shortcut of jumping right to the factors, this only works if there is no coeﬃcient on x2. In the next lesson we will look at how this process changes slightly when we have a number in front of x2. Be careful not to use this shortcut on all factoring problems! World View Note: The ﬁrst person to use letters for unknown values was Francois Vieta in 1591 in France. He used vowels to represent variables we are solving for, just as codes used letters to represent an unknown message. 224 6.3 Practice - Trinomials where a = 1 Factor each completely. 1) p2 + 17p + 72 3) n2 5) x2 − − 9n + 8 9x 10 − 7) b2 + 12b + 32 9) x2 + 3x 70 − 8n + 15 11) n2 − 13) p2 + 15p + 54 15) n2 − 15n + 56 17) u2 8uv + 15v2 − 19) m2 + 2mn 8n2 − 11xy + 18y2 21) x2 − 23) x2 + xy − 25) x2 + 4xy 12y2 12y2 − 27) 5a2 + 60a + 100 29) 6a2 + 24a 192 − 31) 6x2 + 18xy + 12y2 33) 6x2 + 96xy + 378y2 2) x2 + x 4) x2 + x 72 30 − − 6) x2 + 13x + 40 8) b2 17b + 70 − 10) x2 + 3x 12) a2 6a − 14) p2 + 7p 18 27 30 − − − 16) m2 18) m2 − − 15mn + 50n2 3mn 40n2 − 20) x2 + 10xy + 16y2 22) u2 − 9uv + 14v2 24) x2 + 14xy + 45y2 26) 4x2 + 52x + 168 28) 5n2 − 45n + 40 30) 5v2 + 20v − 32) 5m2 + 30mn 25 90n2 162n2 − − 34) 6m2 36mn − 225 6.4 Factoring - Trinomials where a 1 Objective: Factor trinomials using the ac method when the coeﬃcient of x2 is not one. When factoring trinomials we used the ac method to split the middle term and then factor by grouping. The ac method gets it’s name from the general trinomial equation, a x2 + b x + c, where a, b, and c are the numbers in front of x2, x and the constant at the end respectively. World View Note: It was French philosopher Rene Descartes who ﬁrst used letters from the beginning of the alphabet to represent values we know (a, b, c) and letters from the end to represent letters we don’t know and are solving for (x, y, z). The ac method is named ac because we multiply a c to ﬁnd out what we want to multiply to. In the previous lesson we always multiplied to just c because there was no number in front of x2. This meant the number was 1 and we were multiplying to 1c or just c. Now we will have a number in front of x2 so we will be looking for numbers that multiply to ac and add to b. Other than this, the process will be the same. · Example 294. 3x2 + 11x + 6 Multiply to ac or (3)(6) = 18, add to 11 3x2 + 9x + 2x + 6 The numbers are 9 and 2, split the middle term 3x(x + 3) + 2(x + 3) Factor by grouping (x + 3)(3x + 2) Our Solution When a = 1, or no coeﬃcient in front of x2, we were able to use a shortcut, using the numbers that split the middle term in the factors. The previous example illustrates an important point, the shortcut does not work when a 1. We must go through all the steps of grouping in order to factor the problem. Example 295. 15 Multiply to ac or (8)( 15 The numbers are 3) − Factor by grouping 15) = 120, add to 2 − − − 12 and 10, split the middle term 8x2 4x(2x 8x2 2x − 12x + 10x 3) + 5(2x − − − − − (2x − 3)(4x + 5) Our Solution 226 Example 296. 10x2 5x(2x 10x2 − 25x − 1(2x 5)(5x − 5) − (2x − − − − 27x + 5 Multiply to ac or (10)(5) = 50, add to 2x + 5 The numbers are 27 2, split the middle term 25 and − − Factor by grouping 5) 1) Our Solution − The same process works with two variables in the problem Example 297. 4x2 4x2 + 4xy 4x(x + y) xy − − 5xy − − 5y(x + y) − (x + y)(4x − 5y2 Multiply to ac or (4)( − 5y2 The numbers are 4 and 5) = 20, add to 5, split the middle term − − 1 Factor by grouping − 5y) Our Solution As always, when factoring we will ﬁrst look for a GCF before using any other method, including the ac method. Factoring out the GCF ﬁrst also has the added bonus of making the numbers smaller so the ac m
ethod becomes easier. Example 298. 18x3 + 33x2 3x[6x2 + 11x 4x − − − − 2(2x + 5)] 30x GCF = 3x, factor this out ﬁrst 10] Multiply to ac or (6)( 10) = − 10] The numbers are 15 and Factor by grouping − 3x[6x2 + 15x 3x[3x(2x + 5) − 3x(2x + 5)(3x 2) Our Solution − 60, add to 11 4, split the middle term − As was the case with trinomials when a = 1, not all trinomials can be factored. If there is no combinations that multiply and add correctly then we can say the trinomial is prime and cannot be factored. Example 299. 3x2 + 2x 3(7) and 7 Multiply to ac or (3)( 7) = − 7(3) Only two ways to multiply to − 21, add to 2 21, it doesn′t add to 2 − − − Prime, cannot be factored Our Solution − 227 6.4 Practice - Trinomials where a 1 Factor each completely. 1) 7x2 48x + 36 − 3) 7b2 + 15b + 2 5) 5a2 − 7) 2x2 28 13a − 5x + 2 − 9) 2x2 + 19x + 35 11) 2b2 b 3 − − 13) 5k2 + 13k + 6 15) 3x2 − 17x + 20 17) 3x2 + 17xy + 10y2 19) 5x2 + 28xy 49y2 − 21) 6x2 − 23) 21k2 39x 21 − − 90 87k − 60x + 16 25) 14x2 − 27) 6x2 + 29x + 20 29) 4k2 17k + 4 − 31) 4x2 + 9xy + 2y2 33) 4m2 9mn − 35) 4x2 + 13xy + 3y2 − 9n2 2) 7n2 4) 7v2 6) 5n2 − − − 44n + 12 24v 16 − 4n 20 − 8) 3r2 4r 4 − − 10) 7x2 + 29x 30 − 26k + 24 12) 5k2 − 14) 3r2 + 16r + 21 16) 3u2 + 13uv 10v2 18) 7x2 2xy − − 20) 5u2 + 31uv − 5y2 28v2 − 22) 10a2 54a − 24) 21n2 + 45n 36 54 − − 26) 4r2 + r − 28) 6p2 + 11p 3 7 − 30) 4r2 + 3r 7 − 32) 4m2 + 6mn + 6n2 34) 4x2 − 36) 18u2 6xy + 30y2 3uv − 36v2 − 37) 12x2 + 62xy + 70y2 38) 16x2 + 60xy + 36y2 39) 24x2 − 52xy + 8y2 40) 12x2 + 50xy + 28y2 228 6.5 Factoring - Factoring Special Products Objective: Identify and factor special products including a diﬀerence of squares, perfect squares, and sum and diﬀerence of cubes. When factoring there are a few special products that, if we can recognize them, can help us factor polynomials. The ﬁrst is one we have seen before. When multiplying special products we found that a sum and a diﬀerence could multiply to a diﬀerence of squares. Here we will use this special product to help us factor Diﬀerence of Squares: a2 b2 = (a + b)(a b) − − If we are subtracting two perfect squares then it will always factor to the sum and diﬀerence of the square roots. Example 300. x2 (x + 4)(x − − Example 301. Subtracting two perfect squares, the square roots are x and 4 16 4) Our Solution 9a2 − (3a + 5b)(3a − 25b2 5b) Our Solution Subtracting two perfect squares, the square roots are 3a and 5b It is important to note, that a sum of squares will never factor. It is always prime. This can be seen if we try to use the ac method to factor x2 + 36. Example 302. x2 + 36 No bx term, we use 0x. x2 + 0x + 36 Multiply to 36, add to 0 12, 4 9, 6 36, 2 18, 3 1 · · · Prime, cannot factor Our Solution · · 6 No combinations that multiply to 36 add to 0 229 It turns out that a sum of squares is always prime. Sum of Squares: a2 + b2 = Prime A great example where we see a sum of squares comes from factoring a diﬀerence of 4th powers. Because the square root of a fourth power is a square ( a4√ = a2), we can factor a diﬀerence of fourth powers just like we factor a diﬀerence of squares, to a sum and diﬀerence of the square roots. This will give us two factors, one which will be a prime sum of squares, and a second which will be a diﬀerence of squares which we can factor again. This is shown in the following examples. Example 303. a4 (a2 + b2)(a2 (a2 + b2)(a + b)(a b4 Diﬀerence of squares with roots a2 and b2 b2) The ﬁrst factor is prime, the second is a diﬀerence of squares! b) Our Solution − − − Example 304. x4 (x2 + 4)(x2 (x2 + 4)(x + 2)(x − − − 16 Diﬀerence of squares with roots x2 and 4 4) The ﬁrst factor is prime, the second is a diﬀerence of squares! 2) Our Solution Another factoring shortcut is the perfect square. We had a shortcut for multiplying a perfect square which can be reversed to help us factor a perfect square Perfect Square: a2 + 2ab + b2 = (a + b)2 A perfect square can be diﬃcult to recognize at ﬁrst glance, but if we use the ac method and get two of the same numbers we know we have a perfect square. Then we can just factor using the square roots of the ﬁrst and last terms and the sign from the middle. This is shown in the following examples. Example 305. x2 − (x − 6x + 9 Multiply to 9, add to The numbers are 3, the same! Perfect square 3)2 Use square roots from ﬁrst and last terms and sign from the middle 6 − 3 and − − 230 Example 306. 4x2 + 20xy + 25y2 Multiply to 100, add to 20 The numbers are 10 and 10, the same! Perfect square (2x + 5y)2 Use square roots from ﬁrst and last terms and sign from the middle World View Note: The ﬁrst known record of work with polynomials comes from the Chinese around 200 BC. Problems would be written as “three sheafs of a good crop, two sheafs of a mediocre crop, and one sheaf of a bad crop sold for 29 dou. This would be the polynomial (trinomial) 3x + 2y + z = 29. Another factoring shortcut has cubes. With cubes we can either do a sum or a diﬀerence of cubes. Both sum and diﬀerence of cubes have very similar factoring formulas Sum of Cubes: a3 + b3 = (a + b)(a2 ab + b2) − Diﬀerence of Cubes: a3 b3 = (a − − b)(a2 + ab + b2) Comparing the formulas you may notice that the only diﬀerence is the signs in between the terms. One way to keep these two formulas straight is to think of SOAP. S stands for Same sign as the problem. If we have a sum of cubes, we add ﬁrst, a diﬀerence of cubes we subtract ﬁrst. O stands for Opposite sign. If we have a sum, then subtraction is the second sign, a diﬀerence would have addition for the second sign. Finally, AP stands for Always Positive. Both formulas end with addition. The following examples show factoring with cubes. Example 307. m3 (m 3)(m2 3m 9) Use formula, use SOAP to ﬁll in signs 27 We have cube roots m and 3 − (m − 3)(m2 + 3m + 9) Our Solution Example 308. 125p3 + 8r3 We have cube roots 5p and 2r (5p 2r)(25p2 10r 4r2) Use formula, use SOAP to ﬁll in signs (5p + 2r)(25p2 − 10r + 4r2) Our Solution The previous example illustrates an important point. When we ﬁll in the trinomial’s ﬁrst and last terms we square the cube roots 5p and 2r. Often students forget to square the number in addition to the variable. Notice that when done correctly, both get cubed. 231 Often after factoring a sum or diﬀerence of cubes, students want to factor the second factor, the trinomial further. As a general rule, this factor will always be prime (unless there is a GCF which should have been factored out before using cubes rule). The following table sumarizes all of the shortcuts that we can use to factor special products Factoring Special Products Diﬀerence of Squares Sum of Squares Perfect Square Sum of Cubes Diﬀerence of Cubes − − b) b2 = (a + b)(a a2 a2 + b2 = Prime a2 + 2ab + b2 = (a + b)2 a3 + b3 = (a + b)(a2 a3 b3 = (a − ab + b2) b)(a2 + ab + b2) − − As always, when factoring special products it is important to check for a GCF ﬁrst. Only after checking for a GCF should we be using the special products. This is shown in the following examples Example 309. 72x2 2(36x2 2(6x + 1)(6x − − − 2 GCF is 2 1) Diﬀerence of Squares, square roots are 6x and 1 1) Our Solution Example 310. 24xy + 3y GCF is 3y 48x2y − 3y(16x2 − 8x + 1) Multiply to 16 add to 8 The numbers are 4 and 4, the same! Perfect Square 3y(4x − 1)2 Our Solution Example 311. 128a4b2 + 54ab5 GCF is 2ab2 2ab2(64a3 + 27b3) Sum of cubes! Cube roots are 4a and 3b 2ab2(4a + 3b)(16a2 − 12ab + 9b2) Our Solution 232 6.5 Practice - Factoring Special Products Factor each completely. 16 25 4 1) r2 3) v2 − − 5) p2 − 7) 9k2 − 9) 3x2 − 11) 16x2 13) 18a2 4 27 36 − 50b2 − 2a + 1 15) a2 − 17) x2 + 6x + 9 6x + 9 19) x2 − 21) 25p2 10p + 1 − 23) 25a2 + 30ab + 9b2 25) 4a2 27) 8x2 20ab + 25b2 − 24xy + 18y2 − m3 64 29) 8 − 31) x3 − 33) 216 − 35) 125a3 u3 64 − 37) 64x3 + 27y3 39) 54x3 + 250y3 41) a4 43) 16 45) x4 81 z4 y4 − − − 47) m4 81b4 − 9 1 2) x2 − 4) x2 − 6) 4v2 − 8) 9a2 − 10) 5n2 1 1 20 − 12) 125x2 + 45y2 14) 4m2 + 64n2 16) k2 + 4k + 4 18) n2 − 8n + 16 20) k2 4k + 4 − 22) x2 + 2x + 1 24) x2 + 8xy + 16y2 26) 18m2 24mn + 8n2 − 28) 20x2 + 20xy + 5y2 30) x3 + 64 32) x3 + 8 34) 125x3 36) 64x3 216 − 27 − 38) 32m3 108n3 − 40) 375m3 + 648n3 42) x4 256 − 44) n4 − 46) 16a4 48) 81c4 1 − − b4 16d4 233 6.6 Factoring - Factoring Strategy Objective: Idenﬁty and use the correct method to factor various polynomials. With so many diﬀerent tools used to factor, it is easy to get lost as to which tool to use when. Here we will attempt to organize all the diﬀerent factoring types we have seen. A large part of deciding how to solve a problem is based on how many terms are in the problem. For all problem types we will always try to factor out the GCF ﬁrst. Factoring Strategy (GCF First!!!!!) 2 terms: sum or diﬀerence of squares or cubes: b2 = (a + b)(a a2 − b) − a2 + b2 = Prime a3 + b3 = (a + b)(a2 ab + b2) − b3 = (a a3 − − b)(a2 + ab + b2) 3 terms: ac method, watch for perfect square! a2 + 2ab + b2 = (a + b)2 Multiply to ac and add to b 4 terms: grouping • • • We will use the above strategy to factor each of the following examples. Here the emphasis will be on which strategy to use rather than the steps used in that method. Example 312. 4x2 + 56xy + 196y2 GCF ﬁrst, 4 4(x2 + 14xy + 49y2) Three terms, try ac method, multiply to 49, add to 14 7 and 7, perfect square! 234 4(x + 7y)2 Our Solution Example 313. 5x 2y + 15xy 35x2 105x GCF ﬁrst, 5x − 5x(xy + 3y 5x[y(x + 3) − 7x 21) − 7(x + 3)] − − 5x(x + 3)(y Four terms, try grouping (x + 3) match! 7) Our Solution − Example 314. Example 315. 100x2 − 100(x2 100(x + 4)(x − − 400 GCF ﬁrst, 100 4) Two terms, diﬀerence of squares 4) Our Solution 108x3y2 39x2y2 + 3xy2 GCF ﬁrst, 3xy2 3xy2(36x2 3xy2[9x(4x − 3xy2(36x2 − 9x − 1(4x 1)(9x − 1) − 3xy2(4x − − − − 13x + 1) Thee terms, ac method, multiply to 36, add to 4x + 1) 1)] 1) Our Solution − − Factor by grouping 4, split middle term 9 and 13 − World View Note: Variables originated in ancient Greece where Aristotle would use a single capital letter to represent a number. Example
316. 5 + 625y3 GCF ﬁrst, 5 5(1 + 125y3) Two terms, sum of cubes 5(1 + 5y)(1 − 5y + 25y2) Our Solution It is important to be comfortable and conﬁdent not just with using all the factoring methods, but decided on which method to use. This is why practice is very important! 235 6.6 Practice - Factoring Strategy Factor each completely. 45yh − 2) 2x2 − 11x + 15 4) 16x2 + 48xy + 36y2 6) 20uv 60u3 5xv + 15xu2 − 8) 2x3 + 5x2y + 3y2x − 1) 24az 18ah + 60yz − 9uv + 4v2 − 2x3 + 128y3 3) 5u2 5) − 7) 5n3 + 7n2 6n − 9) 54u3 16 − 11) n2 n − 4xy + 3y2 13) x2 − 15) 9x2 − 17) m2 − 19) 36b2c 25y2 4n2 16xd − − 24b2d + 24xc 21) 128 + 54x3 23) 2x3 + 6x2y 20y2x − 25) n3 + 7n2 + 10n 27) 27x3 64 − 29) 5x2 + 2x 31) 3k3 33) mn − − 27k2 + 60k 12x + 3m 4xn − 35) 16x2 8xy + y2 − 37) 27m2 48n2 − 39) 9x3 + 21x2y 41) 2m2 + 6mn 60y2x 20n2 − − 128x3 22x 15 − 10) 54 − 12) 5x2 − 14) 45u2 16) x3 − 18) 12ab − 27y3 − 20) 3m3 6m2n − 22) 64m3 + 27n3 − 150uv + 125v2 18a + 6nb 9n − 24n2m 24) 3ac + 15ad2 + x2c + 5x2d2 26) 64m3 n3 − 28) 16a2 9b2 − 10x + 12 30) 2x2 − 32) 32x2 − 34) 2k2 + k 18y2 10 − 36) v2 + v 38) x3 + 4x2 40) 9n3 − 42) 2u2v2 3n2 11uv3 + 15v4 − 236 6.7 Factoring - Solve by Factoring Objective: Solve quadratic equation by factoring and using the zero product rule. When solving linear equations such as 2x 5 = 21 we can solve for the variable directly by adding 5 and dividing by 2 to get 13. However, when we have x2 (or a higher power of x) we cannot just isolate the variable as we did with the linear equations. One method that we can use to solve for the varaible is known as the zero product rule − Zero Product Rule: If ab = 0 then either a = 0 or b = 0 The zero product rule tells us that if two factors are multiplied together and the answer is zero, then one of the factors must be zero. We can use this to help us solve factored polynomials as in the following example. Example 317. 2x 3)(5x + 1) = 0 One factor must be zero − + 3 + 3 − (2x 3 = 0 or 5x + 1 = 0 1 − 1 − 5 1 or − 5 1 2x = 3 or 5x = 2 x = 3 2 − 2 5 Set each factor equal to zero Solve each equation Our Solution For the zero product rule to work we must have factors to set equal to zero. This means if the problem is not already factored we will factor it ﬁrst. Example 318. Factor using the ac method, multiply to 12, add to 1 3 and 4, split the middle term − 3 = 0 3 = 0 The numbers are 3) = 0 − Factor by grouping 4x2 4x2 + x 3x + 4x − 3) + 1(4x − (4x 3 = 0 or (4x 4x − 3)(x + 1) = 0 One factor must be zero Set each factor equal to zero 237 + 3 + 3 1 4x = 3 or x = − 4 4 Solve each equation 1 1 − − x = 3 4 or − 1 Our Solution Another important part of the zero product rule is that before we factor, the equation must equal zero. If it does not, we must move terms around so it does equal zero. Generally we like the x2 term to be positive. Example 319. x2 = 8x 15 − 8x + 15 8x + 15 = 0 − − 8x + 15 x2 Set equal to zero by moving terms to the left Factor using the ac method, multiply to 15, add to − (x 5)(x − 5 = 0 or x x − + 5 + 5 3) = 0 The numbers are − 3 = 0 − + 3 + 3 3 − Set each factor equal to zero Solve each equation 5 and − 8 − x = 5 or x = 3 Our Solution Example 320. x2 − (x − 7x + 3x x2 4x 7)(x + 3) = 21 = − 21 = − + 9 + 9 12 = 0 − − − 4x − x2 9 Not equal to zero, multiply ﬁrst, use FOIL 9 Combine like terms 9 Move 9 to other side so equation equals zero − Factor using the ac method, mutiply to 12, add to 4 − − x − − − 6)(x + 2) = 0 The numbers are 6 and (x 6 = 0 or x + 2 = 0 2 2 Our Solution 2 Set each factor equal to zero Solve each equation 2 − x = 6 or − − + 6 + 6 − − Example 321. 3x2 + 4x 3x2 − 4x + 5 5 = 7x2 + 4x 3x2 − − − − 0 = 4x2 14 − 4x + 5 9 − Set equal to zero by moving terms to the right Factor using diﬀerence of squares 238 0 = (2x + 3)(2x − 2x + 3 = 0 or 2x 3 3 − 2x = or 2x = or 3) One factor must be zero Set each factor equal to zero Solve each equation Our Solution Most problems with x2 will have two unique solutions. However, it is possible to have only one solution as the next example illustrates. Example 322. 4x2 = 12x 9 − 12x + 9 − 4x2 12x + 9 − (2x − 12x + 9 = 0 3)2 = 0 2x − 3 = 0 − + 3 + 3 2x = 3 2 2 3 2 x = Set equal to zero by moving terms to left Factor using the ac method, multiply to 36, add to 6 and 6, a perfect square! − − Set this factor equal to zero Solve the equation 12 − Our Solution As always it will be important to factor out the GCF ﬁrst if we have one. This GCF is also a factor and must also be set equal to zero using the zero product rule. This may give us more than just two solution. The next few examples illustrate this. Example 323. 4x2 = 8x 8x Set equal to zero by moving the terms to left 8x Be careful, on the right side, they are not like terms! − 4x2 4x(x 4x = 0 or x 4 4 Factor out the GCF of 4x − 8x = 0 − 2) = 0 One factor must be zero − 2 = 0 − + 2 + 2 Set each factor equal to zero Solve each equation x = 0 or 2 Our Solution 239 Example 324. 2x3 − 2x(x2 2x(x 14x2 + 24x = 0 7x + 12) = 0 − 3)(x − 3 = 0 or x 2x = 0 or or 3 or 4 Example 325. Factor out the GCF of 2x Factor with ac method, multiply to 12, add to 3 and 7 − 4) = 0 The numbers are − 4 = 0 − + 4 + 4 4 − Set each factor equal to zero Solve each equation Our Solutions − 6x2 + 21x 3(2x2 + 7x 2x − − − − 1(2x + 9)] = 0 3(2x2 + 9x Factor out the GCF of 3 Factor with ac method, multiply to 27 = 0 9) = 0 9) = 0 The numbers are 9 and 2 − Factor by grouping 3[x(2x + 9) − 3(2x + 9)(x 3 = 0 or 2x + 9 = 0 or x 3 0 1) = 0 One factor must be zero − 1 = 0 − + 1 + 1 9 or x = 1 − 2 Set each factor equal to zero Solve each equation − 9 9 − 2x = 2 18, add to 7 − x = 9 2 − or 1 Our Solution In the previous example, the GCF did not have a variable in it. When we set this factor equal to zero we got a false statement. No solutions come from this factor. Often a student will skip setting the GCF factor equal to zero if there is no variables in the GCF. Just as not all polynomials cannot factor, all equations cannot be solved by factoring. If an equation does not factor we will have to solve it using another method. These other methods are saved for another section. World View Note: While factoring works great to solve problems with x2, Tartaglia, in 16th century Italy, developed a method to solve problems with x3. He kept his method a secret until another mathematician, Cardan, talked him out of his secret and published the results. To this day the formula is known as Cardan’s Formula. A question often asked is if it is possible to get rid of the square on the variable by taking the square root of both sides. While it is possible, there are a few properties of square roots that we have not covered yet and thus it is common to break a rule of roots that we are not aware of at this point. The short reason we want to avoid this for now is because taking a square root will only give us one of the two answers. When we talk about roots we will come back to problems like these and see how we can solve using square roots in a method called completing the square. For now, never take the square root of both sides! 240 6.7 Practice - Solve by Factoring Solve each equation by factoring. 1) (k − 7)(k + 2) = 0 1)(x + 4) = 0 3) (x − 5) 6x2 150 = 0 − 7) 2n2 + 10n 28 = 0 − 9) 7x2 + 26x + 15 = 0 11) 5n2 13) x2 − 4x − 9n 2 = 0 − 8 = − 15) x2 5x 1 = − 17) 49p2 + 371p − 8 5 − − 163 = 5 − 19) 7x2 + 17x 20 = 8 − − 21) 7r2 + 84 = 49r − 6x = 16 23) x2 − 25) 3v2 + 7v = 40 2) (a + 4)(a − 4) (2x + 5)(x 6) p2 + 4p − 3) = 0 7) = 0 − 32 = 0 8) m2 − 10) 40r2 12) 2b2 m 30 = 0 − 285r − 3b − 2 = 0 − − 280 = 0 14) v2 16) a2 − − 8v 3 = − 6a + 6 = 3 2 − − 18) 7k2 + 57k + 13 = 5 20) 4n2 − 13n + 8 = 5 22) 7m2 − 224 = 28m 24) 7n2 28n = 0 − 26) 6b2 = 5 + 7b 28) 9n2 + 39n = 36 30) a2 + 7a − − 9 = − 3 + 6a 27) 35x2 + 120x = 45 − 23 = 6k 29) 4k2 + 18k − 7 − 32) x2 + 10x + 30 = 6 31) 9x2 − 46 + 7x = 7x + 8x2 + 3 33) 2m2 + 19m + 40 = 2m − 35) 40p2 + 183p 168 = p + 5p2 − 34) 5n2 + 41n + 40 = 2 − 80 = 3x 36) 24x2 + 11x − 241 Chapter 7 : Rational Expressions 7.1 Reduce Rational Expressions ..................................................................243 7.2 Multiply and Divide ................................................................................248 7.3 Least Common Denominator ..................................................................253 7.4 Add and Subtract ...................................................................................257 7.5 Complex Fractions ..................................................................................262 7.6 Proportions .............................................................................................268 7.7 Solving Rational Equations ....................................................................274 7.8 Application: Dimensional Analysis .........................................................279 242 7.1 Rational Expressions - Reduce Rational Expressions Objective: Reduce rational expressions by dividing out common factors. Rational expressions are expressions written as a quotient of polynomials. Examples of rational expressions include: x2 x2 − − x 12 − 9x + 20 and 3 − 2 x and a b b a − − and 3 2 As rational expressions are a special type of fraction, it is important to remember with fractions we cannot have zero in the denominator of a fraction. For this reason, rational expressions may have one more excluded values, or values that the variable cannot be or the expression would be undeﬁned. Example 326. State the excluded value(s): Denominator cannot be zero Factor Set each factor not equal to zero Subtract 5 from second equation x2 1 − 3x2 + 5x 3x2 + 5x 0 x(3x + 5) 0 x 0 or 3x + 5 0 5 5 − − 3x − 5 Divide by or − 3 Our Solution Second equation is solved This means we can use any value for x in the equation except for 0 and − 5 3 . We 243 can however, evaluate any other value in the expression. World View Note: The number zero was not widely accepted in mathematical thought around the world for many years. It was the Mayans of Central America who ﬁrst used zero
to aid in the use of their base 20 system as a place holder! Rational expressions are easily evaluated by simply substituting the value for the variable and using order of operations. Example 327. x2 4 − x2 + 6x + 8 when x = 6 − Substitute − 5 in for each variable 6)2 ( − 6)2 + 6( ( − 36 36 + 6( 4 6) + 8 4 6) + 8 − − − − 4 − 36 + 8 36 − 36 Exponents ﬁrst Multiply Add and subtract 32 8 Reduce 4 Our Solution Just as we reduced the previous example, often a rational expression can be reduced, even without knowing the value of the variable. When we reduce we divide out common factors. We have already seen this with monomials when we discussed properties of exponents. If the problem only has monomials we can reduce the coeﬃcients, and subtract exponents on the variables. Example 328. 15x4y2 25x2y6 Reduce , subtract exponents. Negative exponents move to denominator 3x2 5y4 Our Solution 244 However, if there is more than just one term in either the numerator or denominator, we can’t divide out common factors unless we ﬁrst factor the numerator and denominator. Example 329. Example 330. 16 28 8x2 − 28 8(x2 2) − 7 2(x2 − 2) Denominator has a common factor of 8 Reduce by dividing 24 and 8 by 4 Our Solution 9x 18x 3 6 − − Numerator has a common factor of 3, denominator of 6 3(3x 6(3x 1) 1) − − Divide out common factor (3x 1) and divide 3 and 6 by 3 − 1 2 Our Solution Example 331. x2 25 − x2 + 8x + 15 Numerator is diﬀerence of squares, denominator is factored using ac (x + 5)(x 5) (x + 3)(x + 5) − Divide out common factor (x + 5) x 5 − x + 3 Our Solution It is important to remember we cannot reduce terms, only factors. This means if between the parts we want to reduce we cannot. In the prethere are any + or − vious example we had the solution x x + 3 , we cannot divide out the x’s because ) not factors (separated by multiplication). they are terms (separated by + or − 5 − 245 7.1 Practice - Reduce Rational Expressions Evaluate 1) 4v + 2 6 when v = 4 3) 5) x 3 − x2 4x + 3 when x = − b + 2 − b2 + 4b + 4 when b = 0 4 4) 2) b 3b 3 9 when b = − − a + 2 2 − a2 + 3a + 2 when a = 1 − 3 when n = 4 − 6 6) n2 − n n − State the excluded values for each. 7) 3k2 + 30k k + 10 9) 15n2 10n + 25 11) 10m2 + 8m 10m 13) r2 + 3r + 2 5r + 10 15) b2 + 12b + 32 b2 + 4b 32 − Simplify each expression. 17) 21x2 18x 19) 24a 40a2 21) 32x3 8x4 23) 18m − 60 24 25) 20 4p + 2 27) x + 1 x2 + 8x + 7 29) 32x2 28x2 + 28x 31) n2 + 4n n2 12 − 7n + 10 33) v2 − 9v + 54 4v − 35) 12x2 30x2 42x 42x − − 10 − 10a + 4 37) 6a 39) 2n2 + 19n − 9n + 90 10 8) 10) 27p 18p2 36p − x + 10 8x2 + 80x 12) 10x + 16 6x + 20 14) 6n2 21n − 6n2 + 3n 16) 10v2 + 30v 35v2 5v − 18) 12n 4n2 20) 21k 24k2 22) 90x2 20x 24) 10 81n3 + 36n2 − − 9 81 26) n 9n 28) 28m + 12 36 30) 49r + 56 56r 32) b2 + 14b + 48 b2 + 15b + 56 34) 30x 90 − 50x + 40 12k + 32 64 − k2 − 38) 9p + 18 p2 + 4p + 4 40) 3x2 5x2 29x + 40 30x 80 − − − 246 − 60 36) k2 41) 8m + 16 20m 12 − 42) 56x 48 − 24x2 + 56x + 32 43) 2x2 3x2 10x + 8 7x + 4 − − 45) 7n2 32n + 16 16 − 4n − 47) n2 2n + 1 − 6n + 6 49) 7a2 6a2 26a 45 − 34a + 20 − − 44) 50b 80 − 50b + 20 46) 35v + 35 21v + 7 48) 56x 48 − 24x2 + 56x + 32 50) 4k3 9k3 2k2 2k − 18k2 + 9k − − 247 7.2 Rational Expressions - Multiply & Divide Objective: Multiply and divide rational expressions. Multiplying and dividing rational expressions is very similar to the process we use to multiply and divide fractions. Example 332. 15 49 · 14 45 1 7 · 2 3 2 21 First reduce common factors from numerator and denominator (15 and 7) Multiply numerators across and denominators across Our Solution The process is identical for division with the extra ﬁrst step of multiplying by the reciprocal. When multiplying with rational expressions we follow the same process, ﬁrst divide out common factors, then multiply straight across. 248 Example 333. 25x2 9y8 · 24y4 55x7 5 3y4 · 8 11x5 40 33x5y4 Reduce coeﬃcients by dividing out common factors (3 and 5) Reduce, subtracting exponents, negative exponents in denominator Multiply across Our Solution Division is identical in process with the extra ﬁrst step of multiplying by the reciprocal. Example 334. a4b2 a ÷ b4 4 a4b2 a · 4 b4 Multiply by the reciprocal Subtract exponents on variables, negative exponents in denominator a3 1 · 4 b2 Multiply across 4a3 b2 Our Solution Just as with reducing rational expressions, before we reduce a multiplication problem, it must be factored ﬁrst. Example 335. x2 − x2 + x 9 x2 8x + 16 − 3x + 9 20 · Factor each numerator and denominator − (x + 3)(x (x 3) − 4)(x + 5) · − (x 4)(x − − 3(x + 3) 4 Divide out common factors (x + 3) and (x 4) − Multiply across 249 (x 3)(x − − 3(x + 5) 4) Our Solution Again we follow the same pattern with division with the extra ﬁrst step of multiplying by the reciprocal. Example 336. 12 8 ÷ x − 2x − 5x2 + 15x x2 + x 2 − Multiply by the reciprocal x2 + x 2 5x2 + 15x − Factor each numerator and denominator x2 x2 − − x2 x2 12 8 · x − 2x − − − 4)(x + 3) (x − (x + 2)(x 4) · − (x + 2)(x − 5x(x + 3) 1) Divide out common factors: (x 4) and (x + 3) and (x + 2) − Multiply across Our Solution 1 1 · 1 1 x x − 5x − 5x We can combine multiplying and dividing of fractions into one problem as shown below. To solve we still need to factor, and we use the reciprocal of the divided fraction. Example 337. a2 + 7a + 10 a2 + 6a + 5 · a + 1 a2 + 4a + 4 ÷ a 1 − a + 2 Factor each expression (a + 5)(a + 2) (a + 5)(a + 1) · (a + 1) (a + 2)(a + 2) ÷ (a 1) − (a + 2) Reciprocal of last fraction (a + 5)(a + 2) (a + 5)(a + 1) · (a + 1) (a + 2)(a + 2) · (a + 2) (a 1) − 1 − 1 a Divide out common factors (a + 2), (a + 2), (a + 1), (a + 5) Our Solution World View Note: Indian mathematician Aryabhata, in the 6th century, published a work which included the rational expression n(n + 1)(n + 2) for the sum of the ﬁrst n squares (11 + 22 + 32 + . + n2) 6 250 7.2 Practice - Multiply and Divide Simplify each expression. 1) 8x2 9 · 9 2 7 5n 3) 9n 2n · 5) 5x2 4 · 7) 7 (m 6) − m 6 − 6 5 · 5m(7m 7(7m 5) 5) − − 9) 7r 7r(r + 10) ÷ r (r 6 6)2 − − 11) 25n + 25 5 4 30n + 30 · 13) x 15) x2 − 10 35x + 21 ÷ 6x − x + 5 − 7 7 35x + 21 x + 5 x 7 − · 17) 8k 24k2 40k ÷ − 15k 1 − 25 19) (n 8) − 21) 4m + 36 m + 9 · 3x 6 12x 23) − − 80 6 − 10n · m 5 − 5m2 24(x + 3) 12 6n − 13n + 42 n2 − − 27) n 6n 7 12 · 29) 27a + 36 9a + 63 ÷ 12x + 32 6x 16 · − − − 31) x2 x2 − 6a + 8 2 7x2 + 14x 7x2 + 21x 33) (10m2 + 100m) 35) 7p2 + 25p + 12 6p + 48 18m3 20m2 36m2 40m − − · 3p 21p2 · − 8 − 44p 32 − 2) 8x 3x ÷ 4) 9m 5m2 · 4 7 7 2 8 10 6) 10p 8) 5 ÷ 7 10(n + 3) ÷ 10) 6x(x + 4) x 3 · n 2 (n + 3)(n − 2) − (x 3)(x − 6x(x 6) − 6) 5 12 − − − b − 9 b 12) b2 14) v − 1 − 4 b − − b2 12 ÷ 4 11v + 10 v2 · − 8a + 80 8 6 · p − 12p + 32 ÷ 8 − 16) 1 − a 18) p2 x2 · 2r 7r + 42 20) x2 − x 7x + 10 2 − 22) 2r r + 6 ÷ 12n − n + 7 24) 2n2 1 10 p − x + 10 x − − 20 54 − ÷ (2n + 6) 35v v 20 − 9 − ÷ 6x3 + 6x2 x2 + 5x 24 − · 7k2 8k2 28) x2 + 11x + 24 6x3 + 18x2 k 7 − k 30) k2 − − 12 · 32) 9x3 + 54x2 x2 + 5x 28k 56k − − x2 + 5x − 10x2 14 14 · − 7 9n + 54 10n + 50 34) n − 2n n2 − 36) 7x2 − 49x2 + 7x − 35 ÷ 66x + 80 72 ÷ − 7x2 + 39x 2 49x + 7x − − 70 72 15 25) b + 2 40b2 − 24b(5b 3) − 26) 21v2 + 16v 3v + 4 − 16 37) 10b2 30b + 20 · 30b + 20 2b2 + 10b 39) 7r2 24 53r − − 7r + 2 ÷ 49r + 21 49r + 14 38) 35n2 49n2 12n 32 − 91n + 40 · − − 7n2 + 16n 5n + 4 − 40) 12x + 24 10x2 + 34x + 28 · 15x + 21 5 251 41) x2 2x 1 4 · − − x2 x2 − − x 4 2 ÷ − 2 x2 + x 3x − − 6 42) a3 + b3 a2 + 3ab + 2b2 · 3a2 6b 3a − 3ab + 3b2 ÷ − a2 4b2 − a + 2b 43) x2 + 3x + 9 x2 + x 12 · − x2 + 2x x3 − 8 − 27 ÷ x2 − x2 4 − 6x + 9 44) x2 + 3x 10 − x2 + 6x + 5 · 2x2 x − 2x2 + x 3 6 ÷ − − 8x + 20 6x + 15 252 7.3 Rational Expressions - Least Common Denominators Objective: denominators to match this common denominator. Idenﬁty the least common denominator and build up As with fractions, the least common denominator or LCD is very important to working with rational expressions. The process we use to ﬁnd the LCD is based on the process used to ﬁnd the LCD of intergers. Example 338. Find the LCD of 8 and 6 Consider multiples of the larger number 8, 16, 24 . 24 is the ﬁrst multiple of 8 that is also divisible by 6 24 Our Solution When ﬁnding the LCD of several monomials we ﬁrst ﬁnd the LCD of the coeﬃcients, then use all variables and attach the highest exponent on each variable. Example 339. Find the LCD of 4x2y5 and 6x4y3z6 First ﬁnd the LCD of coeﬃcients 4 and 6 12 is the LCD of 4 and 6 12 x4y5z6 Use all variables with highest exponents on each variable 12x4y5z6 Our Solution The same pattern can be used on polynomials that have more than one term. However, we must ﬁrst factor each polynomial so we can identify all the factors to be used (attaching highest exponent if necessary). Example 340. Find the LCD of x2 + 2x (x − 3 and x2 x 12 − − 4)(x + 3) Factor each polynomial LCD uses all unique factors − 1)(x + 3) and (x (x − 1)(x + 3)(x − 4) Our Solution − Notice we only used (x + 3) once in our LCD. This is because it only appears as a factor once in either polynomial. The only time we need to repeat a factor or use an exponent on a factor is if there are exponents when one of the polynomials is factored 253 Example 341. Find the LCD of x2 10x + 25 and x2 14x + 45 − − (x − 5)2 and (x (x 5)(x 5)2(x − − Factor each polynomial LCD uses all unique factors with highest exponent 9) 9) Our Solution − − 5)(x The previous example could have also been done with factoring the ﬁrst polynomial to (x 5) twice in the LCD because it showed up twice in one of the polynomials. However, it is the author’s suggestion to use the exponents in factored form so as to use the same pattern (highest exponent) as used with monomials. 5). Then we would have used (x − − − Once we know the LCD, our goal will be to build up fractions so they have matching denominators. In this lesson we will not be adding and subtracting fractions, just building them up to a common denominator. We can build up a fraction’s denominator by multipliplying the numerator and denoinator by any factors that are not al
ready in the denominator. Example 342. 5a 3a2b = ? 6a5b3 Idenﬁty what factors we need to match denominators 2a3b2 3 · 2 = 6 and we need three more a′s and two more b′s 5a 3a2b 2a3b2 2a3b2 Multiply numerator and denominator by this 10a4b2 6a5b3 Our Solution Example 343. x 2 − x + 4 = ? x2 + 7x + 12 (x + 4)(x + 3) Factor to idenﬁty factors we need to match denominators (x + 3) The missing factor x2 + x 6 − (x + 4)(x + 3) Multiply numerator and denominator by missing factor, FOIL numerator Our Solution 254 As the above example illustrates, we will multiply out our numerators, but keep our denominators factored. The reason for this is to add and subtract fractions we will want to be able to combine like terms in the numerator, then when we reduce at the end we will want our denominators factored. Once we know how to ﬁnd the LCD and how to build up fractions to a desired denominator we can combine them together by ﬁnding a common denominator and building up those fractions. Example 344. Build up each fraction so they have a common denominator 5a 4b3c and 3c 6a2b First identify LCD 12a2b3c Determine what factors each fraction is missing First: 3a2 Second: 2b2c Multiply each fraction by missing factors 5a 4b3c 3a2 3a2 and 3c 6a2b 2b2c 2b2c 15a3 12a2b3c and 6b2c2 12a2b3c Our Solution Example 345. Build up each fraction so they have a common denominator 5x 5x 6 − 6)(x + 1) − x2 (x − and x 2 − x2 + 4x + 3 Factor to ﬁnd LCD (x + 1)(x + 3) Use factors to ﬁnd LCD LCD: (x 6)(x + 1)(x + 3) − First: (x + 3) Second: (x Identify which factors are missing 6) Multiply fractions by missing factors − 5x 6)(x + 1) x + 3 x + 3 (x − and x 2 − (x + 1)(x + 3) x x 6 6 − − Multiply numerators 5x2 + 15x 6)(x + 1)(x + 3) and (x (x − x2 8x + 12 6)(x + 1)(x + 3) − − Our Solution World View Note: When the Egyptians began working with fractions, they expressed all fractions as a sum of unit fraction. Rather than 4 5, they would write 4 + 1 2 + 1 the fraction as the sum, 1 20. An interesting problem with this system is 3 + 1 5 is also equal to the sum 1 this is not a unique solution, 4 6 + 1 10. 5 + 1 255 7.3 Practice - Least Common Denominator Build up denominators. 1) 3 8 = ? 48 3) a x = ? xy 5) 7) 2 3a3b2c = ? 9a5b2c4 2 x + 4 = ? x2 − 16 9) x 4 x + 2 = − ? x2 + 5x + 6 Find Least Common Denominators 11) 2a3, 6a4b2, 4a3b5 13) x2 3x, x 3, x − − 15) x + 2, x 4 − 25, x + 5 17) x2 − 2) a 5 = ? 5a 4) 6) 5 2x2 = ? 8x3y 4 3a5b2c4 = ? 9a5b2c4 8) x + 1 x − 10) x 3 = ? 6x + 9 x2 − 6 x + 3 = − ? 2x x2 − 15 − 12) 5x2y, 25x3y5z 14) 4x − 16) x, x 8, x 2, 4 − 7, x + 1 18) x2 6x + 9 − 9, x2 − − 7x + 10, x2 19) x2 + 3x + 2, x2 + 5x + 6 20) x2 − Find LCD and build up each fraction 2x − − 15, x2 + x 6 − 21) 3a 5b2 , 2 10a3b 23 25) x2 16 − 27) x + 1 x2 36 − , , 3x 8x + 16 x2 − 2x + 3 x2 + 12x + 36 29) 4x x − x2 3x x − 5 − x2 , 2 x, − x − 3 6 6x 5x + 1 3x − − , 10 x 4 − 5 3x + 1 x − − 12 3x 6x + 8 − , , 2x x2 + 4x + 3 x − x2 + x 2 − , 5 x2 + 3x 20 10 − x2 x2 x2 22) 24) 26) 28) 30) 256 7.4 Rational Expressions - Add & Subtract Objective: Add and subtract rational expressions with and without common denominators. Adding and subtracting rational expressions is identical to adding and subtracting with integers. Recall that when adding with a common denominator we add the numerators and keep the denominator. This is the same process used with rational expressions. Remember to reduce, if possible, your ﬁnal answer. Example 346. x − 4 − 2x − x2 + 8 x2 x2 x + 8 2x − − 2x + 4 2x − − Same denominator, add numerators, combine like terms Factor numerator and denominator 8 8 2(x + 2) (x + 2)(x 4) − 2 − 4 x Divide out (x + 2) Our Solution 257 Subtraction with common denominator follows the same pattern, though the subtraction can cause problems if we are not careful with it. To avoid sign errors we will ﬁrst distribute the subtraction through the numerator. Then we can treat it like an addition problem. This process is the same as “add the opposite” we saw when subtracting with negatives. Example 347. 6x 3x 12 6 − − − 12 6 6x 3x − − + − 15x 3x 6 − 6 − 15x + 6 3x 6 − 9x − 3x 6 − 6 − − 3(3x + 2) 3(x 2) − Add the opposite of the second fraction (distribute negative) Add numerators, combine like terms Factor numerator and denominator Divide out common factor of 3 − (3x + 2) x 2 − Our Solution World View Note: The Rhind papyrus of Egypt from 1650 BC gives some of the earliest known symbols for addition and subtraction, a pair of legs walking in the direction one reads for addition, and a pair of legs walking in the opposite direction for subtraction.. When we don’t have a common denominator we will have to ﬁnd the least common denominator (LCD) and build up each fraction so the denominators match. The following example shows this process with integers. Example 348 10 12 3 3 3 12 + The LCD is 12. Build up, multiply 6 by 2 and 4 by 3 Multiply Add numerators 258 13 12 Our Solution The same process is used with variables. Example 349. 7a 3a2b + 4b 6ab4 The LCD is 6a2b4. We will then build up each fraction 2b3 2b3 7a 3a2b + 4b 6ab4 a a Multiply ﬁrst fraction by 2b3 and second by a 4ab 6a2b4 Add numerators, no like terms to combine 14ab3 6a2b4 + 14ab3 + 4ab 6a2b4 2ab(7b3 + 2) 6a2b4 Factor numerator Reduce, dividing out factors 2, a, and b 7b3 + 2 3ab3 Our Solution The same process can be used for subtraction, we will simply add the ﬁrst step of adding the opposite. Example 350. 4 5a − 7b 4a2 Add the opposite 4 5a + − 7b 4a2 LCD is 20a2. Build up denominators 4a 4a 4 5a + − 7b 4a2 5 5 Multiply ﬁrst fraction by 4a, second by 5 16a 35b − 20a2 Our Solution If our denominators have more than one term in them we will need to factor ﬁrst to ﬁnd the LCD. Then we build up each denominator using the factors that are 259 missing on each fraction. Example 351. 3a 6 8a + 4 8 4(2a + 1) 8 + Factor denominators to ﬁnd LCD LCD is 8(2a + 1), build up each fraction 2 2 6 4(2a + 1) + 3a 8 2a + 1 2a + 1 Multiply ﬁrst fraction by 2, second by 2a + 1 12 8(2a + 1) + 6a2 + 3a 8(2a + 1) Add numerators 6a2 + 3a + 12 8(2a + 1) Our Solution With subtraction remember to add the opposite. Example 352 7x + 12 − x2 Add the opposite (distribute negative) 4 − + − x2 − 4)(x x 1 − 7x + 12 3) (x − − Factor denominators to ﬁnd LCD LCD is (x 4)(x − − 3), build up each fraction − + − x2 − x 1 − 7x + 12 Only ﬁrst fraction needs to be multiplied by x 3 − x2 (x 2x − 3)(x 3 4) − − + (x x − 3)(x − − 1 − 4) Add numerators, combine like terms − (x (x (x x2 3x − 3)(x 4 4) − − − 4)(x + 1) 3)(x 4) − − − x + 1 x 3 − Factor numerator Divide out x 4 factor − Our Solution 260 7.4 Practice - Add and Subtract Add or subtract the rational expressions. Simplify your answers 2 a + 3 whenever possible. a + 3 + 4 1 + 2t − 2x2 + 3 7 − 1 − t t 3) t2 + 4t 1) 5) x2 x2 − − 5x + 9 6x + 5 x2 6x + 5 − − 5 8r 7) 5 6r − 9t3 + 5 6t2 9) 8 11) a + 2 13) x 2 − 1 − 4x − 15) 5x + 3y a 4 − 4 2x + 3 x 3x + 4y xy2 2x2y − 2z 3z z + 1 1 − z − 8 4 − x2 − t 3 x + 2 5 12 4 3x + 3 t 4t − − 3 − 2 5x2 + 5x − x2 + 5x + 6 − x x2 + 15x + 56 − 18 17) 19) 21) 23) 25) 27) 29) 31) 33) 35) 6 − − x2 9 − 4 x2 + 2x 3 − 5x x x2 − 2x x2 x2 1 − − x + 1 2x 35 + x + 6 x2 + 7x + 10 37) 4 − a2 − 2z 39) − 2 a a 3 − − − a2 9 − 2z + 3z 2z + 1 − x2 + 3x + 2 + 3x − 2x − 3 1 41) 1 − x2 + 5x + 6 2 x2 + 3x + 2 7 x2 + 13x + 42 3 4z2 − 1 38) 40) 43) x2 2x + 7 2x − 3 − − 3x 2 − x2 + 6x + 5 44) 261 4 a2 + 5a 6 − 2) 4) 6) 3 8) x2 x 2 − − a2 + 3a 6x x 8 − 2 − 7 − x2 6 − a2 + 5a x + 4 xy2 + 3 8 + x 3a2 + 5a + 1 x2y − 12 9a − 1 3 10) x + 5 12) 2a 14) 2c x − 2 c + d cd2 x + 1 d − c2d − 1 + 2 2 5 + 3 25 + x x + 3 + 4 − 4x − 2 x2 4x x x + 5 (x + 3)2 4a 3a − x 6a 20 + 9a 5 + x − x − 5 30 26) x 16) 18) 20) 22) 24) 28) 30) 32) 34) − 2x − 2x − x2 x2 1 − 9 + 4x 2x 3 x2 + 5x + 4 5 x2 + x 6 − 3 5x + 6 x2 x2 − − − x − 1 3 − x2 + 3x + 2 + x + 5 3x + 6 + x x2 4 x2 + 4x + 3 − 4y − 2r y2 r2 2 y + 1 2 y − 1 − s2 + 1 r + s − 4x + 3 + 4x + 5 x2 + 4x − 1 r s − x + 2 8 x2 + 6x + 8 + 2x − x2 + 3x + 2 − 5 − 3 36) 3x + 2 42) x2 − 3x 7.5 Rational Expressions - Complex Fractions Objective: Simplify complex fractions by multiplying each term by the least common denominator. Complex fractions have fractions in either the numerator, or denominator, or usually both. These fractions can be simpliﬁed in one of two ways. This will be illustrated ﬁrst with integers, then we will consider how the process can be expanded to include expressions with variables. The ﬁrst method uses order of operations to simplify the numerator and denominator ﬁrst, then divide the two resulting fractions by multiplying by the reciprocal. Example 353 12 8 12 − 5 6 + 3 6 5 12 4 3 3 4 1 4 5 16 5 12 5 4 Get common denominator in top and bottom fractions Add and subtract fractions, reducing solutions To divide fractions we multiply by the reciprocal Reduce Multiply Our Solution The process above works just ﬁne to simplify, but between getting common denominators, taking reciprocals, and reducing, it can be a very involved process. Generally we prefer a diﬀerent method, to multiply the numerator and denominator of the large fraction (in eﬀect each term in the complex fraction) by the least common denominator (LCD). This will allow us to reduce and clear the small fractions. We will simplify the same problem using this second method. Example 354 LCD is 12, multiply each term 262 2(12) 1(12) 4 5(12) 3 − 6 + 1(12) 2 Reduce each fraction 2(4) 1(3) − 5(2) + 1(6) Multiply 8 3 − 10 + 6 5 16 Add and subtract Our Solution Clearly the second method is a much cleaner and faster method to arrive at our solution. It is the method we will use when simplifying with variables as well. We will ﬁrst ﬁnd the LCD of the small fractions, and multiply each term by this LCD so we can clear the small fractions and simplify. Example 355. 1 x2 1 x 1 1 − − Identify LCD (use highest exponent) LCD = x2 Multiply each term by LCD 1(x2) 1(x2) 1(x2) x2 1(x2) x − − 1(x2) 1(x2) x2 x2 1 x 1 x − − − − Reduce fractions (subtract exponents) Multiply Factor (x + 1)(x x(x − 1) − 1) Divide out (x 1) factor −
x + 1 x Our Solution The process is the same if the LCD is a binomial, we will need to distribute Multiply each term by LCD, (x + 4) 263 3(x + 4) 2(x + 4) x + 4 − 5(x + 4) + 2(x + 4) x + 4 Reduce fractions 2(x + 4) 3 5(x + 4) + 2 − Distribute 2x 3 8 5x + 20 + 2 − − Combine like terms 2x 5 − − 5x + 22 Our Solution The more fractions we have in our problem, the more we repeat the same process. Example 356. ab3 + 1 3 2 ab2 − 4 a2b + ab ab 1 ab − Idenﬁty LCD (highest exponents) LCD = a2b3 Multiply each term by LCD ab3 + 1(a2b3) 3(a2b3) 2(a2b3) ab2 − 4(a2b3) a2b + ab(a2b3) ab 1(a2b3) ab − Reduce each fraction (subtract exponents) 3a + ab2 2ab − 4b2 + a3b4 ab2 Our Solution − World View Note: Sophie Germain is one of the most famous women in mathematics, many primes, which are important to ﬁnding an LCD, carry her name. Germain primes are prime numbers where one more than double the prime 3 + 1 = 7 prime. The number is also prime, for example 3 is prime and so is 2 largest known Germain prime (at the time of printing) is 183027 1 which has 79911 digits! 2265440 − · · Some problems may require us to FOIL as we simplify. To avoid sign errors, if there is a binomial in the numerator, we will ﬁrst distribute the negative through the numerator. Example 357 − Distribute the subtraction to numerator Identify LCD 264 LCD = (x + 3)(x − 3) Multiply each term by LCD (x − 3)(x + 3)(x x + 3 − 3) (x − 3)(x + 3)(x x + 3 − x − 3)(x + 3)(x 3 + ( − + (x + 3)(x + 3)(x 3) − x 3) − 3) − x 3 − (x − (x 3)(x − 3)(x 3) + ( x 3)(x + 3) − 3) + (x + 3)(x + 3) − − − x2 x2 − − x2 6x + 9 6x − 6x + 9 + x2 + 6x − 9 9 − − 12x − 2x2 + 18 12x 2(x2 + 9) − − x2 6x 9 − Reduce fractions FOIL Combine like terms Factor out 2 in denominator Divide out common factor 2 Our Solution If there are negative exponents in an expression we will have to ﬁrst convert these negative exponents into fractions. Remember, the exponent is only on the factor it is attached to, not the whole term. Example 358. 1 2 + 2m− m− m + 4m− 2 1 m2 + 2 m m + 4 m2 1(m2) m2 + 2(m2) m(m2) + 4(m2) m2 m Make each negative exponent into a fraction Multiply each term by LCD, m2 Reduce the fractions 1 + 2m m3 + 4 Our Solution Once we convert each negative exponent into a fraction, the problem solves exactly like the other complex fraction problems. 265 7.5 Practice - Complex Fractions Solve. 1) 3) 5) 7) 9) 11) 13) 15) 17) 19) 21) 1 + 1 x 1 x2 − a2 − a2 + 1 1 a 2 5 − − 3 2a − 6 − − 2a 4 x + 2 10 x2 a2 b2 − 4a2b a + b 16ab2 1 − 1 + 11 10 3 x2 x − x + 18 x2 1 x 2x − 3x − 32 3x − − y2 − 1 + 1 y 25 b2 1 − 4 + 12 2x − 5 + 15 2x − − 10 b − 2a 3x − x 9x2 4 − 6 1 x2 x − x + 3 4 x2 1 1 − − 15 x2 − 4 x2 − 1 2 x − 5 x + 4 12 3x + 10 8 3x + 10 1 x − − − 18 ) 4) 6) 8) 10) 12) 14) 16) 18) 20) 22) 266 23) 25) 27) 29) x − x + 3 4 + 9 2x + 3 5 2x + b2 − 2 b2 + 7 5 3 a2 ab − ab + 3 a2 24) 26) 28) 30 y2 − 1 y2 − 1 2 x2 xy − xy + 2 3 x2 x + 1)2 + 1 (x 1)2 − Simplify each of the following fractional expressions. 31) x− x− 2 y− − 1 + y− 2 1 33) x− 3y x− 2 − − 3 xy− y− 2 2 x− 35) 1 + 9 6x− 9 − x2 − 32) 2 2y + xy− 2 2 x− x− y− − 2 3 4x− 4 1 + x− x− 2 − x− x− 3 + y− 1y− − 1 + y− 2 4 − 34) 36) 2 x− 267 7.6 Rational Expressions - Proportions Objective: Solve proportions using the cross product and use proportions to solve application problems When two fractions are equal, they are called a proportion. This deﬁnition can be generalized to two equal rational expressions. Proportions have an important property called the cross-product. Cross Product: If a b = c d then ad = bc The cross product tells us we can multiply diagonally to get an equation with no fractions that we can solve. Example 359. 20 6 = x 9 Calculate cross product 268 (20)(9) = 6x Multiply 180 = 6x Divide both sides by 6 6 30 = x Our Solution 6 World View Note: The ﬁrst clear deﬁnition of a proportion and the notation for a proportion came from the German Leibniz who wrote, “I write dy: x = dt: a; for dy is to x as dt is to a, is indeed the same as, dy divided by x is equal to dt divided by a. From this equation follow then all the rules of proportion.” If the proportion has more than one term in either numerator or denominator, we will have to distribute while calculating the cross product. Example 360. x + 3 4 = 2 5 Calculate cross product 5(x + 3) = (4)(2) Multiply and distribute Solve Subtract 15 from both sides 5x + 15 = 8 15 15 − − 5x = 7 Divide both sides by 5 − 5 5 7 5 Our Solution x = − This same idea can be seen when the variable appears in several parts of the proportion. Example 361. 4 x = 6 3x + 2 Calculate cross product 4(3x + 2) = 6x Distribute 12x + 8 = 6x Move variables to one side 12x Subtract 12x from both sides 12x 6x Divide both sides by − Our Solution 269 Example 362. 2x 3 − 7x + 4 2 5 3) = 2(7x + 4) Distribute = Calculate cross product 15 = 14x + 8 Move variables to one side Subtract 10x from both sides Subtract 8 from both sides 5(2x − 10x 10x − − − 10x 15 = 4x + 8 8 8 − − − 4 23 = 4x Divide both sides by 4 4 23 4 = x Our Solution − − As we solve proportions we may end up with a quadratic that we will have to solve. We can solve this quadratic in the same way we solved quadratics in the past, either factoring, completing the square or the quadratic formula. As with solving quadratics before, we will generally end up with two solutions. Example 363) = (8)(3) − Calculate cross product FOIL and multiply (k + 3)(k − k2 + k − k2 + k (k + 6)(k k + 6 = 0 or k 6 − k = − 6 − − 24 6 = 24 Make equation equal zero 24 − 30 = 0 5) = 0 5 = 0 Subtract 24 from both sides Factor Set each factor equal to zero Solve each equation − − − + 5 = 5 Add or subtract 6 or k = 5 Our Solutions Proportions are very useful in how they can be used in many diﬀerent types of applications. We can use them to compare diﬀerent quantities and make conclusions about how quantities are related. As we set up these problems it is important to remember to stay organized, if we are comparing dogs and cats, and the number of dogs is in the numerator of the ﬁrst fraction, then the numerator of the second fraction should also refer to the dogs. This consistency of the numerator and denominator is essential in setting up our proportions. Example 364. A six foot tall man casts a shadow that is 3.5 feet long. If the shadow of a ﬂag pole is 8 feet long, how tall is the ﬂag pole? shadow height We will put shadows in numerator, heights in denomintor 270 3.5 6 8 x The man has a shadow of 3.5 feet and a height of 6 feet The ﬂagpole has a shadow of 8 feet, but we don′t know the height 3.5 6 8 x 3.5x = (8)(6) Multiply = This gives us our proportion, calculate cross product 3.5x = 48 Divide both sides by 3.5 3.5 3.5 x = 13.7ft Our Solution Example 365. In a basketball game, the home team was down by 9 points at the end of the game. They only scored 6 points for every 7 points the visiting team scored. What was the ﬁnal score of the game? home visiter x − x 9 6 7 We will put home in numerator, visitor in denominator Don′t know visitor score, but home is 9 points less Home team scored 6 for every 7 the visitor scored This gives our proportion, calculate the cross product − − 7(x 7x 7x − x 9 6 7 = − x 9) = 6x Distribute 63 = 6x Move variables to one side Subtract 7x from both sides 7x x Divide both sides by 1 1 − − 63 = − 1 − − − 63 = x We used x for the visitor score. 9 = 54 Subtract 9 to get the home score 63 − 63 to 54 Our Solution 271 7.6 Practice - Proportions Solve each proportion. 1) 10 a = 6 8 3) 7 6 = 2 k 5) 6 x = 8 2 7) m 1 5 = 8 − 2 2) 7 9 = n 6 4) 8 x = 4 8 6) n − 10 8 = 9 3 8) 8 5 = 3 x − 8 9) 2 9 = 10 p − 4 10) 9 n + 2 = 3 9 11) b − 10 7 = b 4 13) x 5 = x + 2 9 15) 3 10 = a a + 2 17 19) 7 1 = 4 x x − 21) x + 5 5 = 6 x − − 12) 9 4 = r r − 4 14) n 8 = n − 3 4 16 18) n + 8 10 = n − 4 9 20 22 23) m + 3 4 = 11 m − 4 24 25 26) 5 n + 1 = n − 10 4 27 28 29 30 Answer each question. Round your answer to the nearest tenth. Round dollar amounts to the nearest cent. 31) The currency in Western Samoa is the Tala. The exchange rate is approximately S0.70 to 1 Tala. At this rate, how many dollars would you get if you exchanged 13.3 Tala? 32) If you can buy one plantain for S0.49 then how many can you buy with S7.84? 272 33) Kali reduced the size of a painting to a height of 1.3 in. What is the new width if it was originally 5.2 in. tall and 10 in. wide? 34) A model train has a scale of 1.2 in : 2.9 ft. If the model train is 5 in tall then how tall is the real train? 35) A bird bath that is 5.3 ft tall casts a shadow that is 25.4 ft long. Find the length of the shadow that a 8.2 ft adult elephant casts. 36) Victoria and Georgetown are 36.2 mi from each other. How far apart would the cities be on a map that has a scale of 0.9 in : 10.5 mi? 37) The Vikings led the Timberwolves by 19 points at the half. If the Vikings scored 3 points for every 2 points the Timberwolves scored, what was the half time score? 38) Sarah worked 10 more hours than Josh. If Sarah worked 7 hr for every 2 hr Josh worked, how long did they each work? 39) Computer Services Inc. charges S8 more for a repair than Low Cost Computer Repair. If the ratio of the costs is 3 : 6, what will it cost for the repair at Low Cost Computer Repair? 40) Kelsey’s commute is 15 minutes longer than Christina’s. If Christina drives 12 minutes for every 17 minutes Kelsey drives, how long is each commute? 273 7.7 Rational Expressions - Solving Rational Equations Objective: Solve rational equations by identifying and multiplying by the least common denominator. When solving equations that are made up of rational expressions we will solve them using the same strategy we used to solve linear equations with fractions. When we solved problems like the next example, we cleared the fraction by multiplying by the least common denominator (LCD) Example 366. 2 3 x 2(12) 3 x − 5(12) 6 5 6 = 3 4 3(12) 4 − = Multiply each term by LCD, 12 Reduce fractions 2(4)x − 5(2) = 3(3) Multiply 8x 10 = 9 Solve − + 10 + 10 Add 10 to both sides 8x = 19 Divide both sides by 8 8 x
= Our Solution 8 19 8 We will use the same process to solve rational equations, the only diﬀerence is our 274 LCD will be more involved. We will also have to be aware of domain issues. If our LCD equals zero, the solution is undeﬁned. We will always check our solutions in the LCD as we may have to remove a solution from our solution set. Example 367. 5x + 5 x + 2 + 3x = x2 x + 2 Multiply each term by LCD, (x + 2) (5x + 5)(x + 2) x + 2 + 3x(x + 2) = x2(x + 2) x + 2 Reduce fractions 5x + 5 + 3x(x + 2) = x2 Distribute 5x + 5 + 3x2 + 6x = x2 Combine like terms 3x2 + 11x + 5 = x2 Make equation equal zero Subtract x2 from both sides Factor Set each factor equal to zero Solve each equation − − x2 x2 2x2 + 11x + 5 = 0 (2x + 1)(x + 5) = 0 2x + 1 = 0 or x + 5 = 0 5 5 5 1 or x = − 1 − − 1 − − 2x = or − 5 + 2 = 1 2 or − 1 2 − + 2 = 3 2 5 Check solutions, LCD can′t be zero 3 Neither make LCD zero, both are solutions 5 Our Solution − − − The LCD can be several factors in these problems. As the LCD gets more complex, it is important to remember the process we are using to solve is still the same. Example 368x + 1)(x + 2) Multiply terms by LCD, (x + 1)(x + 2) x(x + 1)(x + 2) x + 2 + 1(x + 1)(x + 2) x + 1 = 5(x + 1)(x + 2) (x + 1)(x + 2) Reduce fractions 275 x(x + 1) + 1(x + 2) = 5 Distribute x2 + x + x + 2 = 5 Combine like terms x2 + 2x + 2 = 5 Make equatino equal zero Subtract 6 from both sides Factor Set each factor equal to zero Solve each equation 5 5 − − 3 = 0 − 1 x2 + 2x (x + 3)(x x + 3 = 0 or x 3 − x = 3 + 2)( ( − 3 or x = 1 Check solutions, LCD can′t be zero 3 in (x + 1)(x + 2), it works 2)( 1) = 2 Check − (1 + 1)(1 + 2) = (2)(3) = 6 Check 1 in (x + 1)(x + 2), it works 3 or 1 Our Solution x = − − In the previous example the denominators were factored for us. More often we will need to factor before ﬁnding the LCD Example 369x LCD = (x 11 3x + 2 2) 2) − 1)(x 1)(x − − x2 − − Factor denominator Identify LCD x(x − 1)(x − 2) x − 1 − 1(x − 1)(x − 2) x − 2 = 11(x − 1)(x − 2) (x − 1)(x − 2) Multiply each term by LCD, reduce x(x − x2 2) 1(x 1) = 11 Distribute − − 2x x2 − x + 1 = 11 Combine like terms − 3x + 1 = 11 Make equation equal zero − x2 11 − 3x 11 − 10 = 0 − − (x 5)(x + 2) = 0 5 = 0 or x + 2 = 0 2 2 Check answers, LCD can′t be 0 Subtract 11 from both sides Factor Set each factor equal to zero Solve each equation 2 x = 5 or 5 − 2 ( − − 1)( − 2 − 1)(5 2) = (4)(3) = 12 Check 5 in (x 4) = 12 Check − 2 in (x 3)( − 2) = ( − − − 1)(x − 1)(x − 2), it works 2), it works − − − 276 x = 5 or − 2 Our Solution World View Note: Maria Agnesi was the ﬁrst women to publish a math textbook in 1748, it took her over 10 years to write! This textbook covered everything from arithmetic thorugh diﬀerential equations and was over 1,000 pages! If we are subtracting a fraction in the problem, it may be easier to avoid a future sign error by ﬁrst distributing the negative through the numerator. Example 370 Distribute negative through numerator Identify LCD, 8(x − 3)(x + 2), multiply each term (x − 2)8(x x 3)(x + 2)8(x − x + 2 3)(x + 2) 8(x 5 · = 3)(x + 2) − 8 Reduce 8(x − 2)(x + 2) + 8( − x − 2)(x − 3) = 5(x − 3)(x + 2) 8(x2 − 8x2 x2 + x + 6) = 5(x2 4) + 8( 32 − − − 8x2 + 8x + 48 = 5x2 8x + 16 = 5x2 8x 16 0 = 5x2 − − − − − − − FOIL x 6) Distribute − 5x 30 Combine like terms − 5x 30 Make equation equal zero − Subtract 8x and 16 8x 16 − 13x Factor 46 − 0 = (5x 23)(x + 2) Set each factor equal to zero 23 = 0 or x + 2 = 0 Solve each equation 2 2 − − 2 5x = 23 or x = 5 5 − − 5x − + 23 + 23 x = 2 Check solutions, LCD can′t be 0 3 8 23 5 − 8( 23 5 3)( + ) = 8( − − 5)(0) = 0 Check Check 23 5 in 8(x − 3)(x + 2), it works 2 in 8(x 3)(x + 2), can′t be 0! − − Our Solution 23 5 33 5 or = − 2112 25 x = 23 5 In the previous example, one of the solutions we found made the LCD zero. When this happens we ignore this result and only use the results that make the rational expressions deﬁned. 277 7.7 Practice - Solving Rational Equations Solve the following equations for the given variable: 1) 3x − x 3) x + 20 − 5 = 5x x − 3 = 2x x − 2 4 − 3 3 4 = 4x + 5 6x 4 − − 3x 1 − 3m + 1 = 3 2 7 5 − x = 12 x 3 x − 7) 9) 2x 3x − 3m 2m − 11) 4 1 − − 7 13 − − 15 = 3x + 8 x2 4 − 17 − 2x + 1 + 2x + 1 1 3 2x = 1 − 8x2 − 4x2 1 − 19) 21) x 23) 25) 27 = 5x + 20 6x + 24 − x x 1 − − 2x x + 1 − 2 x + 1 = 4x2 x2 − 1 3 x + 5 = − 8x2 x2 + 6x + 5 29) x x 31 = − x2 − − 4x2 12x + 27 3 6 + x + 5 x + 3 = − − x2 2x2 3x − 33) 4x + 1 x + 3 + 5x x 3 1 = 8x2 x2 + 2x − − 2) x2 + 6 1 + x x x − 2 1 = 2x − − 6) x x − − 4 1 = 12 3 x + 1 − 8) 6x + 5 2x2 2x − 2 1 − x2 = 3x x2 − 1 4 − 5x 15 = = 3x 3x2 − 4x 2x 7 − 2 − 3 3 10) 12) 14) 16) x + 2 3x − 18) x x − − 20) 3x 5x 5 5 + 5x 7x − − 24) 26) 28 2x x − − 5x2 2 = − x2 + x 6 − 4 = x2 3x 2x − 30) x 3 x + 3 = − 2x2 x2 + 4x + 3 2 3 = x2 x2 + 3x − − 18 − 18 3 − 32 = 9x2 x2 − x 2 2 − − 2x x 3 3 = − − − x2 + 3x 3x2 18 − 34) 3x 1 − x + 6 − 278 2 = 1 x2 + x 1 − 6 − 22 2x + 1 = 2x2 1 3x − 2 − 7.8 Rational Expressions - Dimensional Analysis Objective: Use dimensional analysis to preform single unit, dual unit, square unit, and cubed unit conversions. One application of rational expressions deals with converting units. When we convert units of measure we can do so by multiplying several fractions together in a process known as dimensional analysis. The trick will be to decide what fractions to multiply. When multiplying, if we multiply by 1, the value of the expression does not change. One written as a fraction can look like many diﬀerent things as long as the numerator and denominator are identical in value. Notice the numerator and denominator are not identical in appearance, but rather identical in value. Below are several fractions, each equal to one where numerator and denominator are identical in value = 100cm 1m = 1lb 16oz = 1hr 60 min = 60 min 1hr The last few fractions that include units are called conversion factors. We can make a conversion factor out of any two measurements that represent the same distance. For example, 1 mile = 5280 feet. We could then make a conversion 1mi factor 5280ft because both values are the same, the fraction is still equal to one. Similarly we could make a conversion factor 5280ft 1mi . The trick for conversions will 279 be to use the correct fractions. The idea behind dimensional analysis is we will multiply by a fraction in such a way that the units we don’t want will divide out of the problem. We found out when multiplying rational expressions that if a variable appears in the numerator and denominator we can divide it out of the expression. It is the same with units. Consider the following conversion. Example 371. 17.37 miles to feet Write 17.37 miles as a fraction, put it over 1 17.37mi 1 ??ft ??mi 5280ft 1mi 5280ft 1 17.37mi 1 17.37mi 1 17.37 1 To divide out the miles we need miles in the denominator We are converting to feet, so this will go in the numerator Fill in the relationship described above, 1 mile = 5280 feet Divide out the miles and multiply across 91, 713.6ft Our Solution In the previous example, we had to use the conversion factor 5280ft so the miles 1mi would divide out. If we had used 1 mi 5280ft we would not have been able to divide out the miles. This is why when doing dimensional analysis it is very important to use units in the set-up of the problem, so we know how to correctly set up the conversion factor. Example 372. If 1 pound = 16 ounces, how many pounds 435 ounces? 435oz 1 435oz 1 ?? lbs ??oz 435oz 1 1 lbs 16oz Write 435 as a fraction, put it over 1 To divide out oz, put it in the denominator and lbs in numerator Fill in the given relationship, 1 pound = 16 ounces 280 435 1 1 lbs 16 = 435 lbs 16 Divide out oz, multiply across. Divide result 27.1875 lbs Our Solution The same process can be used to convert problems with several units in them. Consider the following example. Example 373. A student averaged 45 miles per hour on a trip. What was the student’s speed in feet per second? 45mi hr 45mi hr 5280ft 1mi 45mi hr 5280ft 1mi 1hr 3600 sec 45 1 5280ft 1 1 3600 sec ′′per′′ is the fraction bar, put hr in denominator To clear mi they must go in denominator and become ft To clear hr they must go in numerator and become sec Divide out mi and hr. Multiply across 237600ft 3600 sec Divide numbers 66 ft per sec Our Solution If the units are two-dimensional (such as square inches - in2) or three-dimensional (such as cubic feet - ft3) we will need to put the same exponent on the conversion factor. So if we are converting square inches (in2) to square ft (ft2), the conversion factor would be squared, the convesion factor. 2 1 ft 12in Example 374. . Similarly if the units are cubed, we will cube Convert 8 cubic feet to yd3 Write 8ft3 as fraction, put it over 1 8ft3 1 To clear ft, put them in denominator, yard in numerator 281 3 3 Because the units are cubed, we cube the conversion factor Evaluate exponent, cubing all numbers and units 8ft3 1 ??yd ??ft 1yd 3ft 8ft3 1 8ft3 1 1yd3 27ft3 Divide out ft3 8 1 1yd3 27 = 8yd3 27 Multiply across and divide 0.296296yd3 Our Solution When calculating area or volume, be sure to use the units and multiply them as well. Example 375. A room is 10 ft by 12 ft. How many square yards are in the room? A = lw = (10ft)(12 ft) = 120ft2 Multiply length by width, also multiply units 120ft2 1 Write area as a fraction, put it over 1 2 120ft2 1 ??yd ??ft 120ft2 1 1yd 3ft Put ft in denominator to clear, square conversion factor Evaluate exponent, squaring all numbers and units 2 120ft2 1 1yd2 9ft2 Divide out ft2 120 1 1yd2 9 = 120yd2 9 Multiply across and divide 13.33yd2 Our solution 282 To focus on the process of conversions, a conversion sheet has been included at the end of this lesson which includes several conversion factors for length, volume, mass and time in both English and Metric units. The process of dimensional analysis can be used to convert other types of units as well. If we can identify relationships that represent the same value we can make them into a conversion factor. Example 3
76. A child is perscribed a dosage of 12 mg of a certain drug and is allowed to reﬁll his prescription twice. If a there are 60 tablets in a prescription, and each tablet has 4 mg, how many doses are in the 3 prescriptions (original + 2 reﬁlls)? Convert 3 Rx to doses 1 Rx = 60 tab, 1 tab = 4 mg, 1 dose = 12mg Identify what the problem is asking Identify given conversion factors 3Rx 1 3Rx 1 60 tab 1Rx 3Rx 1 60 tab 1Rx 4mg 1 tab 3Rx 1 60 tab 1Rx 4mg 1 tab 1 dose 12mg 3 1 60 1 4 1 1 dose 12 Write 3Rx as fraction, put over 1 Convert Rx to tab, put Rx in denominator Convert tab to mg, put tab in denominator Convert mg to dose, put mg in denominator Divide out Rx, tab, and mg, multiply across 720 dose 12 Divide 60 doses Our Solution World View Note: Only three countries in the world still use the English system commercially: Liberia (Western Africa), Myanmar (between India and Vietnam), and the USA. 283 Conversion Factors Length English 12 in = 1 ft 1 yd = 3 ft 1 yd = 36 in 1 mi = 5280 ft Metric 1000 mm = 1 m 10 mm = 1 cm 100 cm = 1 m 10 dm = 1 m 1 dam = 10 m 1 hm = 100 m 1 km = 1000 m English/Metric 2.54 cm = 1 in 1 m = 3.28 ft 1.61 km = 1 mi Volume English 1 c = 8 oz 1 pt = 2 c 1 qt = 2 pt 1 gal = 4 qt Metric 1 mL = 1 cc = 1 cm3 1 L = 1000 mL 1 L = 100 cL 1 L = 10 dL 1000 L = 1 kL English/Metric 16.39 mL = 1 in3 1.06 qt = 1 L 3.79 L = 1gal Area English 1 ft2 = 144 in2 1 yd2 = 9 ft2 1 acre = 43,560 ft2 640 acres = 1 mi2 Metric 1 a = 100 m2 1 ha = 100 a English/Metric 1 ha = 2.47 acres English 1 lb = 16 oz 1 T = 2000 lb Weight (Mass) Metric 1 g = 1000 mg 1 g = 100 cg 1000 g = 1 kg 1000 kg = 1 t English/Metric 28.3 g = 1 oz 2.2 lb = 1 kg Time 60 sec = 1 min 60 min = 1 hr 3600 sec = 1 hr 24 hr = 1 day 284 7.8 Practice - Dimensional Analysis Use dimensional analysis to convert the following: 1) 7 mi. to yards 2) 234 oz. to tons 3) 11.2 mg to grams 4) 1.35 km to centimeters 5) 9,800,000 mm (milimeters) to miles 6) 4.5 ft2 to square yards 7) 435,000 m2 to sqaure kilometers 8) 8 km2 to square feet 9) 0.0065 km3 to cubic meters 10) 14.62 in3 to cubic centimeters 11) 5,500 cm3 to cubic yards 12) 3.5 mph (miles per hour) to feet per second 13) 185 yd. per min. to miles per hour 14) 153 ft/s (feet per second) to miles per hour 15) 248 mph to meters per second 16) 186,000 mph to kilometers per year 17) 7.50 T/yd2 (tons per square yard) to pounds per square inch 18) 16 ft/s2 to kilometers per hour squared 285 Use dimensional analysis to solve the following: 19) On a recent trip, Jan traveled 260 miles using 8 gallons of gas. How many miles per 1-gallon did she travel? How many yards per 1-ounce? 20) A chair lift at the Divide ski resort in Cold Springs, WY is 4806 feet long and takes 9 minutes. What is the average speed in miles per hour? How many feet per second does the lift travel? 21) A certain laser printer can print 12 pages per minute. Determine this printer’s output in pages per day, and reams per month. (1 ream = 5000 pages) 22) An average human heart beats 60 times per minute. If an average person lives to the age of 75, how many times does the average heart beat in a lifetime? 23) Blood sugar levels are measured in miligrams of gluclose per deciliter of blood volume. If a person’s blood sugar level measured 128 mg/dL, how much is this in grams per liter? 24) You are buying carpet to cover a room that measures 38 ft by 40 ft. The carpet cost S18 per square yard. How much will the carpet cost? 25) A car travels 14 miles in 15 minutes. How fast is it going in miles per hour? in meters per second? 26) A cargo container is 50 ft long, 10 ft wide, and 8 ft tall. Find its volume in cubic yards and cubic meters. 27) A local zoning ordinance says that a house’s “footprint” (area of its ground ﬂoor) cannot occupy more than 1 1 3 acre lot, what is the maximum allowed footprint for your house in square feet? in square inches? (1 acre = 43560 ft2) 4 of the lot it is built on. Suppose you own a 28) Computer memory is measured in units of bytes, where one byte is enough memory to store one character (a letter in the alphabet or a number). How many typical pages of text can be stored on a 700-megabyte compact disc? Assume that one typical page of text contains 2000 characters. (1 megabyte = 1,000,000 bytes) 29) In April 1996, the Department of the Interior released a “spike ﬂood” from the Glen Canyon Dam on the Colorado River. Its purpose was to restore the river and the habitants along its bank. The release from the dam lasted a week at a rate of 25,800 cubic feet of water per second. About how much water was released during the 1-week ﬂood? 30) The largest single rough diamond ever found, the Cullinan diamond, weighed 3106 carats; how much does the diamond weigh in miligrams? in pounds? (1 carat - 0.2 grams) 286 Chapter 8 : Radicals 8.1 Square Roots ..........................................................................................288 8.2 Higher Roots ..........................................................................................292 8.3 Adding Radicals .....................................................................................295 8.4 Multiply and Divide Radicals ................................................................298 8.5 Rationalize Denominators ......................................................................303 8.6 Rational Exponents ................................................................................310 8.7 Radicals of Mixed Index .........................................................................314 8.8 Complex Numbers ..................................................................................318 287 8.1 Radicals - Square Roots Objective: Simplify expressions with square roots. Square roots are the most common type of radical used. A square root “unsquares” a number. For example, because 52 = 25 we say the square root of 25 is 5. The square root of 25 is written as 25√ . World View Note: The radical sign, when ﬁrst used was an R with a line through the tail, similar to our perscription symbol today. The R came from the latin, “radix”, which can be translated as “source” or “foundation”. It wasn’t until the 1500s that our current symbol was ﬁrst used in Germany (but even then it was just a check mark with no bar over the numbers! The following example gives several square roots: Example 377. 1√ = 1 4√ = 2 9√ = 3 121√ 625√ = 11 = 25 = Undeﬁned 81√ − 81√ − The ﬁnal example, is currently undeﬁned as negatives have no square root. This is because if we square a positive or a negative, the answer will be positive. Thus we can only take square roots of positive numbers. In another lesson we will deﬁne a method we can use to work with and evaluate negative square roots, but for now we will simply say they are undeﬁned. 8√ on Not all numbers have a nice even square root. For example, if we found our calculator, the answer would be 2.828427124746190097603377448419... and even this number is a rounded approximation of the square root. To be as accurate as possible, we will never use the calculator to ﬁnd decimal approximations of square roots. Instead we will express roots in simplest radical form. We will do this using a property known as the product rule of radicals Product Rule of Square Roots: a b√ · = a√ b√ · We can use the product rule to simplify an expression such as it into two roots, by spliting 5√ , and simplifying the ﬁrst root, 6 5√ . The trick in this 36√ · 5√ 36 · 288 process is being able to translate a problem like . There are several ways this can be done. The most common and, with a bit of practice, the fastest method, is to ﬁnd perfect squares that divide evenly into the radicand, or number under the radical. This is shown in the next example. into · 5√ 36 180√ Example 378. 75√ 3√ 25 · 3√ · 5 3√ 25√ 75 is divisible by 25, a perfect square Split into factors Product rule, take the square root of 25 Our Solution If there is a coeﬃcient in front of the radical to begin with, the problem merely becomes a big multiplication problem. Example 379. 5 63√ 7√ 5 9 · 7√ · 3 7√ · 15 7√ 5 9√ 5 63 is divisible by 9, a perfect square Split into factors Product rule, take the square root of 9 Multiply coeﬃcients Our Solution As we simplify radicals using this method it is important to be sure our ﬁnal answer can be simpliﬁed no more. Example 380. 9√ 72√ 8√ 9 · 8√ · 3 8√ 2√ 3 4 · 2√ · 2 2√ 3 4√ 3 · 72 is divisible by 9, a perfect square Split into factors Product rule, take the square root of 9 But 8 is also divisible by a perfect square, 4 Split into factors Product rule, take the square root of 4 Multiply 289 6 2√ Our Solution. The previous example could have been done in fewer steps if we had noticed that 72 = 36 2, but often the time it takes to discover the larger perfect square is more than it would take to simplify in several steps. · Variables often are part of the radicand as well. When taking the square roots of x8√ = x4, because we variables, we can divide the exponent by 2. For example, divide the exponent of 8 by 2. This follows from the power of a power rule of expoennts, (x4)2 = x8. When squaring, we multiply the exponent by two, so when taking a square root we divide the exponent by 2. This is shown in the following example. Example 381. 5 9√ 2√ · · − · 5 9 p 18x4y6z10 − 2x4y6z10 5 − x4√ z10√ y6 p · · 3x2y3z5 2√ 5 p · 15x2y3z5 2√ − − 18 is divisible by 9, a perfect square Split into factors Product rule, simplify roots, divide exponents by 2 Multiply coeﬃcients Our Solution We can’t always evenly divide the exponent on a variable by 2. Sometimes we have a remainder. If there is a remainder, this means the remainder is left inside the radical, and the whole number part is how many are outside the radical. This is shown in the following example. Example 382. 4 p 20x5y9z6 5x5y9z6 z6√ y9 p · 2x2y4z3 5xy√ · · p x5√ 4√ 5√ · · 20 is divisible by 4, a perfect square Split into factors Simplify, divide exponents by 2, remainder is left inside Our Solution 290
8.1 Practice - Square Roots Simplify. 1) 245√ 3) 36√ 5) 12√ 7) 3 12√ 9) 6 128√ 11) 13) 8 392√ − 192n√ 15) √ 196v2 17) √ 252x2 19) 21) 23) √ 100k4 7 64x4√ 5 36m√ − − − 25) 45x2y2 p p 27) 29) 16x3y3 320x4y4 p 31) 6 80xy2 p 33) 5 245x2y3 35) 37) p − √ 2 180u3v 8 180x4y2z4 − 39) 2 p 80hj4k 41) p 4 − p 54mnp2 2) 125√ 4) 196√ 6) 72√ 8) 5 32√ 10) 7 128√ 12) 14) 7 63√ − 343b√ 16) √ 100n3 18) √ 200a3 20) 22) − − 4 175p4 p 2 128n√ 24) 8 112p2 p √ 72a3b4 26) 28) √ 512a4b2 30) √ 512m4n3 √ 32) 8 98mn 34) 2 72x2y2 72x3y4 36) p 5 − √ 38) 6 50a4bc2 p 40) 42) − − p 8 32xy2z3 32m2p4q p 291 8.2 Radicals - Higher Roots Objective: Simplify radicals with an index greater than two. While square roots are the most common type of radical we work with, we can take higher roots of numbers as well: cube roots, fourth roots, ﬁfth roots, etc. Following is a deﬁnition of radicals. am√ = b if bm = a The small letter m inside the radical is called the index. It tells us which root we are taking, or which power we are “un-doing”. For square roots the index is 2. As this is the most common root, the two is not usually written. World View Note: The word for root comes from the French mathematician Franciscus Vieta in the late 16th century. The following example includes several higher roots. Example 383. 3√ = 5 125 814√ = 3 325√ = 2 3√ 7√ 4 2 = = 64 − 128 − = undeﬁned − − 16 4√ − We must be careful of a few things as we work with higher roots. First its impor814√ = 3. This is tant not to forget to check the index on the root. because 92 = 81 and 34 = 81. Another thing to watch out for is negatives under roots. We can take an odd root of a negative number, because a negative number raised to an odd power is still negative. However, we cannot take an even root of a negative number, this we will say is undeﬁned. In a later section we will discuss how to work with roots of negative, but for now we will simply say they are undeﬁned. 81√ = 9 but We can simplify higher roots in much the same way we simpliﬁed square roots, using the product property of radicals. Product Property of Radicals: abm√ = am√ bm√ · Often we are not as familiar with higher powers as we are with squares. It is important to remember what index we are working with as we try and work our way to the solution. 292 Example 384. We are working with a cubed root, want third powers 543√ 23 = 8 Test 2, 23 = 8, 54 is not divisible by 8. 33 = 27 Test 3, 33 = 27, 54 is divisible by 27! 3√ 273√ · 3 Write as factors Product rule, take cubed root of 27 Our Solution 2 · 23√ 23√ 27 Just as with square roots, if we have a coeﬃcient, we multiply the new coeﬃcients together. Example 385. We are working with a fourth root, want fourth powers 484√ 3 24 = 16 Test 2, 24 = 16, 48 is divisible by 16! 4√ 16 Write as factors Product rule, take fourth root of 16 Multiply coeﬃcients Our Solution 3 4√ 3 3 · 34√ 34√ 34√ 16 · 2 3 6 · We can also take higher roots of variables. As we do, we will divide the exponent on the variable by the index. Any whole answer is how many of that varible will come out of the square root. Any remainder is how many are left behind inside the square root. This is shown in the following examples. Example 386. 5 x25y17z3 y2z3 5 x5y3 p Divide each exponent by 5, whole number outside, remainder inside Our Solution p In the previous example, for the x, we divided 25 remain inside. For the y, we divided 17 inside. For the z, when we divided 3 following example includes integers in our problem. 5 = 5 R 0, so x5 came out, no x’s 5 = 3 R 2, so y3 came out, and y2 remains 5 = 0R 3, all three or z3 remained inside. The Example 387. 3√ 2 3√ 8 2 2ab2 4ab2 40a4b8 5a4b8 5ab2 5ab2 · 3√ 3√ 2 · Looking for cubes that divide into 40. The number 8 works! Take cube root of 8, dividing exponents on variables by 3 Remainders are left in radical. Multiply coeﬃcients Our Solution 293 8.2 Practice - Higher Roots Simplify. 1) 3√ 625 3) 3√ 750 5) 3√ 875 2) 3√ 375 4) 3√ 250 6) 243√ 4 964√ 7) − 4√ 9) 6 112 11) 13) − 4√ 4√ 112 648a2 15) 5√ 224n3 5 17) 224p5 p − − − 3 p 3 p 3 19) 21) 23) 25) 27) 29) 7√ 3 896r 3√ 2 48v7 − 3√ 7 320n6 135x5y3 32x4y4 − − 256x4y6 p 3 31) 7 81x3y7 − p 3√ 33) 2 375u2v8 8 484√ 8) − 10) 3 484√ 4√ 12) 5 243 14) 4√ 64n3 16) 5√ 18) 6√ 96x3 − 256x6 20) 7√ 8 384b8 − 3√ 22) 4 250a6 24) 26) − 3√ 3√ 512n6 64u5v3 28) 3√ 1000a4b5 3 30) 189x3y6 32) 3 56x2y8 4 p − 3√ 34) 8 750xy p − 135xy3 35) 3√ 3 192ab2 3 36) 3 − 3 37) 6 39) 6 41) 7 p 4 p 4 p 54m8n3p7 − 648x5y7z2 128h6j8k8 38) 40) 42) 4 p 6 − 80m4p7q4 p 4√ 6 405a5b8c 4 6 324x7yz7 − − p 294 8.3 Radicals - Adding Radicals Objective: Add like radicals by ﬁrst simplifying each radical. Adding and subtracting radicals is very similar to adding and subtracting with variables. Consider the following example. Example 388. 5x + 3x − 2x Combine like terms 6x Our Solution 5 11√ + 3 11√ − 2 11√ 6 11√ Combine like terms Our Solution 11√ it was just like combining Notice that when we combined the terms with terms with x. When adding and subtracting with radicals we can combine like radicals just as like terms. We add and subtract the coeﬃcients in front of the 295 radical, and the radical stays the same. This is shown in the following example. Example 389. 7 65√ + 4 35√ − 8 9 65√ 35√ + 65√ 35√ 5 − Combine like radicals 7 Our Solution 65√ + 65√ and 4 35√ 8 35√ − We cannot simplify this expression any more as the radicals do not match. Often problems we solve have no like radicals, however, if we simplify the radicals ﬁrst we may ﬁnd we do in fact have like radicals. Example 390. 5 45√ + 6 18√ 5√ 2√ 3 5√ + 6 5 − · · 15 5√ + 18 2√ − 2√ 2 49 2 2 98√ + 20√ 5√ + 4 · · 7 2√ + 2 5√ · 14 2√ + 2 5√ 17 5√ + 4 2√ − Simplify radicals, ﬁnd perfect square factors Take roots where possible Multiply coeﬃcients Combine like terms Our Solution World View Note: The Arab writers of the 16th century used the symbol similar to the greater than symbol with a dot underneath for radicals. This exact process can be used to add and subtract radicals with higher indices Example 391. 4 27 3 12 543√ 2 · − 23√ − 23√ − 9 9 3√ 4 4 · 163√ + 5 9 3√ 8 + 5 2 · 23√ + 5 2 · 23√ + 5 18 23√ + 5 6 − − 93√ 93√ 93√ 93√ 93√ Simplify each radical, ﬁnding perfect cube factors Take roots where possible Multiply coeﬃcients Combine like terms 12 Our Solution 23√ 23√ 18 − 296 8.3 Practice - Adding Radicals Simiplify 1) 2 5√ + 2 5√ + 2 5√ 3 2√ + 3 5√ + 3 5√ 3) − 5) 2 6√ 2 6√ 6√ − − 7) 3 6√ + 3 5√ + 2 5√ − 2) 4) 6) 8) − − − − 2 3√ − 3 6√ 3 6√ 3 3√ − 2 6√ 3√ − − 3 3√ + 2 3√ 2 3√ − 2 3√ − 5√ + 2 3√ 9) 2 2√ 3 18√ 2√ − 10) 54√ − 3 6√ + 3 27√ − 5√ 12) 5√ − − − 14) 2 20√ + 2 20√ 2 54√ 3√ − 3 27√ + 2 3√ 12√ − 16) 18) 20) 22) − − − − 2 2√ − 3 18√ 3 8√ − 5√ − 2√ + 3 8√ + 3 6√ 8√ + 2 8√ + 2 8√ 3 6√ + 2 18√ − 24) 2 6√ 54√ 3 27√ 3√ − − − 3√ 26) 3 135 3√ 81 − − 3 44√ + 3 324 4√ 3√ 135 + 2 644√ 30) 2 64√ + 2 44√ + 3 64√ 4√ 2 243 964√ + 2 964√ 32) − 34) 2 484√ − 4√ 3 405 3 484√ − − 4√ 162 7√ 3 768 7√ + 2 384 + 3 57√ − 3 37√ − 7√ 2 256 7√ 2 256 3 27√ − − 7√ 640 − − 3 6√ 11) − 12√ + 3 3√ − 13) 3 2√ + 2 8√ 3 18√ − 15) 3 18√ 2√ 3 2√ − 3 6√ − 2 18√ − 3 6√ − 3 8√ − − 17) 19) − − 3√ + 3 6√ 20√ + 2 20√ 2 24√ 21) − 23) 3 24√ − 2 6√ + 2 6√ + 2 20√ − 3 27√ + 2 6√ + 2 8√ 2 163√ + 2 163√ + 2 23√ 25) − 4√ 27) 2 243 29) 3 24√ − − 2 24√ 4√ 243 − 4√ + 3 324 31) 4√ 324 − − 33) 2 24√ + 2 34√ + 3 644√ 3 44√ 34√ − 4√ 2 243 34√ − 28) − 3 65√ 35) − 5√ 37) 2 160 − − 645√ + 2 192 5√ 5√ 2 64 − 5√ 2 192 5√ 160 − 39) 6√ 256 2 46√ − − − 6√ 3 320 160 5√ − − 6√ 2 128 − 36) 38) − − 297 8.4 Radicals - Multiply and Divide Radicals Objective: Multiply and divide radicals using the product and quotient rules of radicals. Multiplying radicals is very simple if the index on all the radicals match. The prodcut rule of radicals which we have already been using can be generalized as follows: Product Rule of Radicals: a bm√ · c dm√ = ac bdm√ Another way of stating this rule is we are allowed to multiply the factors outside the radical and we are allowed to multiply the factors inside the radicals, as long as the index matches. This is shown in the following example. Example 392. − 5 14√ 4 6√ · 20 84√ 21√ · 2 21√ · 40 21√ − 20 4 20 − − − Multiply outside and inside the radical Simplify the radical, divisible by 4 Take the square root where possible Multiply coeﬃcients Our Solution The same process works with higher roots Example 393. 3√ 2 18 · 12 3√ 3√ 6 3√ 27 3 · 36 · 3√ 3√ 15 270 10 10 10 12 12 Multiply outside and inside the radical Simplify the radical, divisible by 27 Take cube root where possible Multiply coeﬃcients Our Solution When multiplying with radicals we can still use the distributive property or FOIL just as we could with variables. Example 394. 7 6√ (3 10√ 21 60√ 21 4 21 15√ − · 2 15√ − 42 15√ · − − 35 9 35 5 15√ ) Distribute, following rules for multiplying radicals 35 90√ Simplify each radical, ﬁnding perfect square factors 10√ Take square root where possible · 3 10√ Multiply coeﬃcients · 105 10√ Our Solution − Example 395. ( 5√ − 2 3√ )(4 10√ + 6 6√ ) FOIL, following rules for multiplying radicals 298 4 50√ + 6 30√ 8 30√ · 2√ 4 25 4 + 6 30√ 5 2√ + 6 30√ − − 20 2√ + 6 30√ · − 8 30√ 8 30√ − − 8 30√ 16 2√ − − − 12 9 12 12 18√ 2√ · 3 2√ · 36 2√ 2 30√ − − Simplify radicals, ﬁnd perfect square factors Take square root where possible Multiply coeﬃcients Combine like terms Our Solution World View Note: Clay tablets have been discovered revealing much about Babylonian mathematics dating back from 1800 to 1600 BC. In one of the tables there is an approximation of 2√ accurate to ﬁve decimal places (1.41421) Example 396. (2 5√ − 16 35√ 14 10√ 14 10√ 14 10√ − − 14 10√ − 16 35√ 16 35√ − − 16 35√ 3 6√ )(7 2√ 21 12√ 3√ · 2 3√ · 42 3√ − 21 4 21 8 7√ ) − 24 42√ 24 42√ 24 42√ 24 42√ − − − − − − FOIL, following rules for multiplying radicals Simplify radicals, ﬁnd perfect square factors Take square root where possible Multiply coeﬃcient Our Solution As we are multiplying we always look at our ﬁnal solution to check if all the radicals are simpliﬁed and all like radicals
or like terms have been combined. Division with radicals is very similar to multiplication, if we think about division as reducing fractions, we can reduce the coeﬃcients outside the radicals and reduce the values inside the radicals to get our ﬁnal solution. Quotient Rule of Radicals: a c bm√ dm√ = a c b d m r Example 397. 15 20 3√ 108 23√ Reduce 15 20 and 3√ 108 2√ by dividing by 5 and 2 respectively 3 543√ 4 3√ 3 2 27 4 · 3 · 23√ 3 4 9 23√ 4 Simplify radical, 54 is divisible by 27 Take the cube root of 27 Multiply coeﬃcients Our Solution There is one catch to dividing with radicals, it is considered bad practice to have a radical in the denominator we will rationalize it, or clear out any radicals in the denominator. in the denominator of our ﬁnal answer. If there is a radical 299 We do this by multiplying the numerator and denominator by the same thing. The problems we will consider here will all have a monomial in the denominator. The way we clear a monomial radical in the denominator is to focus on the index. The index tells us how many of each factor we will need to clear the radical. For example, if the index is 4, we will need 4 of each factor to clear the radical. This is shown in the following examples. Example 398. 6√ 5√ 6√ 5√ 5√ 5√ ! Index is 2, we need two ﬁves in denominator, need 1 more Multiply numerator and denominator by 5√ 30√ 5 Our Solution Example 399. 3 4√ 11 24√ 3 114√ 24√ 4√ 23 4√ ! 23 Index is 4, we need four twos in denominator, need 3 more Multiply numerator and denominator by 23 4√ 88 3 4√ 2 Our Solution Example 400. 4 7 23√ 253√ 4 7 23√ 3√ 52 4 7 23√ 3√ 52 53√ 53√ ! 4 4 103√ 5 7 · 103√ 35 The 25 can be written as 52. This will help us keep the numbers small Index is 3, we need three ﬁves in denominator, need 1 more Multiply numerator and denominator by 53√ Multiply out denominator Our Solution 300 The previous example could have been solved by multiplying numerator and 3√ . However, this would have made the numbers very large denominator by and we would have needed to reduce our soultion at the end. This is why re53√ was the better way to simwriting the radical as plify. 3√ and multiplying by just 252 52 We will also always want to reduce our fractions (inside and out of the radical) before we rationalize. Example 401. 6 14√ 12 22√ 7√ 2 11√ 7√ 2 11√ 11√ 11√ ! Reduce coeﬃcients and inside radical Index is 2, need two elevens, need 1 more Multiply numerator and denominator by 11√ 77√ 11 2 · 77√ 22 Multiply denominator Our Solution The same process can be used to rationalize fractions with variables. Example 402. Reduce coeﬃcients and inside radical 6x3y4z 10xy6z3 4 18 4 p8 p 9 4 4 4√ 3x2 5y2z3 Index is 4. We need four of everything to rationalize, three more ﬁves, two more y ′s and one more z. 9 4 4√ 3x2 5y2z3 4 p 4 9 ! 4 p 53y2z 4 p53y2z p 375x2y2z 5yz p4 · Multiply numerator and denominator by 4 53y2z p Multiply denominator 4 9 375x2y2z p20yz Our Solution 301 8.4 Practice - Multiply and Divide Radicals Multiply or Divide and Simplify. 1) 3 5√ 4 16√ · − 3) 12m√ 15m√ · 3√ 3√ 4x3 2x4 · 6√ ( 2√ + 2) 5) 7) 9) 5 15√ (3 3√ + 2) − 11) 5 10√ (5n + 2√ ) 13) (2 + 2 2√ )( 3 + 2√ ) 15) ( 5√ − − 5)(2 5√ 1) − 17) ( 2a√ + 2 3a√ )(3 2a√ + 5a√ ) 19) ( 5 4 3√ )( 3 − − 4 3√ ) − − 12√ 5 100√ 21) 23) 5√ 4 125√ 25) 10√ 6√ 27) 2 4√ 3 3√ 29) 5x2 3x3y3 4 31) p 2p2 p 3p√ 33) 3 103√ 5 273√ 35) 3√ 5 3√ 4 4 4√ 37) 5 5r4 4√ 8r2 5 10√ 2) 4) − 5r3√ · − 15√ · 5 10r2√ 3√ 6) 3 4a4 3√ 10a3 · 10√ ( 5√ + 2√ ) 8) 10) 5 15√ (3 3√ + 2) 12) 15√ ( 5√ 3 3v√ ) − 14) ( − 2 + 3√ )( − 5 + 2 3√ ) 16) (2 3√ + 5√ )(5 3√ + 2 4√ ) 18) ( − 2 2p√ + 5 5√ )( 5p√ + 5p√ ) 20) (5 2√ 1)( − − 2m√ + 5) 22) 24) 26) 15√ 2 4√ 12√ 3√ 2√ 3 5√ 28) 4 3√ 15√ 30) 5 4 3xy4 p 8n2√ 10n√ 32) 34) 153√ 643√ 36) 24√ 2 644√ 38) 4 64m4n2 4√ 302 8.5 Radicals - Rationalize Denominators Objective: Rationalize the denominators of radical expressions. It is considered bad practice to have a radical in the denominator of a fraction. When this happens we multiply the numerator and denominator by the same thing in order to clear the radical. In the lesson on dividing radicals we talked about how this was done with monomials. Here we will look at how this is done with binomials. If the binomial is in the numerator the process to rationalize the denominator is essentially the same as with monomials. The only diﬀerence is we will have to distribute in the numerator. Example 403. 9 3√ − 2 6√ ( 3√ 9) − 2 6√ 6√ 6√ ! Want to clear 6√ in denominator, multiply by 6√ We will distribute the 6√ through the numerator 303 18√ 9 6√ − 6 2 · 9 6√ 2√ 9 · − 12 3 2√ 2√ 9 6√ 3 6√ − 12 − 4 Simplify radicals in numerator, multiply out denominator Take square root where possible Reduce by dividing each term by 3 Our Solution It is important to remember that when reducing the fraction we cannot reduce with just the 3 and 12 or just the 9 and 12. When we have addition or subtraction in the numerator or denominator we must divide all terms by the same number. The problem can often be made easier if we ﬁrst simplify any radicals in the problem. 2 20x5√ 12x2√ − 18x√ Simplify radicals by ﬁnding perfect squares √ 2 4 3x2 4 · · √ 5x3 − 2x√ 9 · 2 · 2x2 5x√ − 3 2x√ 2x 3√ Simplify roots, divide exponents by 2. Multiply coeﬃcients 4x2 5x√ 2x 3√ − 3 2x√ Multiplying numerator and denominator by 2x√ (4x2 5x√ 2x 3√ ) − 3 2x√ 4x2 10x2√ 3 · − 2x 2x√ 2x√ ! 2x 6x√ Distribute through numerator Simplify roots in numerator, multiply coeﬃcients in denominator 4x3 10√ − 6x 2x 6x√ Reduce, dividing each term by 2x 304 2x2 10√ 3x − 6x√ Our Solution As we are rationalizing it will always be important to constantly check our problem to see if it can be simpliﬁed more. We ask ourselves, can the fraction be reduced? Can the radicals be simpliﬁed? These steps may happen several times on our way to the solution. 5 − If the binomial occurs in the denominator we will have to use a diﬀerent strategy 2 , if we were to multiply the denominator by to clear the radical. Consider 3√ 5 3√ . We have 3√ we would have to distribute it and we would end up with 3 not cleared the radical, only moved it to another part of the denominator. So our current method will not work. Instead we will use what is called a conjugate. A conjugate is made up of the same terms, with the opposite sign in the middle. 3√ + So for our example with 5. The advantage of a conjugate is when we multiply them together we have 5)( 3√ + 5), which is a sum and a diﬀerence. We know when we multiply ( 3√ 3√ and 5, with subtraction in the these we get a diﬀerence of squares. Squaring middle gives the product 3 22. Our answer when multiplying conjugates 25 = will no longer have a square root. This is exactly what we want. 5 in the denominator, the conjugate would be 3√ − − − − − Example 404. 2 3√ − 5 2 3√ − 3√ + 5 3√ + 5 ! 5 Multiply numerator and denominator by conjugate Distribute numerator, diﬀerence of squares in denominator 2 3√ + 10 25 3 − 2 3√ + 10 22 − − 5 3√ 11 − Simplify denoinator Reduce by dividing all terms by 2 − Our Solution In the previous example, we could have reduced by dividng by 2, giving the solution 3√ + 5 11 , both answers are correct. − Example 405. 15√ 5√ + 3√ Multiply by conjugate, 5√ 3√ − 305 15√ 5√ + 3√ 5√ 5√ 3√ 3√ ! − − 75√ 5 45√ 3 − − 3√ 25 · 5√ 9 · − 2 Distribute numerator, denominator is diﬀerence of squares Simplify radicals in numerator, subtract in denominator Take square roots where possible 5 3√ 3 5√ − 2 Our Solution Example 406. 2 3x√ 5x3√ 4 − 2 3x√ 5x3√ 4 − 4 + 5x3√ 4 + 5x3√ ! 8 3x√ + 2 15x4√ 5x3 16 − 8 3x√ + 2x2 15√ 5x3 16 − Multiply by conjugate, 4 + 5x3√ Distribute numerator, denominator is diﬀerence of squares Simplify radicals where possible Our Solution The same process can be used when there is a binomial in the numerator and denominator. We just need to remember to FOIL out the numerator. Example 407. 5√ 3√ 3 2 − − 5√ 3√ 2 + 3√ 2 + 3√ ! 3 2 − − Multiply by conjugate, 2 + 3√ FOIL in numerator, denominator is diﬀerence of squares 6 + 3 3√ 2 5√ 3 − 4 − 15√ − Simplify denominator 6 + 3 3√ 2 5√ 1 − − 15√ Divide each term by 1 6 + 3 3√ 2 5√ 15√ − − Our Solution 306 Example 408. 2 5√ 3 7√ − 5 6√ + 4 2√ Multiply by the conjugate, 5 6√ 4 2√ − 3 7√ 2 5√ − 5 6√ + 4 2√ 5 6√ 5 6√ 4 2√ 4 2√ ! − − FOIL numerator, denominator is diﬀerence of squares 10 30√ 10 30√ 10 30√ − − − 8 10√ − 6 25 − · 15 42√ + 12 14√ 16 2 · 8 10√ − 150 − 15 42√ + 12 14√ 32 8 10√ 15 42√ + 12 14√ − 118 Multiply in denominator Subtract in denominator Our Solution The same process is used when we have variables Example 409. 3x 2x√ + 4x3√ 3x√ 5x − Multiply by the conjugate, 5x + 3x√ 3x 2x√ + 4x3√ 3x√ 5x − 5x + 3x√ 5x + 3x√ ! FOIL in numerator, denominator is diﬀerence of squares 15x2 2x√ + 3x 6x2√ + 5x 4x3√ + 12x4√ 25x2 3x − Simplify radicals 15x2 2x√ + 3x2 6√ + 10x2 x√ + 2x2 3√ 25x2 3x − Divide each term by x 15x 2x√ + 3x 6√ + 10x x√ + 2x 3√ 25x 3 − Our Solution World View Note: During the 5th century BC in India, Aryabhata published a treatise on astronomy. His work included a method for ﬁnding the square root of numbers that have many digits. 307 8.5 Practice - Rationalize Denominators Simplify. 1) 4 + 2 3√ 9√ 3) 4 + 2 3√ 5 4√ 5) 2 5 5√ − 4 13√ 7) 2√ 3 3√ − 3√ 9) 5 3 5√ + 2√ 11) 2 5 + 2√ 13) 3 3 3√ 4 − 15) 4 3 + 5√ 17) 19) 21) 23) 4 4 2√ − 4 − 1 1 + 2√ 14√ 7√ 2 − 2√ − ab√ b√ a − a√ − 25) a + ab√ a√ + b√ 27) 2 + 6√ 2 + 3√ 29) a b√ − a + b√ 31) 33) 35) 3 2√ a a b√ 6 − − − 2 3√ b b a√ 5√ 2 − 3 + 5√ − 2) − 4 + 3√ 4 9√ 4) 2 3√ − 2 16√ 2 6) 5√ + 4 4 17√ 8) 5√ 2√ − 3 6√ 10) 5 3√ + 4 5√ 12) 14) 5 − 2√ 2 3√ 4 2√ − 2 16) 2 2 5√ + 2 3√ 18) 4 − 5√ 4 3√ 20) 3 + 3√ 3√ 1 − 22) 2 + 10√ 2√ + 5√ 24) 7√ 14√ − 14√ + 7√ 26) a + ab√ a√ + b√ 28) 2 5√ + 3√ 3√ 1 − 30) 32) a b − a√ + b√ ab − b a√ a b√ 34) 4 2√ + 3 3 2√ + 3√ 36) − 1 + 5√ 2 5√ + 5 2√ 308 37) 5 2√ + 3√ 5 + 5 2√ 38) 3√ + 2√ 2√ 2 3√ − 309 8.6 Radicals - Rational Exponents Objective: Convert between radical notation and exponential notation and simplify expressions with rational exponents using the properties of exponents. When we s
implify radicals with exponents, we divide the exponent by the index. Another way to write division is with a fraction bar. This idea is how we will deﬁne rational exponents. Deﬁnition of Rational Exponents: a n m = ( am√ )n The denominator of a rational exponent becomes the index on our radical, likewise the index on the radical becomes the denominator of the exponent. We can use this property to change any radical expression into an exponential expression. Example 410. ( x5√ )3 = x 3 5 1 ( a7√ )3 3 7 = a− 5 6 ( 3x6√ )5 = (3x) 1 3√ )2 = (xy)− ( xy 2 3 Index is denominator Negative exponents from reciprocals We can also change any rational exponent into a radical expression by using the denominator as the index. Example 411. 5 3 = ( a3√ )5 a (2mn) 2 7√ 7 = ( 2mn )2 4 x− 5 = 1 ( x5√ )4 2 (xy)− 9 = 1 ( xy 9√ )2 Index is denominator Negative exponent means reciprocals World View Note: Nicole Oresme, a Mathematician born in Normandy was the ﬁrst to use rational exponents. He used the notation 1 3• ever his notation went largely unnoticed. 9p to represent 9 3. How- 1 The ability to change between exponential expressions and radical expressions allows us to evaluate problems we had no means of evaluating before by changing to a radical. Example 412. 4 27− 3 Change to radical, denominator is index, negative means reciprocal 1 3√ )4 ( 27 Evaluate radical 310 1 (3)4 1 81 Evaluate exponent Our solution The largest advantage of being able to change a radical expression into an exponential expression is we are now allowed to use all our exponent properties to simplify. The following table reviews all of our exponent properties. Properties of Exponents aman = am+n (ab)m = ambm am an = am n − (am)n = amn a b m = am bm a0 = 1 a− m = 1 am 1 m = am a− a b m − = bm am When adding and subtracting with fractions we need to be sure to have a common denominator. When multiplying we only need to multiply the numerators together and denominators together. The following examples show several diﬀerent problems, using diﬀerent properties to simplify the rational exponents. Example 413 Need common denominator on a′s(6) and b′s(10) 4 5 1 2 a 6 b 10 a 6 b 10 Add exponents on a′s and b′s 5 7 a 6 b 10 Our Solution Example 414. Example 415 Multiply 3 4 by each exponent x 4 y 10 Our Solution x2y 2 3 2x 7 · 2 y0 x 1 2 y 5 6 In numerator, need common denominator to add exponents 311 4 x 2 y 4 6 x 2x 7 · 2 y0 1 2 y 5 6 Add exponents in numerator, in denominator, y0 = 1 5 2x 2 y 9 6 7 2 x Subtract exponents on x, reduce exponent on y 2x− 1y 3 2 Negative exponent moves down to denominator 3 2 2y x Our Solution Example 416. 1 25x 3 y 2 5 1 2 − 4 9x 5 y− 3 2  Using order of operations, simplify inside parenthesis ﬁrst Need common denominators before we can subtract exponents    1 2 − 4 10 15 10   Subtract exponents, be careful of the negative: 15 4 10 10 − 19 10 4 10 15 10 − + = = 5 25x 15 y 12 15 y− 9x   19 10 7 15 y 25x− 9 9 7 25x− 15 y 1 2 − ! 1 2 19 10 ! 1 2 9 1 7 25 2 x− 30 y 3x 5y 19 20 7 30 19 20 The negative exponent will ﬂip the fraction The exponent 1 2 goes on each factor 1 1 Evaluate 9 2 and 25 2 and move negative exponent Our Solution It is important to remember that as we simplify with rational exponents we are using the exact same properties we used when simplifying integer exponents. The only diﬀerence is we need to follow our rules for fractions as well. It may be worth reviewing your notes on exponent properties to be sure your comfortable with using the properties. 312 8.6 Practice - Rational Exponents Write each expression in radical form. 3 5 1) m 3) (7x) 3 2 3 4 2) (10r)− 4) (6b) − 4 3 Write each expression in exponential form. 5) 7) 1 ( 6x√ )3 1 ( n4√ )7 Evaluate. 2 3 9) 8 3 2 11) 4 6) v√ 8) 5a√ 1 4 10) 16 12) 100− 3 2 Simplify. Your answer should contain only positive exponents. 3 2 xy 1 2)− · 1 13) yx 1 3 1 2b 15) (a 17) a2b0 3a4 19) uv u · · 21) (x0y 1 3) 3 2)3 (v 3 2x0 23) a 7 4 b 3 4b−1 · 3b−1 25) − 5 4 3y − 1 3 y−1 2y · 7 4 27) m 3 2n−2 4 (mn 3)−1 ! 1 2)0 30) y0 3 4 y−1) 1 3 (x 29) (m2n 3 4 n 31) (x y)−−2 x 1 2 33) (uv2) − 1 4v2 v 32) 34) 313 1 v− 2 3 14) 4v 16) (x 5 · 3 y− 2)0 1 3 1 2 y 18) 2x 4 3 y 2x − 7 4 xy2)0 20) (x · 4v2 22) u− 5 · 2x−2y − 5 3 4 y 24) x − 5 1 3 26) ab 4a 2b · − 1 2b − 2 3 3 2 28) (y x − 1 2) 1 2 3 2 y xy · − 5 4 3 (u 2)− y0) − 4 3 (x − 2 3 x−2y y4 · 1 3 y −2 y 5 − 3 3 y3) (x 2 ! 8.7 Radicals - Radicals of Mixed Index Objective: Reduce the index on a radical and multiply or divide radicals of diﬀerent index. Knowing that a radical has the same properties as exponents (written as a ratio) allows us to manipulate radicals in new ways. One thing we are allowed to do is reduce, not just the radicand, but the index as well. This is shown in the following example. Example 417. 8 x6y2 Rewrite as rational exponent (x6y2) p 1 5 Multiply exponents 6 2 x 8 y 8 Reduce each fraction 3 1 4 All exponents have denominator of 4, this is our new index x 4 y x3y 4 Our Solution p What we have done is reduced our index by dividing the index and all the exponents by the same number (2 in the previous example). If we notice a common factor in the index and all the exponnets on every factor we can reduce by dividing by that common factor. This is shown in the next example Example 418. 24√ a6b9c15 8√ a2b3c5 Index and all exponents are divisible by 3 Our Solution We can use the same process when there are coeﬃcients in the problem. We will ﬁrst write the coeﬃcient as an exponential expression so we can divide the exponet by the common factor as well. Example 419. 9√ 9√ 8m6n3 23m6n3 3√ 2m2n Write 8 as 23 Index and all exponents are divisible by 3 Our Solution We can use a very similar idea to also multiply radicals where the index does not match. First we will consider an example using rational exponents, then identify the pattern we can use. 314 Example 420. 3√ ab2 4√ a2b Rewrite as rational exponents (ab2) 1 3(a2b) 1 4 Multiply exponents To have one radical need a common denominator, 12 4 8 6 3 12 Write under a single radical with common index, 12 a 12 b 12√ 12 b 12 a a4b8a6b3 12√ a10b11 Add exponents Our Solution To combine the radicals we need a common index (just like the common denominator). We will get a common index by multiplying each index and exponent by an integer that will allow us to build up to that desired index. This process is shown in the next example. Example 421. 4√ a2b3 6√ a2b 12√ a6b9a4b2 12√ a10b11 Common index is 12. Multiply ﬁrst index and exponents by 3, second by 2 Add exponents Our Solution Often after combining radicals of mixed index we will need to simplify the resulting radical. Example 422. 5 x3y4 3 x2y p 15 p p x9y12x10y5 x19y17 15 x4y2 p xy 15 Common index: 15. Multiply ﬁrst index and exponents by 3, second by 5 Add exponents Simplify by dividing exponents by index, remainder is left inside Our Solution p Just as with reducing the index, we will rewrite coeﬃcients as exponential expressions. This will also allow us to use exponent properties to simplify. Example 423. 3 4x2y 22x2y p 3 4 8xy3 23xy3 4 p p 12 p 12 p 28x8y429x3y9 217x11y13 12 25x11y 32x11y 2y p 2y 12 p Rewrite 4 as 22 and 8 as 23 Common index: 12. Multiply ﬁrst index and exponents by 4, second by 3 Add exponents (even on the 2) Simplify by dividing exponents by index, remainder is left inside Simplify 25 Our Solution p 315 If there is a binomial in the radical then we need to keep that binomial together through the entire problem. Example 424. 3 9x(y + z)2 3x(y + z) 3 32x(y + z)2 3x(y + z) p p 33x3(y + z)334x2(y + z)4 p 6 37x5(y + z)7 6 3x5(y + z) p 3(y + z) 6 p p Rewrite 9 as 32 Common index: 6. Multiply ﬁrst group by 3, second by 2 Add exponents, keep (y + z) as binomial Simplify, dividing exponent by index, remainder inside Our Solution p World View Note: Originally the radical was just a check mark with the rest of the radical expression in parenthesis. In 1637 Rene Descartes was the ﬁrst to put a line over the entire radical expression. The same process is used for dividing mixed index as with multiplying mixed index. The only diﬀerence is our ﬁnal answer cannot have a radical over the denominator. Example 425. 6 x4y3z2 8 p x7y2z p x16y12z8 x21y6z3 24 r Common index is 24. Multiply ﬁrst group by 4, second by 3 Subtract exponents 24 x− 5y6z5 Negative exponent moves to denominator p y6z5 x5 24 r 24 y6z5 x5 r 24 x19 x19 r ! Cannot have denominator in radical, need 12x′s, or 7 more Multiply numerator and denominator by 12√ x7 x19y6z5 x 24 p Our Solution 316 8.7 Practice - Radicals of Mixed Index Reduce the following radicals. 8 1) 16x4y6 12 p3) 64x4y6z8 6 p5) q 12 7) 16x2 9y4 x6y9 8 p9) x6y4z2 9 p11) 8x3y6 p Combine the following radicals. 4 2) 9x2y6 4 p4) q 15 6) 25x3 16x5 x9y12z6 10 p8) 64x8y4 4 p10) 25y2 16 p12) 81x8y12 p 53√ 6√ x√ 7y3√ 14) 16) 73√ 54√ 5√ y3√ 3z a2b3c 6 24) x2yz3 5 x2yz2 13) 15) 17) 3√ x√ x 2 − xy√ 5 19) x2y 21) p 4 xy2 3 x2y 23) 25) 5√ p 4√ p a2bc2 4√ a√ a3 27) 5√ b3√ b2 29) xy3 3 x2y 31) 4√ p 9ab3 3a4b√ p 3 33) 3xy2z 4 9x3yz2 35) p 27a5(b + 1) p 3 81a(b + 1)4 37) 3√ a2 p a4√ p 39) 41) 43) 45) 4 x2y3 3√ p xy ab3c√ a2b3c−1 5√ 4 (3x 5 p(3x − 1)3 1)3 p 3 − (2x + 1)2 5 p(2x + 1)2 p 18) 3x4√ y + 4√ 20) ab√ 5√ 2a2b2 22) 5√ a2b3 4√ a2b 26) 3√ p x2 6√ x5 p 28) 4√ a3 3√ a2 30) 5√ a3b ab√ 32) 2x3y3 3 4xy2 34) √ p a4b3c4 3√ p ab2c 8x (y + z)5 3 4x2(y + z)2 p 3√ x2 p x5√ 5√ 3√ a4b2 ab2 5 x3y4z9 p xy−2z p 3 (2 + 5x)2 4 p (2 + 5x) 3x)3 3x)2 p 4 (5 3 p(5 p − − 36) 38) 40) 42) 44) 46) 317 8.8 Radicals - Complex Numbers Objective: Add, subtract, multiply, rationalize, and simplify expressions using complex numbers. World View Note: When mathematics was ﬁrst used, the primary purpose was for counting. Thus they did not originally use negatives, zero, fractions or irrational numbers. However, the ancient Egyptians quickly developed the need for “a part” and so they made up a new type of number, the ratio or fraction. The Ancient Greeks did not believe in irrational numbers (people were killed for believing otherwise)
. The Mayans of Central America later made up the number zero when they found use for it as a placeholder. Ancient Chinese Mathematicians made up negative numbers when they found use for them. In mathematics, when the current number system does not provide the tools to solve the problems the culture is working with, we tend to make up new ways for dealing with the problem that can solve the problem. Throughout history this has been the case with the need for a number that is nothing (0), smaller than zero (negatives), between integers (fractions), and between fractions (irrational numbers). This is also the case for the square roots of negative numbers. To work with the square root of negative numbers mathematicians have deﬁned what are called imaginary and complex numbers. Deﬁnition of Imaginary Numbers: i2 = 1 (thus i = 1√ ) − − Examples of imaginary numbers include 3i, number is one that contains both a real and imaginary part, such as 2 + 5i. i and 3i 5√ . A complex 6i, − 3 5 With this deﬁnition, the square root of a negative number is no longer undeﬁned. We now are allowed to do basic operations with the square root of negatives. First we will consider exponents on imaginary numbers. We will do this by manipulating our deﬁnition of i2 = 1. If we multiply both sides of the deﬁnition − by i, the equation becomes i3 = i. Then if we multiply both sides of the equation again by i, the equation becomes i4 = 1) = 1, or simply i4 = 1. Multiplying again by i gives i5 = i. One more time gives i6 = i2 = 1. And if this pattern continues we see a cycle forming, the exponents on i change we cycle through simpliﬁed answers of i, i, 1. As there are 4 diﬀerent possible − answers in this cycle, if we divide the exponent by 4 and consider the remainder, we can simplify any exponent on i by learning just the following four values: i2 = 1, − − − − − − ( Cyclic Property of Powers of i i0 = 1 i = i i2 = i3 = − − 1 i 318 Example 426. Example 427. i35 Divide exponent by 4 8R3 Use remainder as exponent on i i3 Simplify i Our Solution − i124 Divide exponent by 4 31R0 Use remainder as exponent on i i0 Simplify 1 Our Solution When performing operations (add, subtract, multilpy, divide) we can handle i just like we handle any other variable. This means when adding and subtracting complex numbers we simply add or combine like terms. Example 428. (2 + 5i) + (4 − 6 − 7i) Combine like terms 2 + 4 and 5i 2i Our Solution 7i − It is important to notice what operation we are doing. Students often see the parenthesis and think that means FOIL. We only use FOIL to multiply. This problem is an addition problem so we simply add the terms, or combine like terms. For subtraction problems the idea is the same, we need to remember to ﬁrst distribute the negative onto all the terms in the parentheses. Example 429. (4 8i) − 4 − 8i − (3 − 5i) Distribute the negative − 3 + 5i Combine like terms 4 1 3i Our Solution − 3 and − − 8i + 5i Addition and subtraction can be combined into one problem. Example 430. (5i) − (3 + 8i) + ( 8i 3 5i − − 4 + 7i) Distribute the negative 4 + 7i Combine like terms 5i 7 + 4i Our Solution − − − − 8i + 7i and 3 4 − − Multiplying with complex numbers is the same as multiplying with variables with one exception, we will want to simplify our ﬁnal answer so there are no exponents on i. 319 Example 431. Example 432. (3i)(7i) Multilpy coeﬃcients and i′s 21i2 Simplify i2 = 21( − − 1) Multiply 21 Our Solution 1 − 5i(3i 15i2 1) 15 Simplify i2 = 7) Distribute − 35i − 35i Multiply − 35i Our Solution − − − 1 − 15( Example 433. (2 6 + 10i − − 12i − 6 + 10i 6 + 10i 4i)(3 + 5i) 20i2 12i FOIL Simplify i2 = 1 − − 1) Multiply − 20( 12i + 20 Combine like terms 6 + 20 and 10i 26 2i Our Solution − − 12i − − Example 434. (3i)(6i)(2 18i2(2 36i2 3i) Multiply ﬁrst two monomials − 3i) Distribute − 54i3 − i) Multiply 54( 36 + 54i Our Solution Simplify i2 = 1 and i3 = − − − i 36( − 1) − − Remember when squaring a binomial we either have to FOIL or use our shortcut to square the ﬁrst, twice the product and square the last. The next example uses the shortcut Example 435. (4 5i)2 Use perfect square shortcut − 42 = 16 Square the ﬁrst 2(4)( − (5i)2 = 25i2 = 25( − 16 − 5i) = 1) = 40i 9 − − − − − 40i Twice the product 25 25 Combine like terms 40i Our Solution Square the last, simplify i2 = 1 − 320 − Dividing with complex numbers also has one thing we need to be careful of. If i is 1√ , and it is in the denominator of a fraction, then we have a radical in the denominator! This means we will want to rationalize our denominator so there are no i’s. This is done the same way we rationalized denominators with square roots. Example 436. 7 + 3i 5i − 7 + 3i 5i − i i 7i + 3i2 5i2 − 7i + 3( 5( − 1) − 1) − Just a monomial in denominator, multiply by i Distribute i in numerator Simplify i2 = 1 − Multiply 7i 3 − 5 Our Solution The solution for these problems can be written several diﬀerent ways, for example i, The author has elected to use the solution as written, but it is − important to express your answer in the form your instructor prefers. 5 + 7 3 + 7i 5 or − 3 5 Example 437. 6i 2 − 4 + 8i Binomial in denominator, multiply by conjugate, 4 8i − 6i 2 − 4 + 8i 8i 8i 4 4 − − FOIL in numerator, denominator is a diﬀerence of squares 8 − 16i 16 24i + 48i2 64i2 − − Simplify i2 = 1 − 8 − 16i 16 24i + 48( 1) 64( − − − 1) − Multiply 8 − 16i 24i − 16 + 64 − 48 Combine like terms 8 48 and 16i − − − 24i and 16 + 64 40i − 40 − 80 − i 1 − 2 Reduce, divide each term by 40 Our Solution 321 Using i we can simplify radicals with negatives under the root. We will use the product rule and simplify the negative as a factor of negative one. This is shown in the following examples. Example 438. √ Example 439. 16√ − 16 1 · 4i Our Solution − Consider the negative as a factor of Take each root, square root of − 1 is i 1 − 24√ − 6 4 1 · · 2i 6√ √ − Find perfect square factors, including Square root of Our Solution 1 1 is i, square root of 4 is 2 − − When simplifying complex radicals it is important that we take the radical (as an i) before we combine radicals. − 1 out of the Example 440. Simplify the negatives, bringing i out of radicals 1, also multiply radicals 3√ − 6√ − (i 6√ )(i 3√ ) Multiply i by i is i2 = Simplify the radical Take square root of 9 Our Solution 18√ − 2√ 9 · 3 2√ − − − If there are fractions, we need to make sure to reduce each term by the same number. This is shown in the following example. Example 441. 15 − 200 Simplify the radical ﬁrst √ − − 20 √ √ 1 − · 15 − − − 20 3 200 − 100 2 · 10i 2√ 10i 2√ 2i 2√ − 4 Find perfect square factors, including Take square root of Put this back into the expression 1 and 100 − 1 − All the factors are divisible by 5 Our Solution By using i = we will be able to simplify and solve problems that we could not simplify and solve before. This will be explored in more detail in a later section. 1√ − 322 8.8 Practice - Complex Numbers Simplify. 8 + 4i) ( − 2) (3i) (7i) − 1) 3 − 3) (7i) 5) ( − 7) (3 − 9) (i) − 11) (6i)( 13) ( − (3 2i) − (3 + 7i) − 6i) − 3i) + ( − (2 + 3i) 7 8i) − 6 − 8i) − 5i)(8i) 4) 5 + ( 6 6i) − 8i) − (7i) (5 3i) − − i) + (1 − 5i) − 4i) + (8 − − 4i) 6) ( − 8) ( − 10) (5 4 − 12) (3i)( 14) (8i)( 8i) 4i) − − 7i)2 15) ( − 17) (6 + 5i)2 19) ( 21) ( − − 7 4i)( 8 + 6i) − − 4 + 5i)(2 7i) − 4 + 2i) 23) ( 8 6i)( − − − 25) (1 + 5i)(2 + i) 27) − 9 + 5i i 29) − 9i 10 6i − 31) − 6i 3 − 4i i 33) 10 − − i 35) 37) 39) 41) 4i 10 + i − 8 6i 7 − 7 − 7i 10 5i 6 − i − 16) ( − i)(7i)(4 3i) − 18) (8i)( − 20) (3i)( 2i)( − 3i)(4 2 8i) − 4i) − − 8(4 22) − 24) ( − 26) ( − 6i)(3 − 2 + i)(3 8i) 2( 2 6i) − 2i) − − (7i)(4i) − 5i) − − 3 + 2i 3i − 28) − 30) − 32) − 4 + 2i 3i 5 + 9i 9i 34) 10 5i 36) 38) 40) 42) 9i 5i 1 − 4 4 + 6i 9 8 − − 6i 8i 7i 6 − 323 43) 81√ − 45) 10√ − 2√ − 47 3 + 27√ − 6 49) 8 16√ − − 4 51) i73 53) i48 55) i62 57) i154 44) 45√ − 46) 12√ − 2√ − 48) − 4 8√ − − 4 − 50) 6 + 32√ − 4 52) i251 54) i68 56) i181 58) i51 324 Chapter 9 : Quadratics 9.1 Solving with Radicals .............................................................................326 9.2 Solving with Exponents ..........................................................................332 9.3 Complete the Square ..............................................................................337 9.4 Quadratic Formula .................................................................................343 9.5 Build Quadratics From Roots ................................................................348 9.6 Quadratic in Form .................................................................................352 9.7 Application: Rectangles .........................................................................357 9.8 Application: Teamwork ..........................................................................364 9.9 Simultaneous Products ...........................................................................370 9.10 Application: Revenue and Distance ......................................................373 9.11 Graphs of Quadratics ............................................................................380 325 9.1 Quadratics - Solving with Radicals Objective: Solve equations with radicals and check for extraneous solutions. Here we look at equations that have roots in the problem. As you might expect, to clear a root we can raise both sides to an exponent. So to clear a square root we can rise both sides to the second power. To clear a cubed root we can raise both sides to a third power. There is one catch to solving a problem with roots in it, sometimes we end up with solutions that do not actually work in the equation. This will only happen if the index on the root is even, and it will not happen all the time. So for these problems it will be required that we check our answer in the original problem. If a value does not work it is called an extraneous solution and not included in the ﬁnal solution. When solving a radical problem with an even index: check answers! Example 442. 7x + 2 √ √
( 7x + 2 )2 = 42 7x + 2 = 16 2 2 = 4 Even index! We will have to check answers Square both sides, simplify exponents Solve Subtract 2 from both sides − − 7x = 14 Divide both sides by 7 7 7 x = 2 Need to check answer in original problem 7(2) + 2 √ 14 + 2 = 4 Multiply = 4 Add p 16√ = 4 Square root 4 = 4 True! It works! x = 2 Our Solution Example 443. 3√ 3√ ( x − = x 1 − )3 = ( 1 = 1 x 4 Odd index, we don′t need to check answer − 4)3 Cube both sides, simplify exponents 64 Solve − − − 326 + 1 + 1 Add 1 to both sides 63 Our Solution x = − Example 444. 4√ 4√ ( 3x + 6 = 3x + 6 ) = ( 3x + 6 = 81 6 6 − 3 Even index! We will have to check answers − 3)4 Rise both sides to fourth power Solve Subtract 6 from both sides 3 − − 3x = 75 Divide both sides by 3 3 x = 25 Need to check answer in original problem = = 814√ = 3 = 3 Multiply 3 Add 3 Take root 3 False, extraneous solution − − − − No Solution Our Solution 4 p 3(25) + 6 4√ 75 + 6 If the radical is not alone on one side of the equation we will have to solve for the radical before we raise it to an exponent Example 445. √ x + 4x + 1 x √ 4x + 1 = 5 − = 5 Even index! We will have to check solutions x x Isolate radical by subtracting x from both sides Square both sides )2 = (5 − − − x)2 Evaluate exponents, recal (a order terms − − − − − − 12)(x 10x + x2 Re 10x + 25 Make equation equal zero 4x 1 14x + 24 2 Subtract 4x and 1 from both sides Factor Set each factor equal to zero Solve each equation √ ( 4x + 1 4x + 1 = 25 4x + 1 = x2 4x 1 0 = x2 − − 0 = (x 12 = 0 or x − x − + 12 + 12 b)2 = a2 − − 2ab + b2 x = 12 or x = 2 Need to check answers in original problem (12) + 4(12) + 1 = 5 Check x = 5 ﬁrst p 327 = 5 Add √ 12 + 48 + 1 12 + 49√ = 5 Take root 12 + 7 = 5 Add 19 = 5 False, extraneous root (2) + 4(2) + 1 2 + 8 + 1√ p = 5 Check x = 2 = 5 Add 2 + 9√ = 5 Take root 2 + 3 = 5 Add 5 = 5 True! It works x = 2 Our Solution The above example illustrates that as we solve we could end up with an x2 term or a quadratic. In this case we remember to set the equation to zero and solve by factoring. We will have to check both solutions if the index in the problem was even. Sometimes both values work, sometimes only one, and sometimes neither works. World View Note: The babylonians were the ﬁrst known culture to solve quadratics in radicals - as early as 2000 BC! If there is more than one square root in a problem we will clear the roots one at a time. This means we must ﬁrst isolate one of them before we square both sides. Example 446. √ 3x x√ = 0 Even index! We will have to check answers Isolate ﬁrst root by adding x√ to both sides Square both sides − − 8 + x√ + x√ = x√ 3x 8 8 √ − √ ( 3x − )2 = ( x√ )2 Evaluate exponents 3x Solve Subtract 3x from both sides 3x 8 = x 3x 2x Divide both sides by − Need to check answer in original 3(4) √ 12 p 8 − 8 − 4√ − − − 4√ = 0 Multiply 4√ = 0 Subtract 4√ = 0 Take roots 328 2 − Subtract 2 = 0 0 = 0 True! It works x = 4 Our Solution When there is more than one square root in the problem, after isolating one root and squaring both sides we may still have a root remaining in the problem. In this case we will again isolate the term with the second root and square both sides. When isolating, we will isolate the term with the square root. This means the square root can be multiplied by a number after isolating. Example 447. √ 2x + 1 x√ = 1 Even index! We will have to check answers − + x√ + x√ = x√ + 1 √ 2x + 1 Isolate ﬁrst root by adding x√ to both sides Square both sides √ ( 2x + 1 )2 = ( x√ + 1)2 Evaluate exponents, recall (a + b)2 = a2 + 2ab + b2 2x + 1 = x + 2 x√ + 1 x 1 x = 2 x√ − − − − x 1 Isolate the term with the root Subtract x and 1 from both sides Square both sides (x)2 = (2 x√ )2 Evaluate exponents − x2 x(x x = 0 or x x2 = 4x Make equation equal zero Subtract x from both sides 4x Factor Set each factor equal to zero Solve − − − + 4 + 4 Add 4 to both sides of second equation 4x − 4x = 0 4) = 0 4 = 0 x = 0 or x = 4 Need to check answers in original 2(0) + 1 − 1√ p 2(4) + 1 − 8 + 1√ 9√ p = 1 Check x = 0 ﬁrst (0) 0√ = 1 Take roots − Subtract 0 = 1 1 = 1 True! It works Check x = 4 (4) 4√ = 1 Add 4√ = 1 Take roots Subtract 2 = 1 1 = 1 True! It works 329 x = 0 or 4 Our Solution Example 448. √ 3x + 9 x + 4√ − + x + 4√ 3x + 9 √ − = + x + 4√ 1 1 Even index! We will have to check answers Isolate the ﬁrst root by adding x + 4√ Square both sides = x + 4√ √ ( 3x + 9 3x + 9 = x + 4 )2 = ( x + 4√ − 2 x + 4√ − 3x + 9 = x + 5 x x 5 5 − 2x + 4 = − − − 2 x + 4√ − 2 x + 4√ − 1)2 Evaluate exponents + 1 Combine like terms Isolate the term with radical Subtract x and 5 from both sides Square both sides − (2x + 4)2 = ( 2 x + 4√ 4x2 + 16x + 16 = 4(x + 4) Distribute 4x2 + 16x + 16 = 4x + 16 Make equation equal zero )2 Evaluate exponents − − − − − 4x 4x 16 16 4x2 + 12x = 0 4x(x + 3) = 0 4x = 0 or x + 3 = 0 4 3 3 Check solutions in original Subtract 4x and 16 from both sides Factor Set each factor equal to zero Solve 3 x = 0 or x = − 4 − − 3(0) + 9 − p 9√ p − 3 (0) + 4 = 4√ = 2 = 1 = − 3( 3√ − − p p 3) + 4 = 3) + 4 = 1 Check x = 0 ﬁrst 1 Take roots 1 1 Subtract False, extraneous solution 3 − 1 Check x = 1 Add 1 Take roots Subtract 1 1 True! It works 3 Our Solution 330 9.1 Practice - Solving with Radicals Solve. 1) √ 2x + 3 3) √ 6x ) 3 + x = 6x + 13 √ 7) √ 3 3x − 9) √ 4x + 5 − − 1 = 2x x + 4√ = 2 11) √ 2x + 4 13) √ 2x + 6 15) √ 6 2x − − − − x + 3√ = 1 x + 4√ = 1 √ 2x + 3 = 3 2) √ 5x + 1 4 = 0 − 4) x + 2√ x√ = 2 − 6) x 1 = 7 x√ − − 8) √ 2x + 2 = 3 + 2x √ 1 − 10) √ 3x + 4 12) √ 7x + 2 14) √ 4x 3 − 16) √ 2 3x − − − − − x + 2√ = 2 √ 3x + 6 = 6 √ 3x + 1 = 1 √ 3x + 7 = 3 331 9.2 Quadratics - Solving with Exponents Objective: Solve equations with exponents using the odd root property and the even root property. Another type of equation we can solve is one with exponents. As you might expect we can clear exponents by using roots. This is done with very few unexpected results when the exponent is odd. We solve these problems very straight forward using the odd root property Odd Root Property: if an = b, then a = bn√ when n is odd Example 449. 5√ = 325√ x5 x5 = 32 Use odd root property Simplify roots x = 2 Our Solution However, when the exponent is even we will have two results from taking an even root of both sides. One will be positive and one will be negative. This is because both 32 = 9 and ( 3)2 = 9. so when solving x2 = 9 we will have two solutions, one − positive and one negative: x = 3 and 3 − Even Root Property: if an = b, then a = bn√ when n is even ± Example 450. x4 = 16 Use even root property ( ) ± 332 4√ = x4 164√ ± x = ± Simplify roots 2 Our Solution World View Note: In 1545, French Mathematicain Gerolamo Cardano published his book The Great Art, or the Rules of Algebra which included the solution of an equation with a fourth power, but it was considered absurd by many to take a quantity to the fourth power because there are only three dimensions! Example 451. (2x + 4)2 = 36 Use even root property ( ) ± (2x + 4)2 = ± 2x + 4 = 2x + 4 = 6 or 2x + 4 = p 4 − 4 4 − 2x = 2 or 2x = 2 − 2 2 x = 1 or x = − 36√ Simplify roots Subtract 4 from both sides 6 To avoid sign errors we need two equations ± 6 One equation for + , one equation for − 4 − 10 Divide both sides by 2 − 2 5 Our Solutions − In the previous example we needed two equations to simplify because when we 6. If the roots did took the root, our solutions were two rational numbers, 6 and not simplify to rational numbers we can keep the in the equation. − ± Example 452. (6x − 6x p ) ± 45√ 3 5√ 9)2 = 45 Use even root property ( (6x − 9)2 Simplify roots = ± Use one equation because root did not simplify to rational 9 = − ± + 9 + 9 Add 9 to both sides 6x = 9 Divide both sides by 6 6 Simplify, divide each term by 3 3 5√ ± 6 3 5√ ± 2 Our Solution 333 When solving with exponents, it is important to ﬁrst isolate the part with the exponent before taking any roots. Example 453. (x + 4)3 6 = 119 − + 6 + 6 Isolate part with exponent (x + 4)3 = 125 Use odd root property (x + 4)3 = 125 Our Solution Simplify roots Solve Subtract 4 from both sides − − 3 p Example 454. (6x + 1)2 + 6 = 10 6 6 − − Isolate part with exponent Subtract 6 from both sides (6x + 1)2 = 4 Use even root property ( ) ± 4√ Simplify roots (6x + 1)2 = ± 6x + 1 = 6x + 1 = 2 or 6x + 1 = p 1 − 1 − 1 − 6x = 1 or 6x = 6 6 x = or x = 6 1 6 Solve each equation Subtract 1 from both sides 2 To avoid sign errors, we need two equations ± 2 − 1 − 3 Divide both sides by 6 − 6 1 2 Our Solution − When our exponents are a fraction we will need to ﬁrst convert the fractional exponent into a radical expression to solve. Recall that a . Once we have done this we can clear the exponent using either the even ( ) or odd root property. Then we can clear the radical by raising both sides to an exponent (remember to check answers if the index is even). n = ( an√ ) ± m m Example 455. 2 (4x + 1) 5√ ( 4x + 1 5 = 9 Rewrite as a radical expression )2 = 9 Clear exponent ﬁrst with even root property ( 5√ ( 4x + 1 )2 = 9√ ± q Simplify roots 334 ) ± = 3 Clear radical by raising both sides to 5th power 5√ 5√ ( 4x + 1 4x + 1 )5 = ( 4x + 1 = 4x + 1 = 243 or 4x + 1 = 1 1 1 ± 3)5 243 243 1 ± ± − − − − 4x = 242 or 4x = 4 4 − 4 121 2 x = Simplify exponents Solve, need 2 equations! Subtract 1 from both sides − 244 Divide both sides by 4 4 61 Our Solution , − Example 456. 3 − − )3 2 3 q − 4√ 4√ ( 3x 3x (3x 4√ ( 3x 2) 2 4 = 64 Rewrite as radical expression )3 = 64 Clear exponent ﬁrst with odd root property 4√ ( 3x 3√ = 64 Simplify roots 3x = 4 Even Index! Check answers. 2 − )4 = 44 Raise both sides to 4th power 2 2 = 256 − Solve − + 2 + 2 Add 2 to both sides 3x = 258 Divide both sides by 3 3 3 x = 86 Need to check answer in radical form of problem )3 = 64 Multiply )3 = 64 Subtract 4 ( 3(86) 4√ ( 258 2 − 2 − 4√ )3 = 64 Evaluate root ( 256 p 43 = 64 Evaluate exponent 64 = 64 True! It works x = 86 Our Solution With rational exponents it is very helpful to convert to radical form to be able to see if we need a because we us
ed the even root property, or to see if we need to check our answer because there was an even root in the problem. When checking we will usually want to check in the radical form as it will be easier to evaluate. ± 335 9.2 Practice - Solving with Exponents Solve. 1) x2 = 75 3) x2 + 5 = 13 5) 3x2 + 1 = 73 7) (x + 2)5 = 243 − 9) (2x + 5)3 6 = 21 − 11) (x 13) (2 2 1) 3 = 16 3 x) 2 = 27 − − 15) (2x − 2 3) 3 = 4 17) (x + 1 2 )− 2 3 = 4 19) (x − 5 1)− 2 = 32 21) (3x − 4 2) 5 = 16 23) (4x + 2) 5 = 3 8 − 2) x3 = 8 − 4) 4x3 6) (x − 2 = 106 − 4)2 = 49 8) (5x + 1)4 = 16 10) (2x + 1)2 + 3 = 21 12) (x − 3 1) 2 = 8 14) (2x + 3) 3 = 16 4 16) (x + 3)− 3 = 4 1 18) (x − 5 1)− 3 = 32 3 20) (x + 3) 2 = 8 − 22) (2x + 3) 2 = 27 3 24) (3 − 4 2x) 3 = 81 − 336 9.3 Quadratics - Complete the Square Objective: Solve quadratic equations by completing the square. When solving quadratic equations in the past we have used factoring to solve for our variable. This is exactly what is done in the next example. Example 457. x2 + 5x + 6 = 0 (x + 3)(x + 2) = 0 x + 3 = 0 or x + 2 = 0 2 2 Our Solutions Factor Set each factor equal to zero Solve each equation 2 − x = − x = − 3 or 3 3 − − − 2x However, the problem with factoring is all equations cannot be factored. Consider the following equation: x2 7 = 0. The equation cannot be factored, however − there are two solutions to this equation, 1 + 2 2√ and 1 2 2√ . To ﬁnd these two solutions we will use a method known as completing the square. When completing the square we will change the quadratic into a perfect square which can easily be solved with the square root property. The next example reviews the square root property. − − Example 458. p (x + 5)2 (x + 5)2 = 18 18√ 3 2 − ± Square root of both sides Simplify each radical Subtract 5 from both sides 3 2√ Our Solution 337 To complete the square, or make our problem into the form of the previous example, we will be searching for the third term in a trinomial. If a quadratic is of the form x2 + b x + c, and a perfect square, the third term, c, can be easily found by the formula 2 . This is shown in the following examples, where we b ﬁnd the number that completes the square and then factor the perfect square. 1 2 · Example 459. x2 + 8x + c c = 2 1 2 · b 2 and our b = 8 8 = 42 = 16 The third term to complete the square is 16 x2 + 8x + 16 Our equation as a perfect square, factor (x + 4)2 Our Solution 1 2 · Example 460. x2 7x + c c = 1 2 · b 2 and our = 49 4 49 4 2 x2 − 11x + 7 2 x − The third term to complete the square is Our equation as a perfect square, factor Our Solution Example 461. x2 + and our b = 8 = 25 36 The third term to complete the square is 338 49 4 25 36 x2 + 5 3 x + x + 5 6 25 36 2 Our equation as a perfect square, factor Our Solution The process in the previous examples, combined with the even root property, is used to solve quadratic equations by completing the square. The following ﬁve steps describe the process used to complete the square, along with an example to demonstrate each step. Problem 3x2 + 18x 1. Separate constant term from variables 2. Divide each term by a 3. Find value to complete the square: b 1 2 · 2 4. Add to both sides of equation 5. Factor Solve by even root property = 32 = 9 3x2 + 18x x2 + 18 3 3 3 x2 + 6x 2 x 6 1 2 · x2 + 6x = 2 + 9 + 9 x2 + 6x + 9 = 11 (x + 3)2 = 11 (x + 3)2 = ± 11√ ± 3 − 11 = − 11√ World View Note: The Chinese in 200 BC were the ﬁrst known culture group to use a method similar to completing the square, but their method was only used to calculate positive roots. The advantage of this method is it can be used to solve any quadratic equation. The following examples show how completing the square can give us rational solutions, irrational solutions, and even complex solutions. Example 462. 2x2 + 20x + 48 = 0 Separate constant term from varaibles 339 − 2x2 + 20x 2 2 48 = − − 48 Subtract 24 48 Divide by a or 2 2 x2 + 10x 24 Find number to complete the square: = − 2 1 2 · b 2 10 = 52 = 25 Add 25 to both sides of the equation 1 2 · x2 + 10x − = 24 + 25 + 25 x2 + 10x + 25 = 1 (x + 5)2 = 1 1x + 5)2 ± 5 p Factor Solve with even root property Simplify roots Subtract 5 from both sides − x = − 5 − 4 or 1 Evaluate 6 Our Solution ± − x = − Example 463. x2 3x − 2 = 0 Separate constant from variables − + 2 + 2 Add 2 to both sides 3x − = 2 No a, ﬁnd number to complete the square 2 = 9 4 Add 9 4 to both sides, 1 2 · b 2 x2 x2 − 3x + = = = 17 4 17 4 17 4 Need common denominator (4) on right Factor Solve using the even root property 3 2 x − s 2 = ± = ± 17 4 r 17√ 2 3 2 − x Simplify roots Add 3 2 to both sides, 340 + 3 2 + 3 2 we already have a common denominator x = 3 ± 17√ 2 Our Solution Example 464. x2 1 2 · 2 3 3x2 = 2x 2x 2x − 2x = 3 − − 3x2 − Separate the constant from the variables Subtract 2x from both sides 7 Divide each term by a or 3 − 3 7 3 − = 1 9 Add to both sides, Find the number to complete the square 21 = − 3 + 1 9 20 = − 9 get common denominator on right x2 − 2 3 x + = − 1 3 2 20 9 20 9 Factor Solve using the even root property 20 − 9 ± r Simplify roots + 2i 5√ 3 Add 1 3 to both sides, Already have common denominator x = 1 ± 2i 5√ 3 Our Solution As several of the examples have shown, when solving by completing the square we will often need to use fractions and be comfortable ﬁnding common denominators and adding fractions together. Once we get comfortable solving by completing the square and using the ﬁve steps, any quadratic equation can be easily solved. 341 9.3 Practice - Complete the Square Find the value that completes the square and then rewrite as a perfect square. 1) x2 3) m2 5) x2 7) y2 − − − − 30x + __ 36m + __ 15x + __ y + __ 2) a2 4) x2 6) r2 8) p2 − − − − 24a + __ 34x + __ 1 9 17p + __ r + __ − 14 = 4 6 − − − − − − − − − − − − − − − 29 16p 12n 24 = 3 = 6 1 = 0 52 = 0 57 = 4 − 10x 6x + 47 = 0 8v + 45 = 0 − 37 = 5 10k + 48 = 0 16x + 55 = 0 16n + 67 = 4 − 21 + 10n 10) n2 8n 12 = 0 12) b2 + 2b + 43 = 0 14) 3x2 16) 8a2 + 16a 18) p2 20) m2 8m 22) 6r2 + 12r 24) 6n2 − − 26) v2 = 14v + 36 28) a2 − 30) 5x2 = 32) 5n2 = − 34) x2 + 8x + 15 = 8 36) n2 + 4n = 12 Solve each equation by completing the square. 9) x2 − 11) v2 13) 6x2 + 12x + 63 = 0 15) 5k2 17) x2 + 10x 19) n2 21) 2x2 + 4x + 38 = 23) 8b2 + 16b 25) x2 = 27) n2 = 29) 3k2 + 9 = 6k 31) 2x2 + 63 = 8x 33) p2 − 35) 7n2 37) 13b2 + 15b + 44 = 39) 5x2 + 5x = 31 41) v2 + 5v + 28 = 0 43) 7x2 45) k2 − 47) 5x2 + 8x 49) m2 = 51) 8r2 + 10r = 53) 5n2 38) 40) 8n2 + 16n = 64 42) b2 + 7b 44) 4x2 + 4x + 25 = 0 46) a2 − 48) 2p2 50) n2 − 52) 3x2 54) 4b2 − 56) 10v2 8p = 55 n + 7 = 7n + 6n2 − 7k + 50 = 3 − 2x2 + 3x − 15 + 9m 3n + 6 + 4n2 5 + 7b2 + 3b 6x + 40 = 0 5a + 25 = 3 − 11x = 8n + 60 = − n = 10n + 15 26 + 10x p + 56 = − 5 = − 5x 33 = 0 40 = 8 56 = 10a 4x2 41 55 18 − − − − − − − − − − − − 55) − − − 8 3r2 + 12r + 49 = 6 6r2 − − 15b + 56 = 3b2 15v = 27 + 4v2 6v − − 342 9.4 Quadratics - Quadratic Formula Objective: Solve quadratic equations by using the quadratic formula. The general from of a quadratic is ax2 + bx + c = 0. We will now solve this formula for x by completing the square Example 465. 1 2 · b a c Separate constant from variables Subtract c from both sides ax2 + bc + c = 0 c − c Divide each term by b2 4a2 Add to both sides, b 2a = ax2 + bx a a b a x2 + 2 = Find the number that completes the square b2 4a2 − c a 4a 4a = b2 4a2 − 4ac 4a2 = b2 4ac − 4a2 Get common denominator on right x2 + b a x + b2 4a2 = b2 4a2 − 4ac 4a2 = b2 4ac − 4a2 Factor x + b 2a 2 = b2 4ac − 4a2 Solve using the even root property x + b 2a s x + b 2a 2 = b2 4ac − 4a2 ± r Simplify roots = ± 4ac √ b2 − 2a Subtract b 2a from both sides x = − b ± √ b2 2a − 4ac Our Solution This solution is a very important one to us. As we solved a general equation by completing the square, we can use this formula to solve any quadratic equation. Once we identify what a, b, and c are in the quadratic, we can substitute those 343 values into x = − known as the quadratic fromula − ± b b2 p2a 4ac and we will get our two solutions. This formula is Quadratic Formula: if ax2 + b x + c = 0 then x = − b ± b2 p2a 4ac − World View Note: Indian mathematician Brahmagupta gave the ﬁrst explicit formula for solving quadratics in 628. However, at that time mathematics was not done with variables and symbols, so the formula he gave was, “To the absolute number multiplied by four times the square, add the square of the middle term; the square root of the same, less the middle term, being divided by twice the to square 4ac + b2 2a as the solution to the equation ax2 + bx = c. translate value.” would This the is − b p We can use the quadratic formula to solve any quadratic, this is shown in the following examples. Example 466. x = − 3 ± x2 + 3x + 2 = 0 4(1)(2) 32 − p2(1√ 9 − 2 1√ 3 ± 2 3 ± 2 4 or − or − − a = 1, b = 3, c = 2, use quadratic formula Evaluate exponent and multiplication Evaluate subtraction under root Evaluate root Evaluate ± to get two answers Simplify fractions 2 Our Solution As we are solving using the quadratic formula, it is important to remember the equation must ﬁst be equal to zero. Example 467. 25x2 = 30x + 11 30x 11 11 − 11 = 0 11) − 30x − 4(25)( − 30x − 25x2 30)2 − − 2(25) − 30 ± x = − ( p First set equal to zero Subtract 30x and 11 from both sides a = 25, b = 30, c = − − 11, use quadratic formula Evaluate exponent and multiplication 344 x = √ 30 ± 900 + 1100 50 x = 30 2000√ 50 ± x = 30 20 5√ 50 ± x = 3 ± 2 5√ 5 Evaluate addition inside root Simplify root Reduce fraction by dividing each term by 10 Our Solution Example 468. 3x2 + 4x + 8 = 2x2 + 6x 2x2 x2 2x2 − − 5 − 6x + 5 − − − 2 6x + 5 2x + 13 = 0 2)2 − − p 2(1(1)(13) − 52 √ 4 2 48√ ± − 2 4i 3√ 2 2i 3 First set equation equal to zero Subtract 2x2 and 6x and add 5 a = 1, b = 2, c = 13, use quadratic formula − Evaluate exponent and multiplication Evaluate subtraction inside root Simplify root Reduce fraction by dividing each term by 2 Our Solution When we use the quadratic formula we don’t necessarily get two unique answers. We can end up with on
ly one solution if the square root simpliﬁes to zero. Example 469. 12 x = ± 4x2 ( − 12)2 − p 2(4) √ 12 x = ± 12x + 9 = 0 4(4)(9) a = 4, b = − 12, c = 9, use quadratic formula Evaluate exponents and multiplication 144 − Evaluate subtraction inside root 0√ Evaluate root − 144 8 12 x = x = ± 8 12 0 ± 8 12 x = 8 3 2 x = Evaluate ± Reduce fraction Our Solution 345 If a term is missing from the quadratic, we can still solve with the quadratic formula, we simply use zero for that term. The order is important, so if the term with x is missing, we have b = 0, if the constant term is missing, we have c = 0. Example 470. x = − 0 ± 3x2 + 7 = 0 02 4(3)(7) − p2(3) 84√ x = ± − 6 2i 21√ 6 i 21, b = 0(missing term), c = 7 Evaluate exponnets and multiplication, zeros not needed Simplify root Reduce, dividing by 2 Our Solution We have covered three diﬀerent methods to use to solve a quadratic: factoring, complete the square, and the quadratic formula. It is important to be familiar with all three as each has its advantage to solving quadratics. The following table walks through a suggested process to decide which method would be best to use for solving a problem. 1. If it can easily factor, solve by factoring 2. If a = 1 and b is even, complete the square 3. Otherwise, solve by the quadratic formula − − x2 5x + 6 = 0 (x 2)(x x = 2 or x = 3 x2 + 2x = 4 − 3) = 0 2 2 = 12 = 1 1 2 · 5√ 5√ x2 + 2x + 1 = 5 (x + 1) x2 − x = 3 x = 3 ± − ± 3x + 4 = 0 ( 3)2 − p 2(1) i 7√ 2 ± ± − 4(1)(4) The above table is mearly a suggestion for deciding how to solve a quadtratic. Remember completing the square and quadratic formula will always work to solve any quadratic. Factoring only woks if the equation can be factored. 346 9.4 Practice - Quadratic Formula Solve each equation with the quadratic formula. 1) 4a2 + 6 = 0 3) 2x2 8x − − 2 = 0 5) 2m2 3 = 0 − 7) 3r2 9) 4n2 11) v2 − − − 2r − 1 = 0 36 = 0 4v 5 = 8 − − 13) 2a2 + 3a + 14 = 6 2) 3k2 + 2 = 0 4) 6n2 1 = 0 − 6) 5p2 + 2p + 6 = 0 8) 2x2 2x 15 = 0 − − 10) 3b2 + 6 = 0 12) 2x2 + 4x + 12 = 8 14) 6n2 3n + 3 = 4 − − 16) 4x2 14 = 2 − − 18) 4n2 + 5n = 7 20) m2 + 4m 48 = 3 − − 3 = 8b 22) 3b2 − 24) 3r2 + 4 = 6r − 26) 6a2 = 5a + 13 − 28) 6v2 = 4 + 6v 30) x2 = 8 32) 2k2 + 6k 16 = 2k − 34) 12x2 + x + 7 = 5x2 + 5x 4 = 7 16 = 2 − 16 = 4 15) 3k2 + 3k 17) 7x2 + 3x − − 19) 2p2 + 6p − 21) 3n2 + 3n = 3 − 7x + 49 23) 2x2 = − 25) 5x2 = 7x + 7 27) 8n2 = 3n − 29) 2x2 + 5x = 8 3 − − 31) 4a2 64 = 0 − 33) 4p2 + 5p 36 = 3p2 − 3n 35) − 37) 7r2 39) 2n2 5n2 − 52 = 2 7n2 − − 36) 7m2 6m + 6 = m − − 12 = 3r − 9 = 4 − − 38) 3x2 3 = x2 − 40) 6b2 = b2 + 7 b − 347 9.5 Quadratics - Build Quadratics From Roots Objective: Find a quadratic equation that has given roots using reverse factoring and reverse completing the square. Up to this point we have found the solutions to quadratics by a method such as factoring or completing the square. Here we will take our solutions and work backwards to ﬁnd what quadratic goes with the solutions. We will start with rational solutions. If we have rational solutions we can use factoring in reverse, we will set each solution equal to x and then make the equation equal to zero by adding or subtracting. Once we have done this our expressions will become the factors of the quadratic. Example 471. The solutions are 4 and x = 4 or x = 4 4 + 2 + 2 − − Set each solution equal to x 2 2 Make each equation equal zero Subtract 4 from ﬁrst, add 2 to second x − − 4 = 0 or x + 2 = 0 These expressions are the factors (x 4)(x + 2) = 0 FOIL − − x2 + 2x x2 − 2x − − − 4x 8 Combine like terms 8 = 0 Our Solution If one or both of the solutions are fractions we will clear the fractions by multiplying by the denominators. Example 472. The solution are and 2 3 or Set each solution equal to x Clear fractions by multiplying by denominators 3x 2 − 3x = 2 or 4x = 3 Make each equation equal zero 2 3 3 = 0 These expressions are the factors 3) = 0 FOIL 8x + 6 = 0 Combine like terms − 2 = 0 or 4x 2)(4x (3x − 12x2 9x − − − − − 3 − − Subtract 2 from the ﬁrst, subtract 3 from the second 348 12x2 − 17x + 6 = 0 Our Solution If the solutions have radicals (or complex numbers) then we cannot use reverse factoring. In these cases we will use reverse completing the square. When there are radicals the solutions will always come in pairs, one with a plus, one with a . We will then set this minus, that can be combined into “one” solution using solution equal to x and square both sides. This will clear the radical from our problem. ± Example 473. The solutions are 3√ and x = 3√ Write as ′′one′′ expression equal to x 3√ − Square both sides ± x2 = 3 Make equal to zero 3 3 = 0 Our Solution − 3 Subtract 3 from both sides x2 − − We may have to isolate the term with the square root (with plus or minus) by adding or subtracting. With these problems, remember to square a binomial we use the formula (a + b)2 = a2 + 2ab + b2 Example 474. The solutions are 2 − 5 2√ and 2 + 5 2√ Write as ′′one′′ expression equal to x Isolate the square root term Subtract 2 from both sides Square both sides ± x = 2 2 2 − − x 2 = − 4x + 4 = 25 ± 5 2√ 5 2√ 2 · x2 − x2 − 4x + 4 = 50 Make equal to zero 50 50 Subtract 50 46 = 0 Our Solution x2 − − 4x − − World View Note: Before the quadratic formula, before completing the square, before factoring, quadratics were solved geometrically by the Greeks as early as 300 BC! In 1079 Omar Khayyam, a Persian mathematician solved cubic equations geometrically! If the solution is a fraction we will clear it just as before by multiplying by the denominator. 349 Example 475. The solutions are 2 + 3√ 4 2 2 and x = 3√ − 4 3√ ± 4 3√ Write as ′′one′′ expresion equal to x Clear fraction by multiplying by 4 ± 4x = 2 2 2 − 2 = Isolate the square root term Subtract 2 from both sides Square both sides − 16x + 4 = 3 Make equal to zero 3√ ± − 4x 3 3 Subtract 3 − − 16x + 1 = 0 Our Solution 16x2 16x2 − − The process used for complex solutions is identical to the process used for radicals. Example 476. The solutions are 4 5i and 4 + 5i Write as ′′one′′ expression equal to x − ± 5i − x x = 4 4 4 − 5i 4 = 8x + 16 = 25i2 8x + 16 = − ± − x2 x2 − − Isolate the i term Subtract 4 from both sides Square both sides i2 = 1 25 Make equal to zero − + 25 + 25 Add 25 to both sides 8x + 41 = 0 Our Solution x2 − Example 477. The solutions are 5i 3 − 2 and x = 3 + 5i 2 3 5i ± 2 Write as ′′one′′ expression equal to x Clear fraction by multiplying by denominator ± 5i 2x = 3 3 3 − − 2x 5i 3 = 12x + 9 = 5i2 ± − Isolate the i term Subtract 3 from both sides Square both sides i2 = 1 25 Make equal to zero − 4x2 4x2 − − 12x + 9 = − + 25 + 25 Add 25 to both sides 12x + 34 = 0 Our Solution 4x2 − 350 9.5 Practice - Build Quadratics from Roots From each problem, ﬁnd a quadratic equation with those numbers as its solutions. 1) 2, 5 3) 20, 2 5) 4, 4 7) 0, 0 9) − 11) 3 4 4, 11 , 1 4 13) 1 2 , 1 3 15) 3 7 , 4 17) 19) 21) 23) 25) 27) 29) 1 3 , 5 6 6, 1 9 5 1 5 11√ 3√ 4 i 13√ − − ± ± ± ± ± 31) 2 33) 1 35) 6 6√ 3i i 3√ ± ± ± 37) − 6√ 1 ± 2 39) 6 ± i 2√ 8 2) 3, 6 4) 13, 1 6) 0, 9 8) − 10) 3, 2, 5 − 1 − , 5 7 12) 5 8 14) 1 2 , 2 3 16) 2, 2 9 18) 5 3 , 1 2 − 20) 22) 24) 26) 28) 30) 32) 34√ 2 3√ 11i 5i 2√ 2√ 4i i 5√ 3 2 9 ± ± ± 36) − 38) 2 ± 3 5i 40) − 2 i 15√ ± 2 351 9.6 Quadratics - Quadratic in Form Objective: Solve equations that are quadratic in form by substitution to create a quadratic equation. We have seen three diﬀerent ways to solve quadratics: factoring, completing the square, and the quadratic formula. A quadratic is any equation of the form 0 = a x2 + bx + c, however, we can use the skills learned to solve quadratics to solve problems with higher (or sometimes lower) powers if the equation is in what is called quadratic form. Quadratic Form: 0 = axm + bxn + c where m = 2n An equation is in quadratic form if one of the exponents on a variable is double the exponent on the same variable somewhere else in the equation. If this is the case we can create a new variable, set it equal to the variable with smallest exponent. When we substitute this into the equation we will have a quadratic equation we can solve. World View Note: Arab mathematicians around the year 1000 were the ﬁrst to use this method! Example 478. − x4 13x2 + 36 = 0 Quadratic form, one exponent, 4, double the other, 2 y = x2 New variable equal to the variable with smaller exponent y2 = x4 y2 13y + 36 = 0 − 9)(y (y 4) = 0 − − 9 = 0 or or y = 4 9 = x2 or 4 = x2 4√ = x2√ 3, Square both sides Substitute y for x2 and y2 for x4 Solve. We can solve this equation by factoring Set each factor equal to zero Solve each equation Solutions for y, need x. We will use y = x2 equation Substitute values for y Solve using the even root property, simplify roots 2 Our Solutions ± √ = x2√ or ± When we have higher powers of our variable, we could end up with many more solutions. The previous equation had four unique solutions. 352 Example 479. 1 − 2 a− 1 a− 6 = 0 Quadratic form, one exponent, 2, is double the other, 1 Make a new variable equal to the variable with lowest exponent − b2 = a− b2 b 6 = 0 − (b 3)(b + 2) = 0 3 = 0 or Raise both sides to 2 = a− 1 = a 2)− 1 1 3 2 Square both sides 2 and b for a− Substitute b2 for a− Solve. We will solve by factoring Set each factor equal to zero Solve each equation Solutions for b, still need a, substitute into b = a− 1 power Simplify negative exponents 2 b = 3 or b = 1 or 3 = a− 1 = a or ( Our Solution − + − Just as with regular quadratics, these problems will not always have rational solutions. We also can have irrational or complex solutions to our equations. Example 480. 2x4 + x2 = 6 Make equation equal to zero 6 − 2x4 + x2 Subtract 6 from both sides 6 − 6 = 0 Quadratic form, one exponent, 4, double the other, 2 − y = x2 New variable equal variable with smallest exponent y2 = x4 6 = 0 − (2y 3)(y + 2) = 0 3 = 0 or y + 2 = 0 2 2 Square both sides Solve. We will factor this equation Set each factor equal to zero Solve each equation 2y2 + y − − 2y − + 3 + 3 2 2y = 3 or = x2 or 3 2 − 2 = x2 − 2√ = x2√ ± −
6√ 2 x = ± 3 2 ± r = x2√ or i 2√ , ± Our Solution 353 or y = 2 We have y, still need x. Substitute into y = x2 Square root of each side Simplify each root, rationalize denominator When we create a new variable for our substitution, it won’t always be equal to just another variable. We can make our substitution variable equal to an expression as shown in the next example. Example 481. 3(x 7)2 − − 2(x − 7) + 5 = 0 Quadratic form y = x 7 Deﬁne new variable Square both sides Substitute values into original Factor Set each factor equal to zero Solve each equation − 3y2 y2 = (x − 7)2 − 2y + 5 = 0 − 5)(y + 1) = 0 (3y 5 = 0 or y + 1 = 0 1 1 1 3y = 5 or 3y y = 3 5 3 7 or = x 5 3 21 3 + − + 7 or y = − 1 We have y, we still need x. 1 = x − + 7 7 Substitute into y = x 7 − + 7 Add 7. Use common denominator as needed − x = 26 3 , 6 Our Solution Example 482. (x2 − 6x)2 = 7(x2 7(x2 6x) + 12 6x)2 − 7(x2 − − − − 7(x2 (x2 − − 12 Make equation equal zero 6x) 6x) + 12 Move all terms to left − − 6x) + 12 = 0 Quadratic form y = x2 y2 = (x2 y2 − 3)(y (y − 3 = 0 or y − 6x)2 − 7y + 12 = 0 4 6x Make new variable Square both sides Substitute into original equation Solve by factoring Set each factor equal to zero Solve each equation We have y, still need x. Solve each equation, complete the square 6x 6 = 32 = 9 Add 9 to both sides of each equation y = 3 or y = 4 6x or 4 = x3 2 − 3 = x2 − 1 2 · 12 = x2 − 6x + 9 or 13 = x2 6x + 9 Factor − 354 ± 12√ = (x 12 = (x 3)2 − 2 3√ = x p + 3 ± 3)2 or 13 = (x − 13√ = (x or ± 3 or − + 3 ± − − 13√ = x p + 3 13√ , 3 ± 3)2 Use even root property 3)2 Simplify roots 3 Add 3 to both sides Our Solution The higher the exponent, the more solution we could have. This is illustrated in the following example, one with six solutions. Example 483. x6 − y − + 1 + 1 9x3 + 8 = 0 Quadratic form, one exponent, 6, double the other, 3 y2 − 1)(y (y − 1 = 0 or y y = x3 New variable equal to variable with lowest exponent y2 = x6 9y + 8 = 0 8 or y = 8 x3 = 1 or x3 = 8 1 8 8 − 8 = 0 Square both sides Substitute y2 for x6 and y for x3 Solve. We will solve by factoring. Set each factor equal to zero Solve each equation Solutions for y, we need x. Substitute into y = x3 Set each equation equal to zero 1 Factor each equation, diﬀerence of cubes First equation factored. Set each factor equal to zero First equation is easy to solve x x3 (x − − − − 1 = 0 or x3 − 1)(x2 + x + 1) = 0 − 1 = 0 or x2 + 12 − p2(1) 2)(x2 + 2x + 4) = 0 (x − 2 = 0 or x2 + 2x + 4 = 0 i 3√ ± 2 4(1)(1) = 1 4(1)(4) = 1 i 3√ i 3 22 − p2(1) x = 1, 2, 2 − ± First solution Quadratic formula on second factor Factor the second diﬀerence of cubes Set each factor equal to zero. First equation is easy to solve Our fourth solution Quadratic formula on second factor Our ﬁnal six solutions 355 9.6 Practice - Quadratic in Form Solve each of the following equations. Some equations will have 2) y4 4) y4 6) b4 − − − 8) y4 − 10) x6 14) x− 9y2 + 20 = 0 29y2 + 100 = 0 10b2 + 9 = 0 40y2 + 144 = 0 − − 2 35x3 + 216 = 0 2y2 = 24 1 x− 12 = 0 − 20 = 21y− 1 − 2 − 7x2 + 12 = 0 216 = 19z3 12) y4 1 3 16) 5y− 3 = 8 18) x4 − 2x2 − 3 = 0 20) x4 + 7x2 + 10 = 0 complex roots. 1) x4 − 5x2 + 4 = 0 3) m4 7m2 8 = 0 − − 50a2 + 49 = 0 5) a4 7) x4 − − 25x2 + 144 = 0 9) m4 − 20m2 + 64 = 0 11) z6 − 13) 6z4 z2 = 12 − 2 3 15) x 17) y− 35 = 2x − 6 + 7y− 19) x4 − 21) 2x4 23) x4 − 25) 8x6 27) x8 − 5x2 + 2 = 0 − 9x2 + 8 = 0 9x3 + 1 = 0 − 17x4 + 16 = 0 22) 2x4 x2 3 = 0 − − 10x3 + 16 = 0 24) x6 − 26) 8x6 + 7x3 1 = 0 − 4(x 28) (x 1)2 − − 1) = 5 − 30) (x + 1)2 + 6(x + 1) + 9 = 0 32) (m 1)2 − 34) (a + 1)2 + 2(a − 5(m 1) = 14 − 1) = 15 − 1) = 3 36) 2(x 38) (x2 1)2 3)2 − − − − (x − 2(x2 − 3) = 3 40) (x2 + x + 3)2 + 15 = 8(x2 + x + 3) 42) (x2 + x)2 − 8(x2 + x) + 12 = 0 44) (2x2 + 3x)2 = 8(2x2 + 3x) + 9 29) (y + b)2 31) (y + 2)2 33) (x 3)2 − − − − 35) (r 1)2 − − 37) 3(y + 1)2 4(y + b) = 21 6(y + 2) = 16 2(x 8(r 3) = 35 − 1) = 20 − 14(y + 1) = 5 − 39) (3x2 − 2x)2 + 5 = 6(3x2 2x) − 1 5(3x + 1) 3 = 88 2(x2 + 2x) = 3 41) 2(3x + 1) 2 3 43) (x2 + 2x)2 45) (2x2 x)2 − − − − 4(2x2 − x) + 3 = 0 46) (3x2 − 4x)2 = 3(3x2 4x) + 4 − 356 9.7 Quadratics - Rectangles Objective: Solve applications of quadratic equations using rectangles. An application of solving quadratic equations comes from the formula for the area of a rectangle. The area of a rectangle can be calculated by multiplying the width by the length. To solve problems with rectangles we will ﬁrst draw a picture to represent the problem and use the picture to set up our equation. Example 484. The length of a rectangle is 3 more than the width. If the area is 40 square inches, what are the dimensions? 40 x + 3 x We do not know the width, x. Length is 4 more, or x + 4, and area is 40. 357 x(x + 3) = 40 Multiply length by width to get area x2 + 3x = 40 Distribute 40 40 Make equation equal zero − − x2 + 3x − 40 = 0 − (x 5)(x + 8) = 0 5 = 0 or x + 8 = 0 8 8 Our x is a width, cannot be negative. Factor Set each factor equal to zero Solve each equation 8 x = 5 or 5) + 3 = 8 5 in by 8in Our Solution Length is x + 3, substitute 5 for x to ﬁnd length The above rectangle problem is very simple as there is only one rectangle involved. When we compare two rectangles, we may have to get a bit more creative. Example 485. If each side of a square is increased by 6, the area is multiplied by 16. Find the side of the original square. x2 x x Square has all sides the same length Area is found by multiplying length by width 16x2 x + 6 Each side is increased by 6, x + 6 Area is 16 times original area (x + 6)(x + 6) = 16x2 Multiply length by width to get area x2 + 12x + 36 = 16x2 FOIL 16x2 − − 16x2 Make equation equal zero − 15x2 + 12x + 36 = 0 Divide each term by 15x2 12)2 12x − 4(15)( 36 = 0 36) − Solve using the quadratic formula 1, changes the signs − Evaluate 12 ± x = ( p − − − 2(15) 16 x = ± 2304√ 30 16 48 ± 30 60 30 x = Can′t have a negative solution, we will only add x = = 2 Our x is the original square 358 2 Our Solution Example 486. The length of a rectangle is 4 ft greater than the width. If each dimension is increased by 3, the new area will be 33 square feet larger. Find the dimensions of the original rectangle. x(x + 4) x We don′t know width, x, length is 4 more, x + 4 x + 4 Area is found by multiplying length by width x(x + 4) + 33 x + 3 Increase each side by 3. width becomes x + 3, length x + 3)(x + 7) = x(x + 4) + 33 x2 + 10x + 21 = x2 + 4x + 33 x2 x2 − − Area is 33 more than original, x(x + 4) + 33 Set up equation, length times width is area Subtract x2 from both sides − − 10x + 21 = 4x + 33 Move variables to one side 4x Subtract 4x from each side Subtract 21 from both sides 4x 6x + 21 = 33 21 21 6x = 12 Divide both sides by 6 6 − − 6 x = 2 (2) + 4 = 6 2 ft by 6ft Our Solution x is the width of the original x + 4 is the length. Substitute 2 to ﬁnd From one rectangle we can ﬁnd two equations. Perimeter is found by adding all the sides of a polygon together. A rectangle has two widths and two lengths, both the same size. So we can use the equation P = 2l + 2w (twice the length plus twice the width). Example 487. The area of a rectangle is 168 cm2. The perimeter of the same rectangle is 52 cm. What are the dimensions of the rectangle? x We don′t know anything about length or width y xy = 168 Use two variables, x and y Length times width gives the area. 2x + 2y = 52 Also use perimeter formula. 2x − 2y = 2x + 52 Divide each term by 2 Solve by substitution, isolate y 2x − − 359 2 2 2 x + 26 x( Substitute into area equation − y = x + 26) = 168 Distribute − x2 + 26x = 168 Divide each term by − x2 26 168 26x = − 2 − = 132 = 169 1 2 · x2 Solve by completing the square. 1, changing all the signs − Find number to complete the square: 1 2 · b 2 − 26x + 324 = 1 Add 169 to both sides (x Factor x Square root both sides 13)2 = 1 − 1 13 = − + 13 + 13 x = 13 ± ± 1 Evaluate y = y = − − x = 14 or 12 Two options for ﬁrst side. (14) + 26 = 12 (12) + 26 = 14 Substitute 14 into y = − Substitute 12 into y = − Both are the same rectangle, variables switched! x + 26 x + 26 12 cm by 14cm Our Solution World View Note: Indian mathematical records from the 9th century demonstrate that their civilization had worked extensivly in geometry creating religious alters of various shapes including rectangles. Another type of rectangle problem is what we will call a “frame problem”. The idea behind a frame problem is that a rectangle, such as a photograph, is centered inside another rectangle, such as a frame. In these cases it will be important to rememember that the frame extends on all sides of the rectangle. This is shown in the following example. Example 488. An 8 in by 12 in picture has a frame of uniform width around it. The area of the frame is equal to the area of the picture. What is the width of the frame? 8 12 12 + 2x Draw picture, picture if 8 by 10 If frame has width x, on both sides, we add 2x 8 + 2x 8 2 · 12 = 96 Area of the picture, length times width · 96 = 192 Frame is the same as the picture. Total area is double this. (12 + 2x)(8 + 2x) = 192 Area of everything, length times width 96 + 24x + 16x + 4x2 = 192 FOIL 4x2 + 40x + 96 = 192 Combine like terms 192 192 Make equation equal to zero by subtracting 192 − 4x2 + 40x − 96 = 0 − Factor out GCF of 4 360 x − − 4(x2 + 10x 4(x 24) = 0 2)(x + 12) = 0 2 = 0 or x + 12 = 0 12 12 Can′t have negative frame width. Factor trinomial Set each factor equal to zero Solve each equation 12 x = 2 or − − + 2 + 2 − − 2 inches Our Solution Example 489. A farmer has a ﬁeld that is 400 rods by 200 rods. He is mowing the ﬁeld in a spiral pattern, starting from the outside and working in towards the center. After an hour of work, 72% of the ﬁeld is left uncut. What is the size of the ring cut around the outside? 200 − 2x 200 Draw picture, outside is 200 by 400 If frame has width x on both sides, subtract 2x from each side to get center 400 2x − 400 400 200 = 80000 Area of entire ﬁeld, length times width (0.72) = 57600 Area of center, multiply by 28% as decimal · 2x) = 57600 Are
a of center, length times width 80000 · 2x)(200 − (400 − 800x − 4x2 − − 80000 400x + 4x2 = 57600 FOIL 1200x + 80000 = 57600 Combine like terms 57600 57600 Make equation equal zero − − 4x2 − 4(x2 4(x 1200x + 22400 = 0 300x + 5600) = 0 280)(x 20) = 0 − 280 = 0 or x 20 = 0 − + 20 + 20 − − − + 280 + 280 x Factor out GCF of 4 Factor trinomial Set each factor equal to zero Solve each equation x = 280 or 20 The ﬁeld is only 200 rods wide, Can′t cut 280 oﬀ two sides! 20 rods Our Solution For each of the frame problems above we could have also completed the square or use the quadratic formula to solve the trinomials. Remember that completing the square or the quadratic formula always will work when solving, however, factoring only works if we can factor the trinomial. 361 9.7 Practice - Rectangles 1) In a landscape plan, a rectangular ﬂowerbed is designed to be 4 meters longer than it is wide. If 60 square meters are needed for the plants in the bed, what should the dimensions of the rectangular bed be? 2) If the side of a square is increased by 5 the area is multiplied by 4. Find the side of the original square. 3) A rectangular lot is 20 yards longer than it is wide and its area is 2400 square yards. Find the dimensions of the lot. 4) The length of a room is 8 ft greater than it is width. If each dimension is increased by 2 ft, the area will be increased by 60 sq. ft. Find the dimensions of the rooms. 5) The length of a rectangular lot is 4 rods greater than its width, and its area is 60 square rods. Find the dimensions of the lot. 6) The length of a rectangle is 15 ft greater than its width. If each dimension is decreased by 2 ft, the area will be decreased by 106 ft2. Find the dimensions. 7) A rectangular piece of paper is twice as long as a square piece and 3 inches wider. The area of the rectangular piece is 108 in2. Find the dimensions of the square piece. 8) A room is one yard longer than it is wide. At 75c per sq. yd. a covering for the ﬂoor costs S31.50. Find the dimensions of the ﬂoor. 9) The area of a rectangle is 48 ft2 and its perimeter is 32 ft. Find its length and width. 10) The dimensions of a picture inside a frame of uniform width are 12 by 16 inches. If the whole area (picture and frame) is 288 in2, what is the width of the frame? 11) A mirror 14 inches by 15 inches has a frame of uniform width. If the area of the frame equals that of the mirror, what is the width of the frame. 12) A lawn is 60 ft by 80 ft. How wide a strip must be cut around it when mowing the grass to have cut half of it. 13) A grass plot 9 yards long and 6 yards wide has a path of uniform width around it. If the area of the path is equal to the area of the plot, determine the width of the path. 14) A landscape architect is designing a rectangular ﬂowerbed to be border with 28 plants that are placed 1 meter apart. He needs an inner rectangular space in the center for plants that must be 1 meter from the border of the bed and 362 that require 24 square meters for planting. What should the overall dimensions of the ﬂowerbed be? 15) A page is to have a margin of 1 inch, and is to contain 35 in2 of painting. How large must the page be if the length is to exceed the width by 2 inches? 16) A picture 10 inches long by 8 inches wide has a frame whose area is one half the area of the picture. What are the outside dimensions of the frame? 17) A rectangular wheat ﬁeld is 80 rods long by 60 rods wide. A strip of uniform width is cut around the ﬁeld, so that half the grain is left standing in the form of a rectangular plot. How wide is the strip that is cut? 18) A picture 8 inches by 12 inches is placed in a frame of uniform width. If the area of the frame equals the area of the picture ﬁnd the width of the frame. 19) A rectangular ﬁeld 225 ft by 120 ft has a ring of uniform width cut around the outside edge. The ring leaves 65% of the ﬁeld uncut in the center. What is the width of the ring? 20) One Saturday morning George goes out to cut his lot that is 100 ft by 120 ft. He starts cutting around the outside boundary spiraling around towards the center. By noon he has cut 60% of the lawn. What is the width of the ring that he has cut? 21) A frame is 15 in by 25 in and is of uniform width. The inside of the frame leaves 75% of the total area available for the picture. What is the width of the frame? 22) A farmer has a ﬁeld 180 ft by 240 ft. He wants to increase the area of the ﬁeld by 50% by cultivating a band of uniform width around the outside. How wide a band should he cultivate? 23) The farmer in the previous problem has a neighber who has a ﬁeld 325 ft by 420 ft. His neighbor wants to increase the size of his ﬁeld by 20% by cultivating a band of uniform width around the outside of his lot. How wide a band should his neighbor cultivate? 24) A third farmer has a ﬁeld that is 500 ft by 550 ft. He wants to increase his ﬁeld by 20%. How wide a ring should he cultivate around the outside of his ﬁeld? 25) Donna has a garden that is 30 ft by 36 ft. She wants to increase the size of the garden by 40%. How wide a ring around the outside should she cultivate? 26) A picture is 12 in by 25 in and is surrounded by a frame of uniform width. The area of the frame is 30% of the area of the picture. How wide is the frame? 363 9.8 Quadratics - Teamwork Objective: Solve teamwork problems by creating a rational equation to model the problem. If it takes one person 4 hours to paint a room and another person 12 hours to paint the same room, working together they could paint the room even quicker, it turns out they would paint the room in 3 hours together. This can be reasoned by the following logic, if the ﬁrst person paints the room in 4 hours, she paints 1 4 of the room each hour. If the second person takes 12 hours to paint the room, he paints 1 4 + 1 12 of the room each hour. So together, each hour they paint 1 12 of the room. Using a common denominator of 12 gives: 3 12 = 1 3. This means each hour, working together they complete 1 3 is completed each hour, it follows that it will take 3 hours to complete the entire room. 12 + 1 12 = 4 3 of the room. If 1 This pattern is used to solve teamwork problems. If the ﬁrst person does a job in A, a second person does a job in B, and together they can do a job in T (total). We can use the team work equation. Teamwork Equation: 1 A + 1 B = 1 T Often these problems will involve fractions. Rather than thinking of the ﬁrst fraction as 1 A , it may be better to think of it as the reciprocal of A’s time. World View Note: When the Egyptians, who were the ﬁrst to work with fractions, wrote fractions, they were all unit fractions (numerator of one). They only used these type of fractions for about 2000 years! Some believe that this cumbersome style of using fractions was used for so long out of tradition, others believe the Egyptians had a way of thinking about and working with fractions that has been completely lost in history. Example 490. Adam can clean a room in 3 hours. If his sister Maria helps, they can clean it in 2 2 5 hours. How long will it take Maria to do the job alone? 2 2 5 = Adan: 3, Maria: x, Total: 12 5 5 12 Together time, 2 2 5 , needs to be converted to fraction Clearly state times for each and total, using x for Maria 364 1 3 + 1 x = 5 12 Using reciprocals, add the individual times gives total 1(12x) 3 + 1(12x) x = 5(12x) 12 Multiply each term by LCD of 12x 4x + 12 = 5x Reduce each fraction 4x 4x Move variables to one side, subtracting 4x − − 12 = x Our solution for x It takes Maria 12 hours Our Solution Somtimes we only know how two people’s times are related to eachother as in the next example. Example 491. Mike takes twice as long as Rachel to complete a project. Together they can complete the project in 10 hours. How long will it take each of them to complete the project alone? Mike: 2x, Rachel: x, Total: 10 Clearly deﬁne variables. If Rachel is x, Mike is 2x 1 2x + 1 x = 1 10 Using reciprocals, add individal times equaling total 1(10x) 2x + 1(10x) x = 1(10x) 10 Multiply each term by LCD, 10x 5 + 10 = x Combine like terms 15 = x We have our x, we said x was Rachel′s time 2(15) = 30 Mike is double Rachel, this gives Mike′s time. Mike: 30 hr, Rachel: 15hr Our Solution With problems such as these we will often end up with a quadratic to solve. Example 492. Brittney can build a large shed in 10 days less than Cosmo can. If they built it together it would take them 12 days. How long would it take each of them working alone? Britney: x − 10, Cosmo: x, Total: 12 1 12 1 x 10 = + x 1 − 365 If Cosmo is x, Britney is x 10 − Using reciprocals, make equation 1(12x(x x − − 10 10)) + 1(12x(x x − 10)) = 1(12x(x 12 − 10)) Multiply by LCD: 12x(x 10) − 12x + 12(x − 12x + 12x − 24x − 24x + 120 10) = x(x 120 = x2 120 = x2 − − − − 0 = x2 0 = (x 30 = 0 or x x − + 30 + 30 − − − 10) Reduce fraction 10x Distribute 10x Combine like terms 24x + 120 Move all terms to one side 34x + 120 30)(x 4 Factor Set each factor equal to zero Solve each equation x = 30 or x = 4 This, x, was deﬁned as Cosmo. 30 10 = 20 or 4 10 = 6 Find Britney, can′t have negative time − − − Britney: 20 days, Cosmo: 30 days Our Solution In the previous example, when solving, one of the possible times ended up negative. We can’t have a negative amount of time to build a shed, so this possibility is ignored for this problem. Also, as we were solving, we had to factor x2 34x + 120. This may have been diﬃcult to factor. We could have also chosen to complete the square or use the quadratic formula to ﬁnd our solutions. − It is important that units match as we solve problems. This means we may have to convert minutes into hours to match the other units given in the problem. Example 493. An electrician can complete a job in one hour less than his apprentice. Together they do the job in 1 hour and 12 minutes. How long would it take each of them working alone? 1 hr 12 min = 1 12 60 hr Change 1 hour 12 minutes to mixed number 1 12 60 = 1 1 5 = Electrician: x − 1, Appre
ntice: x, Total Reduce and convert to fraction Clearly deﬁne variables Using reciprocals, make equation 1(6x(x x − − 1 1)) + 1(6x(x x − 1)) = 5(6x(x 6 − 1) Multiply each term by LCD 6x(x 1) − 366 6x + 6(x 1) = 5x(x 6 = 5x2 6 = 5x2 − − − − 6x + 6x 12x 12x + 6 0 = 5x2 − − − 1) Reduce each fraction 5x Distribute 5x Combine like terms 5x 0 = (5x 2 = 0 or x − − − 2)(x Factor Set each factor equal to zero Solve each equation 12x + 6 Move all terms to one side of equation 17x + 6 3 5x = 2 or x = 3 5 − + 2 + 2 or x = 3 Subtract 1 from each to ﬁnd electrician or 3 − Electrician: 2 hr, Apprentice: 3 hours Our Solution 1 = 2 Ignore negative. Very similar to a teamwork problem is when the two involved parts are working against each other. A common example of this is a sink that is ﬁlled by a pipe and emptied by a drain. If they are working against eachother we need to make one of the values negative to show they oppose eachother. This is shown in the next example.. Example 494. A sink can be ﬁlled by a pipe in 5 minutes but it takes 7 minutes to drain a full sink. If both the pipe and the drain are open, how long will it take to ﬁll the sink? Sink: 5, Drain: 7, Total: x Deﬁne variables, drain is negative 1 5 − 1 7 = 1 x 1(35x) 5 − 1(35x) 7 = 1(35x) x Using reciprocals to make equation, Subtract because they are opposite Multiply each term by LCD: 35x 7x − 5x = 35 Reduce fractions 2x = 35 Combine like terms 2 x = 17.5 Our answer for x Divide each term by 2 2 17.5 min or 17 min 30 sec Our Solution 367 9.8 Practice - Teamwork 1) Bills father can paint a room in two hours less than Bill can paint it. Working together they can complete the job in two hours and 24 minutes. How much time would each require working alone? 2) Of two inlet pipes, the smaller pipe takes four hours longer than the larger pipe to ﬁll a pool. When both pipes are open, the pool is ﬁlled in three hours and forty-ﬁve minutes. If only the larger pipe is open, how many hours are required to ﬁll the pool? 3) Jack can wash and wax the family car in one hour less than Bob can. The two working together can complete the job in 1 1 each require if they worked alone? 5 hours. How much time would 4) If A can do a piece of work alone in 6 days and B can do it alone in 4 days, how long will it take the two working together to complete the job? 5) Working alone it takes John 8 hours longer than Carlos to do a job. Working together they can do the job in 3 hours. How long will it take each to do the job working alone? 6) A can do a piece of work in 3 days, B in 4 days, and C in 5 days each working alone. How long will it take them to do it working together? 7) A can do a piece of work in 4 days and B can do it in half the time. How long will it take them to do the work together? 8) A cistern can be ﬁlled by one pipe in 20 minutes and by another in 30 minutes. How long will it take both pipes together to ﬁll the tank? 9) If A can do a piece of work in 24 days and A and B together can do it in 6 days, how long would it take B to do the work alone? 10) A carpenter and his assistant can do a piece of work in 3 3 4 days. If the carpenter himself could do the work alone in 5 days, how long would the assistant take to do the work alone? 11) If Sam can do a certain job in 3 days, while it takes Fred 6 days to do the same job, how long will it take them, working together, to complete the job? 12) Tim can ﬁnish a certain job in 10 hours. It take his wife JoAnn only 8 hours to do the same job. If they work together, how long will it take them to complete the job? 13) Two people working together can complete a job in 6 hours. If one of them works twice as fast as the other, how long would it take the faster person, working alone, to do the job? 14) If two people working together can do a job in 3 hours, how long will it take the slower person to do the same job if one of them is 3 times as fast as the other? 15) A water tank can be ﬁlled by an inlet pipe in 8 hours. It takes twice that long for the outlet pipe to empty the tank. How long will it take to ﬁll the tank if both pipes are open? 368 16) A sink can be ﬁlled from the faucet in 5 minutes. It takes only 3 minutes to empty the sink when the drain is open. If the sink is full and both the faucet and the drain are open, how long will it take to empty the sink? 17) It takes 10 hours to ﬁll a pool with the inlet pipe. It can be emptied in 15 hrs with the outlet pipe. If the pool is half full to begin with, how long will it take to ﬁll it from there if both pipes are open? 18) A sink is 1 4 full when both the faucet and the drain are opened. The faucet alone can ﬁll the sink in 6 minutes, while it takes 8 minutes to empty it with the drain. How long will it take to ﬁll the remaining 3 4 of the sink? 19) A sink has two faucets, one for hot water and one for cold water. The sink can be ﬁlled by a cold-water faucet in 3.5 minutes. If both faucets are open, the sink is ﬁlled in 2.1 minutes. How long does it take to ﬁll the sink with just the hot-water faucet open? 20) A water tank is being ﬁlled by two inlet pipes. Pipe A can ﬁll the tank in 4 1 2 hrs, while both pipes together can ﬁll the tank in 2 hours. How long does it take to ﬁll the tank using only pipe B? 21) A tank can be emptied by any one of three caps. The ﬁrst can empty the tank in 20 minutes while the second takes 32 minutes. If all three working together could empty the tank in 8 8 to empty the tank? 59 minutes, how long would the third take 22) One pipe can ﬁll a cistern in 1 1 3 hrs. Three pipes working together ﬁll the cistern in 42 minutes. How long would it take the third pipe alone to ﬁll the tank? 2 hours while a second pipe can ﬁll it in 2 1 23) Sam takes 6 hours longer than Susan to wax a ﬂoor. Working together they can wax the ﬂoor in 4 hours. How long will it take each of them working alone to wax the ﬂoor? 24) It takes Robert 9 hours longer than Paul to rapair a transmission. If it takes 5 hours to do the job if they work together, how long will it take each them 2 2 of them working alone? 25) It takes Sally 10 1 2 minutes longer than Patricia to clean up their dorm room. If they work together they can clean it in 5 minutes. How long will it take each of them if they work alone? 26) A takes 7 1 2 minutes longer than B to do a job. Working together they can do the job in 9 minutes. How long does it take each working alone? 27) Secretary A takes 6 minutes longer than Secretary B to type 10 pages of manuscript. If they divide the job and work together it will take them 8 3 4 minutes to type 10 pages. How long will it take each working alone to type the 10 pages? 28) It takes John 24 minutes longer than Sally to mow the lawn. If they work together they can mow the lawn in 9 minutes. How long will it take each to mow the lawn if they work alone? 369 9.9 Quadratics - Simultaneous Products Objective: Solve simultaneous product equations using substitution to create a rational equation. When solving a system of equations where the variables are multiplied together we can use the same idea of substitution that we used with linear equations. When we do so we may end up with a quadratic equation to solve. When we used substitution we solved for a variable and substitute this expression into the other equation. If we have two products we will choose a variable to solve for ﬁrst and divide both sides of the equations by that variable or the factor containing the variable. This will create a situation where substitution can easily be done. Example 495. xy = 48 (x + 3)(y 2) = 54 − To solve for x, divide ﬁrst equation by x, second by x + 3 y = 48 x and y 2 = − 54 x + 3 Substitute 48 x for y in the second equation 48 x − 2 = 54 x + 3 Multiply each term by LCD: x(x + 3) 48x(x + 3) x − 2x(x + 3) = 54x(x + 3) x + 3 Reduce each fraction 48(x + 3) 48x + 144 2x(x + 3) = 54x Distribute 2x2 6x = 54x Combine like terms − − − 2x2 + 42x + 144 = 54x Make equation equal zero − 2 − 54x 54x Subtract 54x from both sides 12x + 144 = 0 Divide each term by GCF of − − 2x2 − − x2 + 6x x − − + 6 + 6 12 x = 6 or x = Factor Set each factor equal to zero Solve each equation 72 = 0 − 6)(x + 12) = 0 (x 6 = 0 or x + 12 = 0 12 − 12 − 12y = 48 12 12 4 Our solutions for y, 6y = 48 or − 6 − − y = 8 or y = − 4) Our Solutions as ordered pairs (6, 8) or ( Substitute each solution into xy = 48 Solve each equation 12, − 6 − − 370 These simultaneous product equations will also solve by the exact same pattern. We pick a variable to solve for, divide each side by that variable, or factor containing the variable. This will allow us to use substitution to create a rational expression we can use to solve. Quite often these problems will have two solutions. Example 496. xy = 35 (x + 6)(y − − 2) = 5 To solve for x, divide the ﬁrst equation by x, second by x + 6 35 y = − x and y 2 = − 5 x + 6 35 Substitute − x for y in the second equation − 35 x − 2 = 5 x + 6 Multiply each term by LCD: x(x + 6) − 35x(x + 6) x − 2x(x + 6) = 5x(x + 6) x + 6 Reduce fractions − − 35(x + 6) 35x 210 2x2 − − 12x = 5x Combine like terms 210 = 5x Make equation equal zero 2x(x + 6) = 5x Distribute 2x2 47x 5x 52x − − 5x − 210 = 0 Divide each term by 2 − − − − − − Factor Set each factor equal to zero Solve each equation − 2x2 − x2 + 26x + 105 = 0 (x + 5)(x + 21) = 0 x + 5 = 0 or x + 21 = 0 21 21 35 21 5 3 − − − − y = 7 or y = 21 − 5 or x = 21y = 21 5 − x = − 35 or 5 − − − − − 5 5, 7) or ( − 21, 5 3 − − 5y = 5 − Substitute each solution into xy = Solve each equation 35 − Our solutions for y Our Solutions as ordered pairs The processes used here will be used as we solve applications of quadratics including distance problems and revenue problems. These will be covered in another section. World View Note: William Horner, a British mathematician from the late 18th century/early 19th century is credited with a method for solving simultaneous equations, however, Chinese mathematician Chu Shih-chieh in 1303 solved these equations with exponents as high as
14! 371 9.9 Practice - Simultaneous Product Solve. 1) 3) xy = 72 (x + 2)(y − xy = 150 4) = 128 6)(y + 1) = 64 (x − 5) xy = 45 (x + 2)(y + 1) = 70 7) xy = 90 5)(y + 1) = 120 (x − 9) 11) xy = 12 (x + 1)(y − xy = 45 4) = 16 5)(y + 3) = 160 (x − 2) 4) 6) 8) 10) (x xy = 180 1)(y − − xy = 120 1 2) = 205 (x + 2)(y 3)=120 − xy = 65 (x 8)(y + 2) = 35 − xy = 48 (x 6)(y + 3) = 60 − xy = 60 (x + 5)(y + 3) = 150 12) xy = 80 5)(y + 5) = 45 (x − 372 9.10 Quadratics - Revenue and Distance Objective: Solve revenue and distance applications of quadratic equations. A common application of quadratics comes from revenue and distance problems. Both are set up almost identical to each other so they are both included together. Once they are set up, we will solve them in exactly the same way we solved the simultaneous product equations. Revenue problems are problems where a person buys a certain number of items for a certain price per item. If we multiply the number of items by the price per item we will get the total paid. To help us organize our information we will use the following table for revenue problems Number Price Total First Second The price column will be used for the individual prices, the total column is used for the total paid, which is calculated by multiplying the number by the price. Once we have the table ﬁlled out we will have our equations which we can solve. This is shown in the following examples. Example 497. A man buys several ﬁsh for S56. After three ﬁsh die, he decides to sell the rest at a proﬁt of S5 per ﬁsh. His total proﬁt was S4. How many ﬁsh did he buy to begin with? Buy Sell Buy Sell Number Price Total n p 56 Using our table, we don′t know the number he bought, or at what price, so we use varibles n and p. Total price was S56. Number Price Total n − 3 n p p + 5 56 60 When he sold, he sold 3 less (n more (p + 5). Total proﬁt was S4, combined with S56 spent is S60 3), for S5 − np = 56 3)(p + 5) = 60 Find equatinos by multiplying number by price These are a simultaneous product (n − p = 56 n and p + 5 = 60 − n 3 Solving for number, divide by n or (n 3) − 373 56 n + 5 = 60 − n 3 Substitute 56 n for p in second equation 56n(n n − 3) + 5n(n 3) = − 3) 60n(n n − − 3 Multiply each term by LCD: n(n 3) − 56(n 56n − 3) + 5n(n − 168 + 5n2 5n2 + 41n 60n − − 3) = 60n − 15n = 60n 168 = 60n Move all terms to one side Reduce fractions Combine like terms − 5n2 − 19)2 19n − 4(5)( − 2(5) − 60n 168 = 0 168) − Solve with quadratic formula Simplify 19 ± n = ( p − n = 19 3721√ 10 ± 19 = 61 ± 10 We don′t want negative solutions, only do + n = 80 10 = 8 This is our n 8 ﬁsh Our Solution Example 498. A group of students together bought a couch for their dorm that cost S96. However, 2 students failed to pay their share, so each student had to pay S4 more. How many students were in the original group? Number Price Total n p 96 Deal Paid Number Price Total Deal Paid n − 2 n p p + 4 96 96 S96 was paid, but we don′t know the number or the price agreed upon by each student. There were 2 less that actually paid (n − and they had to pay S4 more (p + 4). The total here is still S96. 2) np = 96 2)(p + 4) = 96 Equations are product of number and price This is a simultaneous product (n − p = 96 n and p + 4 = 96 n + 4 = 96 − 96 − n n 2 2 Solving for number, divide by n and n 2 − Substitute 96 n for p in the second equation 374 96n(n n − 2) + 4n(n 2) = − 2) 96n(n n − − 2 Multiply each term by LCD: n(n 2) − 96(n 96n 2) + 4n(n − 192 + 4n2 − 4n2 + 88n 96n − 4n2 − − − 2) = 96n 8n = 96n 192 = 96n 96n 8n 192 = 0 + 192 + 192 8n = 192 4 4 − − − − 4n2 4 Reduce fractions Distribute Combine like terms Set equation equal to zero Solve by completing the square, Separate variables and constant Divide each term by a or 4 2 1 2 · n2 2n = 48 Complete the square: b − 2 2 n2 · = 12 = 1 Add to both sides of equation 1 2 2n + 1 = 49 − 1)2 = 49 ( We don′t want a negative solution n = 1 + 7 = 8 Factor Square root of both sides Add 1 to both sides ± ± 8 students Our Solution The above examples were solved by the quadratic formula and completing the square. For either of these we could have used either method or even factoring. Remember we have several options for solving quadratics. Use the one that seems easiest for the problem. Distance problems work with the same ideas that the revenue problems work. The only diﬀerence is the variables are r and t (for rate and time), instead of n and p (for number and price). We already know that distance is calculated by multiplying rate by time. So for our distance problems our table becomes the following: rate time distance First Second Using this table and the exact same patterns as the revenue problems is shown in the following example. Example 499. 375 Greg went to a conference in a city 120 miles away. On the way back, due to road construction he had to drive 10 mph slower which resulted in the return trip taking 2 hours longer. How fast did he drive on the way to the conference? There Back There Back r rate time distance r t 120 We do not know rate, r, or time, t he traveled on the way to the conference. But we do know the distance was 120 miles. time distance rate r t 10 t + 2 − 120 120 Coming back he drove 10 mph slower (r 10) and took 2 hours longer (t + 2). The distance was still 120 miles. − rt = 120 Equations are product of rate and time 10)(t + 2) = 120 We have simultaneous product equations (r − t = 120 r and t + 2 = 120 r + 2 = 120 r r 10 − 120 10 − Solving for rate, divide by r and r 10 − Substitute 120 r for t in the second equation 120r(r r − 10) + 2r(r 10) = − 120r(r r − 10) − 10 Multiply each term by LCD: r(r 10) − Reduce each fraction Distribute Combine like terms 120r Make equation equal to zero Divide each term by 2 Factor Set each factor equal to zero Solve each equation 120(r 120r − − 20r = 120r − 20r = 120r − 1200 = 120r 10) + 2r2 − 1200 + 2r2 − 2r2 + 100r 120r − 2r2 20r 1200 = 0 − − r2 10r 600 = 0 − − 30)(r + 20) = 0 30 = 0 and r + 20 = 0 20 − 20 − 30 mph 20 r = 30 and r = − + 30 + 30 (r − − r Can′t have a negative rate Our Solution World View Note: The world’s fastest man (at the time of printing) is Jamaican Usain Bolt who set the record of running 100 m in 9.58 seconds on August 16, 2009 in Berlin. That is a speed of over 23 miles per hour! Another type of simultaneous product distance problem is where a boat is traveling in a river with the current or against the current (or an airplane ﬂying with the wind or against the wind). If a boat is traveling downstream, the current will push it or increase the rate by the speed of the current. If a boat is traveling 376 upstream, the current will pull against it or decrease the rate by the speed of the current. This is demonstrated in the following example. Example 500. A man rows down stream for 30 miles then turns around and returns to his original location, the total trip took 8 hours. If the current ﬂows at 2 miles per hour, how fast would the man row in still water? 8 rate time distance down up rate down r + 2 up 2 8 r − t 8 t − 30 30 time distance t t − 30 30 Write total time above time column We know the distance up and down is 30. Put t for time downstream. Subtracting 8 t becomes time upstream − Downstream the current of 2 mph pushes the boat (r + 2) and upstream the current pulls the boat (r 2) − (r + 2)t = 30 Multiply rate by time to get equations t) = 30 We have a simultaneous product 2)(8 − (r − t = 30 r + 2 and 8 t = − 30 r + 2 = 8 − 30 − 30 − 2 2 r r Solving for rate, divide by r + 2 or r 2 − Substitute 30 r + 2 for t in second equation 8(r + 2)(r 2) − − 30(r + 2)(r r + 2 2) = − 30(r + 2)(r r 2 − 2) − Multiply each term by LCD: (r + 2)(r 2) − 8(r + 2)(r − 8r2 − 2) − 32 8r2 − 30(r 2) = 30(r + 2) Reduce fractions 30r + 60 = 30r + 60 Multiply and distribute 30r + 28 = 30r + 60 Make equation equal zero 30r 30r − − − 60 60 − − − 8r2 60r 32 = 0 − − 2r2 15r 8 = 0 − − (2r + 1)(r 8) = 0 − 2r + 1 = 0 or or r = 8 − 2 1 2 1 1 − 2r = 2 or r = 8 r = − − 8 mph Divide each term by 4 Factor Set each factor equal to zero Solve each equation Can′t have a negative rate Our Solution 377 9.10 Practice - Revenue and Distance 1) A merchant bought some pieces of silk for S900. Had he bought 3 pieces more for the same money, he would have paid S15 less for each piece. Find the number of pieces purchased. 2) A number of men subscribed a certain amount to make up a deﬁcit of S100 but 5 men failed to pay and thus increased the share of the others by S1 each. Find the amount that each man paid. 3) A merchant bought a number of barrels of apples for S120. He kept two barrels and sold the remainder at a proﬁt of S2 per barrel making a total proﬁt of S34. How many barrels did he originally buy? 4) A dealer bought a number of sheep for S440. After 5 had died he sold the remainder at a proﬁt of S2 each making a proﬁt of S60 for the sheep. How many sheep did he originally purchase? 5) A man bought a number of articles at equal cost for S500. He sold all but two for S540 at a proﬁt of S5 for each item. How many articles did he buy? 6) A clothier bought a lot of suits for S750. He sold all but 3 of them for S864 making a proﬁt of S7 on each suit sold. How many suits did he buy? 7) A group of boys bought a boat for S450. Five boys failed to pay their share, hence each remaining boys were compelled to pay S4.50 more. How many boys were in the original group and how much had each agreed to pay? 8) The total expenses of a camping party were S72. If there had been 3 fewer persons in the party, it would have cost each person S2 more than it did. How many people were in the party and how much did it cost each one? 9) A factory tests the road performance of new model cars by driving them at two diﬀerent rates of speed for at least 100 kilometers at each rate. The speed rates range from 50 to 70 km/hr in the lower range and from 70 to 90 km/hr in the higher range. A driver plans to test a car on an available speedway by driving it for 120 kilometers at a speed in the lower range and t
hen driving 120 kilometers at a rate that is 20 km/hr faster. At what rates should he drive if he plans to complete the test in 3 1 2 hours? 10) A train traveled 240 kilometers at a certain speed. When the engine was replaced by an improved model, the speed was increased by 20 km/hr and the travel time for the trip was decreased by 1 hour. What was the rate of each engine? 11) The rate of the current in a stream is 3 km/hr. A man rowed upstream for 3 kilometers and then returned. The round trip required 1 hour and 20 minutes. How fast was he rowing? 378 12) A pilot ﬂying at a constant rate against a headwind of 50 km/hr ﬂew for 750 kilometers, then reversed direction and returned to his starting point. He completed the round trip in 8 hours. What was the speed of the plane? 13) Two drivers are testing the same model car at speeds that diﬀer by 20 km/hr. The one driving at the slower rate drives 70 kilometers down a speedway and returns by the same route. The one driving at the faster rate drives 76 kilometers down the speedway and returns by the same route. Both drivers leave at the same time, and the faster car returns 1 slower car. At what rates were the cars driven? 2 hour earlier than the 14) An athlete plans to row upstream a distance of 2 kilometers and then return to his starting point in a total time of 2 hours and 20 minutes. If the rate of the current is 2 km/hr, how fast should he row? 15) An automobile goes to a place 72 miles away and then returns, the round trip occupying 9 hours. His speed in returning is 12 miles per hour faster than his speed in going. Find the rate of speed in both going and returning. 16) An automobile made a trip of 120 miles and then returned, the round trip occupying 7 hours. Returning, the rate was increased 10 miles an hour. Find the rate of each. 17) The rate of a stream is 3 miles an hour. If a crew rows downstream for a distance of 8 miles and then back again, the round trip occupying 5 hours, what is the rate of the crew in still water? 18) The railroad distance between two towns is 240 miles. If the speed of a train were increased 4 miles an hour, the trip would take 40 minutes less. What is the usual rate of the train? 19) By going 15 miles per hour faster, a train would have required 1 hour less to travel 180 miles. How fast did it travel? 20) Mr. Jones visits his grandmother who lives 100 miles away on a regular basis. Recently a new freeway has opend up and, although the freeway route is 120 miles, he can drive 20 mph faster on average and takes 30 minutes less time to make the trip. What is Mr. Jones rate on both the old route and on the freeway? 21) If a train had traveled 5 miles an hour faster, it would have needed 1 1 2 hours less time to travel 150 miles. Find the rate of the train. 22) A traveler having 18 miles to go, calculates that his usual rate would make him one-half hour late for an appointment; he ﬁnds that in order to arrive on time he must travel at a rate one-half mile an hour faster. What is his usual rate? 379 9.11 Quadratics - Graphs of Quadratics Objective: Graph quadratic equations using the vertex, x-intercepts, and y-intercept. Just as we drew pictures of the solutions for lines or linear equations, we can draw a picture of solution to quadratics as well. One way we can do that is to make a table of values. Example 501. y = x2 − 4x + 3 Make a table of values x y 0 1 2 3 4 We will test 5 values to get an idea of shape y = (0)2 + 4(0) + 3 = 0 y = (1)2 4(1) + 3 = 1 − y = (2)2 4(23)2 4(34)2 4(4) + 3 = 16 − − 0 + 3 = 3 Plug 0 in for x and evaluate − 4 + 3 = 0 Plug 1 in for x and evaluate − 1 Plug 2 in for x and evaluate 8 + 3 = 12 + 3 = 0 Plug 3 in for x and evaluate 16 + 3 = 3 Plug 4 in for x and evaluate − Our completed table. Plot points on graph Plot the points (0, 3), (1, 0), (2, (3, 0), and (4, 3). − 1), Connect curve. the dots with a smooth Our Solution When we have x2 in our equations, the graph will no longer be a straight line. Quadratics have a graph that looks like a U shape that is called a parabola. World View Note: The ﬁrst major female mathematician was Hypatia of Egypt who was born around 370 AD. She studied conic sections. The parabola is one type of conic section. 380 The above method to graph a parabola works for any equation, however, it can be very tedious to ﬁnd all the correct points to get the correct bend and shape. For this reason we identify several key points on a graph and in the equation to help us graph parabolas more eﬃciently. These key points are described below. Point A: y-intercept: Where the graph crosses the vertical y-axis. Points B and C: x-intercepts: Where the graph crosses the horizontal x-axis Point D: Vertex: The point where the graph curves and changes directions. B C A D We will use the following method to ﬁnd each of the points on our parabola. To graph the equation y = ax2 + b x + c, ﬁnd the following points 1. y-intercept: Found by making x = 0, this simpliﬁes down to y = c 2. x-intercepts: Found by making y = 0, this means solving 0 = ax2 + bx + c 3. Vertex: Let x = − 2a to ﬁnd x. Then plug this value into the equation to ﬁnd b y. After ﬁnding these points we can connect the dots with a smooth curve to ﬁnd our graph! Example 502. y = x2 + 4x + 3 Find the key points y = 3 y = c is the y intercept − 0 = x2 + 4x + 3 To ﬁnd x intercept we solve the equation − 0 = (x + 3)(x + 1) x + 3 = 0 and x + 1 = 0 1 1 Our x Factor Set each factor equal to zero Solve each equation intercepts 1 3 and (1) 2)2 + 4 To ﬁnd the vertex, ﬁrst use x = − 2a = − 2) + 3 Plug this answer into equation to ﬁnd y 8 + 3 Evaluate 1 The y 1) Vertex as a point coordinate − − y = 2, − ( − − − coordinate − 381 Graph the y-intercept at 3, the x3 and intercepts at 1, and the vertex at ( 1). Connect the dots with a smooth curve in a U shape to get our parabola. − − 2, − − Our Solution If the a in y = ax2 + bx + c is a negative value, the parabola will end up being an upside-down U. The process to graph it is identical, we just need to be very careful of how our signs operate. Remember, if a is negative, then ax2 will also be negative because we only square the x, not the a. Example 503. y = − 3x2 + 12x y = 9 9 − − Find key points intercept is y = c y − − 3)(x 3x2 + 12x 0 = − 3(x2 0 = − 3(x 0 = − 3 = 0 and x − 4x + 3) 1 and x = 1 Our x − − + 3 + 3 x 9 To ﬁnd x intercept solve this equation − Factor out GCF ﬁrst, then factor rest Set each factor with a varaible equal to zero Solve each equation intercepts − y = 12 3) = − − 12 x = − 6 2( − 3(2)2 + 12(2) 3(4) + 24 12 + 24 − y = − y = − b = 2 To ﬁnd the vertex, ﬁrst use x = − 2a 9 Plug this value into equation to ﬁnd y − 9 Evaluate − 9 − y = 3 (2, 3) Vertex as a point value of vertex − y coordinate − Graph the y-intercept at 9, the xintercepts at 3 and 1, and the vertex at (2, 3). Connect the dots with smooth curve in an upside-down U shape to get our parabola. − Our Solution 382 It is important to remember the graph of all quadratics is a parabola with the same U shape (they could be upside-down). If you plot your points and we cannot connect them in the correct U shape then one of your points must be wrong. Go back and check your work to be sure they are correct! Just as all quadratics (equation with y = x2) all have the same U-shape to them and all linear equations (equations such as y = x) have the same line shape when graphed, diﬀerent equations have diﬀerent shapes to them. Below are some common equations (some we have yet to cover!) with their graph shape drawn. Absolute Value y = x \| \| Cubic y = x3 Quadratic y = x2 Exponential y = ax Square Root y = x√ Logarithmic y = logax 383 9.11 Practice - Graphs of Quadratics Find the vertex and intercepts of the following quadratics. Use this information to graph the quadratic. 1) y = x2 − 3) y = 2x2 2x 8 − 12x + 10 − 2) y = x2 − 4) y = 2x2 2x 3 − 12x + 16 − 18 45 2x2 + 12x − 3x2 + 24x − x2 + 4x + 5 5) y = 7) y = − − 9) y = − 11) y = − 5 x2 + 6x − 2x2 + 16x 13) y = − − 15) y = 3x2 + 12x + 9 24 17) y = 5x2 19) y = − − 5x2 40x + 75 60x 175 − − 10 9 − − 6) y = − 2x2 + 12x 3x2 + 12x 8) y = − 10) y = − x2 + 4x − 2x2 + 16x 3 30 − 12) y = − 14) y = 2x2 + 4x 6 − 16) y = 5x2 + 30x + 45 18) y = 5x2 + 20x + 15 20) y = − 5x2 + 20x 15 − 384 Chapter 10 : Functions 10.1 Function Notation ..................................................................................386 10.2 Operations on Functions ........................................................................393 10.3 Inverse Functions ...................................................................................401 10.4 Exponential Functions ...........................................................................406 10.5 Logarithmic Functions ...........................................................................410 10.6 Application: Compound Interest ............................................................414 10.7 Trigonometric Functions ........................................................................420 10.8 Inverse Trigonometric Functions ............................................................428 385 10.1 Functions - Function Notation Objective: Idenﬁty functions and use correct notation to evaluate functions at numerical and variable values. There are many diﬀerent types of equations that we can work with in algebra. An equation gives the relationship between variables and numbers. Examples of several relationships are below: 3)2 (x − 9 (y + 2)2 4 − = 1 and y = x2 − 2x + 7 and y + x√ 7 = xy − There is a speical classiﬁcation of relationships known as functions. Functions have at most one output for any input. Generally x is the variable that we plug into an equation and evaluate to ﬁnd y. For this reason x is considered an input variable and y is considered an output variable. This means the deﬁnition of a function, in terms of equations in x and y could be said, for any x value there is at most one y value that corresponds with it. A great way to visualize t
his deﬁnition is to look at the graphs of a few relationships. Because x values are vertical lines we will draw a vertical line through the graph. If the vertical line crosses the graph more than once, that means we have too many possible y values. If the graph crosses the graph only once, then we say the relationship is a function. 386 Example 504. Which of the following graphs are graphs of functions? line Drawing a vertical through this graph will only cross the graph once, it is a function. Drawing a vertical line through this graph will cross the graph twice, once at top and once at bottom. This is not a function. line Drawing a vertical through this graph will cross the graph only once, it is a function. We can look at the above idea in an algebraic method by taking a relationship and solving it for y. If we have only one solution then it is a function. Example 505. Is 3x2 y = 5 a function? 3x2 − 3x2 + 5 − 1 1 − 3x2 − y = 1 − y = 3x2 − 5 − − − Solve the relation for y Subtract 3x2 from both sides Divide each term by 1 − Only one solution for y. It is a function Yes! Example 506. Is y2 x = 5 a function? − + x + x y2 = x + 5 y2 = y = ± ± x + 5√ x + 5√ p Solve the relation for y Add x to both sides Square root of both sdies Simplify Two solutions for y (one + , one ) − No! Not a function Once we know we have a function, often we will change the notation used to emphasis the fact that it is a function. Instead of writing y = , we will use function notation which can be written f (x) = . We read this notation “f of x”. So for 387 the above example that was a function, instead of writing y = 3x2 5, we could have written f (x) = 3x2 5. It is important to point out that f (x) does not mean f times x, it is mearly a notation that names the function with the ﬁrst letter (function f ) and then in parenthesis we are given information about what variables are in the function (variable x). The ﬁrst letter can be anything we want it to be, often you will see g(x) (read g of x). − − World View Note: The concept of a function was ﬁrst introduced by Arab mathematician Sharaf al-Din al-Tusi in the late 12th century Once we know a relationship is a function, we may be interested in what values can be put into the equations. The values that are put into an equation (generally the x values) are called the domain. When ﬁnding the domain, often it is easier to consider what cannot happen in a given function, then exclude those values. Example 507. Find the domain: f (x) = 1 3x − x2 + x 6 − 6 0 − 2 x2 + x (x + 3)(x x + 3 0 and x 3 3 − − x − 3, 2 Our Solution With fractions, zero can′t be in denominator Solve by factoring Set each factor not equal to zero Solve each equation The notation in the previous example tells us that x can be any value except for 3 and 2. If x were one of those two values, the function would be undeﬁned. − Example 508. Find the domain: f (x) = 3x2 x With this equation there are no bad values − All Real Numbers or R Our Solution In the above example there are no real numbers that make the function undeﬁned. This means any number can be used for x. Example 509. Find the domain: f (x) = 2x √ Square roots can′t be negative 3 − 388 2x 3 > 0 − + 3 + 3 2x > 3 2 2 x > 3 2 Set up an inequality Solve Our Solution The notation in the above example states that our variable can be 3 number larger than 3 ﬁned (without using imaginary numbers). 2 or any 2. But any number smaller would make the function unde- Another use of function notation is to easily plug values into functions. If we want to substitute a variable for a value (or an expression) we simply replace the variable with what we want to plug in. This is shown in the following examples. Example 510. f (x) = 3x2 − 2) = 3( f ( − f ( − 4x; ﬁnd f ( 2)2 4( − 4( 2) = 3(4) f ( − − − − − 2 in for x in the function Substitute 2) 2) Evaluate, exponents ﬁrst 2) Multiply − 2) = 12 + 8 Add − f ( 2) = 20 Our Solution − Example 511. h(x) = 32x − 6; ﬁnd h(4) 6 h(4) = 32(4) h(4) = 38 − 6 − Substitute 4 in for x in the function Simplify exponent, mutiplying ﬁrst Subtract in exponent h(4) = 32 Evaluate exponent h(4) = 9 Our Solution Example 512. k(a) = 2 \| k( a + 4 ; ﬁnd k( \| 7) = 2 − k( \| − 7) = 2 7( − 7) = 2(3) Multiply 7 in for a in the function Substitute − Add inside absolute values Evaluate absolute value 389 k( − 7) = 6 Our Solution As the above examples show, the function can take many diﬀerent forms, but the pattern to evaluate the function is always the same, replace the variable with what is in parenthesis and simplify. We can also substitute expressions into functions using the same process. Often the expressions use the same variable, it is important to remember each variable is replaced by whatever is in parenthesis. Example 513. g (x) = x4 + 1; ﬁnd g(3x) Replace x in the function with (3x) g(3x) = (3x)4 + 1 g(3x) = 81x4 + 1 Our Solution Simplify exponet Example 514. p(t) = t2 − t; ﬁnd p(t + 1) Replace each t in p(t) with (t + 1) Square binomial p(t + 1) = (t + 1)2 p(t + 1) = t2 + 2t + 1 − − p(t + 1) = t2 + 2t + 1 (t + 1) (t + 1) Distribute negative 1 Combine like terms t − − p(t + 1) = t2 + t Our Solution It is important to become comfortable with function notation and how to use it as we transition into more advanced algebra topics. 390 10.1 Practice - Function Notation Solve. 1) Which of the following is a function? a) c) e) y = 3x 7 − g) y√ + x = 2 b) d) f) y2 x2 = 1 − h) x2 + y2 = 1 Specify the domain of each of the following funcitons. 5x + 1 2) f (x) = − 4) s(t) = 1 t2 6) s(t) = 1 t2 + 1 2 8) f (x) = − x2 3x − 10 y(x) = x x2 − 25 4 − √ 3) f (x) = 5 4x − 5) f (x) = x2 − 7) f (x) = x √ 3x 4 − 16 − √ 9) h(x) = 3x − x2 − 12 25 391 Evaluate each function. 11) g(x) = 4x 4; Find g(0) − 3x + 1 13) f (x) = \| + 1; Find f (0) \| 17) f (t) = 3t 19) f (t) = \| − t + 3 2) 2; Find f ( − ; Find f (10) \| 15) f (n; Find f ( 6) − 16) f (n) = n 12) g(n) = 3 5− n; Find g(2) − 14) f (x) = x2 + 4; Find f ( · 9) − 3; Find f (10) − 1 − 3a 18) f (a) − 3; Find f (2) − 20) w(x) = x2 + 4x; Find w( 5) − 4x + 3; Find w(6) 21) w(n) = 4n + 3; Find w(2) 22) w(x) = − 23) w(n) = 2n+2; Find w( 2) 25) p(n) = n 3 \| − − ; Find p(7) \| 27) p(t) = t3 + t; Find p(4) − n 29) k(n) = 1 ; Find k(3) \| 31) h(x) = x3 + 2; Find h( − \| 33) h(x) = 3x + 2; Find h( 4x) 1 + x) − − 35) h(t) = 2 3t − − 1 + 2; Find h(n2) 37) g(x) = x + 1; Find g(3x) 39) g(x) = 5x; Find g( 3 − − x) 24) p(x) = x + 1; Find p(5) − \| 26) k(a) = a + 3; Find k( \| 1) − 2; Find k(2) · 42t+1 + 1; Find p( 28) k(x) = 2 − 42x − 30) p(t) = 2 − · 2) − 32) h(n) = 4n + 2; Find h(n + 2) 2a+3; Find h( a 4 ) 34) h(a) = 3 − 36) h(x) = x2 + 1; Find h( x 4 ) · 38) h(t) = t2 + t; Find h(t2) 40) h(n) = 5n − 1 + 1; Find h( n 2 ) 392 10.2 Functions - Operations on Functions Objective: Combine functions using sum, diﬀerence, product, quotient and composition of functions. Several functions can work together in one larger function. There are 5 common operations that can be performed on functions. The four basic operations on functions are adding, subtracting, multiplying, and dividing. The notation for these functions is as follows. Addition Subtraction Multiplication Division (f + g)(x) = f (x) + g(x) (f g(x) g)(x) = f (x) − − g)(x) = f (x)g(x) (f · f f (x) g g(x) (x) = When we do one of these four basic operations we can simply evaluate the two functions at the value and then do the operation with both solutions Example 515. f ( − x f (x) = x2 − g(x) = x + 1 ﬁnd (f + g)( 2 − 3) − Evaluate f and g at 3 − 3) = ( 3)2 − f ( − ( 3) 3) = 9 + 3 − − 2 Evaluate f at − 2 − 3) = 10 f ( − g( − 3) = ( g( − 3) + 1 Evaluate g at − 3) + g( (10) + ( − − 3) Add the two functions together 2) Add 8 Our Solution The process is the same regardless of the operation being performed. Example 516. h(x) = 2x k(x) = Find (h 4 − 3x + 1 k)(5) − · Evaluate h and k at 5 h(5) = 2(5) − 4 Evaluate h at 5 393 h(5) = 10 4 h(5) = 6 − k(5) = 3(5) + 1 Evaluate k at 5 − k(5) = − k(5) = 15 + 1 14 − h(5)k(5) Multiply the two results together (6)( 14) Multiply 84 Our Solution − − Often as we add, subtract, multiply, or divide functions, we do so in a way that keeps the variable. If there is no number to plug into the equations we will simply use each equation, in parenthesis, and simplify the expression. Example 517. f (x) = 2x 4 − g(x) = x2 Find (f − − x + 5 Write subtraction problem of functions g)(x) (2x − 2x 4) − f (x) (x2 g(x) Replace f (x) with (2x − − x + 5) Distribute the negative 5 Combine like terms 9 Our Solution − 4 − x2 + x − x2 + 3x − − − 3) and g(x) with (x2 x + 5) − The parenthesis are very important when we are replacing f (x) and g(x) with a variable. In the previous example we needed the parenthesis to know to distribute the negative. Example 518. 5 − Write division problem of functions f (x) = x2 g(x) = x f g Find 4x − 5 − (x) f (x) g(x) Replace f (x) with (x2 4x − − 5) and g(x) with (x 5) − (x2 − (x 4x − − 5) 5) To simplify the fraction we must ﬁrst factor 394 (x 5)(x + 1) 5) − (x − Divide out common factor of x 5 − x + 1 Our Solution Just as we could substitute an expression into evaluating functions, we can substitute an expression into the operations on functions. Example 519. f (x) = 2x 1 − g(x) = x + 4 Write as a sum of functions Find (f + g)(x2) f (x2) + g(x2) Replace x in f (x) and g(x) with x2 1] + [(x2) + 4] Distribute the + does not change the problem [2(x2) − 2x2 − 1 + x2 + 4 Combine like terms 3x2 + 3 Our Solution Example 520. f (x) = 2x 1 − g(x) = x + 4 Write as a product of functions Find (f g)(3x) · [2(3x) (6x 18x2 + 24x − − f (3x)g(3x) Replace x in f (x) and g(x) with 3x 1][(3x) + 4] Multiply our 2(3x) 1)(3x + 4) FOIL 3x 18x2 + 21x 4 Combine like terms 4 Our Solution − − − The ﬁfth operation of functions is called composition of functions. A composition of functions is a function inside of a function. The notation used for composition of functions is: (f ◦ g)(x) = f (g(x)) To calculate a composition of function we will evaluate the inner function and substitute the an
swer into the outer function. This is shown in the following example. 395 Example 521. a(x) = x2 2x + 1 − b(x) = x − b)(3) Find (a ◦ 5 Rewrite as a function in function a(b(3)) Evaluate the inner function ﬁrst, b(3) 2 This solution is put into a, a( 5 = 2) − b(3) = (3) 2)2 − 2( − a( − 2) = ( 2) + 1 Evaluate − a( − − 2) = 4 + 4 + 1 Add − a( − 2) = 9 Our Solution 396 We can also evaluate a composition of functions at a variable. In these problems we will take the inside function and substitute into the outside function. Example 522. x f (x) = x2 − g(x) = x + 3 Find (f g)(x) ◦ Rewrite as a function in function f (g(x)) Replace g(x) with x + 3 f (x + 3) Replace the variables in f with (x + 3) (x + 3)2 (x2 + 6x + 9) − − x2 + 6x + 9 (x + 3) Evaluate exponent (x + 3) Distribute negative 3 Combine like terms x − − x2 + 5x + 6 Our Solution f )(x) as the It is important to note that very rarely is (f following example will show, using the same equations, but compositing them in the opposite direction. g)(x) the same as (g ◦ ◦ Example 523. f (x) = x2 x − g(x) = x + 3 Find (g f )(x) ◦ Rewrite as a function in function g(f (x)) Replace f (x) with x2 x − g(x2 x) Replace the variable in g with (x2 x) − x) + 3 Here the parenthesis don′t change the expression x + 3 Our Solution − (x2 x2 − − World View Note: The term “function” came from Gottfried Wihelm Leibniz, a German mathematician from the late 17th century. 397 10.2 Practice - Operations on Functions Perform the indicated operations. 1) g(a) = a3 + 5a2 f(a) = 2a + 4 Find g(3) + f (3) 3) g(a) = 3a + 3 f (a) = 2a 2 Find (g + f )(9) − 5) g(x) = x + 3 f(x) = Find (g x + 4 f )(3) − − 7) g(x) = x2 + 2 f (x) = 2x + 5 Find (g f )(0) − 9) g(t) = t 3 − 3t3 + 6t h(t) = − Find g(1) + h(1) 11) h(t) = t + 5 g(t) = 3t 5 − g)(5) Find (h · 13) h(n) = 2n g(n) = 3n Find h(0) − − ÷ 2a g(a) = a2 + 3 Find ( f g )(7) 15) f (a) = − 1 5 g(0) 4 − 17) g(x) = − h(x) = 4x Find (g x3 2 − h)(x) − 19) f (x) = 3x + 2 − g(x) = x2 + 5x Find (f g)(x) − 21) g(x) = 4x + 5 h(x) = x2 + 5x Find g(x) h(x) · 2) f (x) = 3x2 + 3x − g(x) = 2x + 5 Find f ( 4) g( 4) − ÷ − 4) g(x) = 4x + 3 h(x) = x3 Find (g − h)( 2x2 − − 4x + 1 6) g(x) = − h(x) = 2x 1 − Find g(5) + h(5) − 1) 8) g(x) = 3x + 1 f (x) = x3 + 3x2 f (2) Find g(2) · 10) f (n) = n 5 g(n) = 4n + 2 Find (f + g)( − 12) g(a) = 3a 2 h(a) = 4a 2 Find (g + h) ( − − 8) − 10) − 14) g(x) = x2 2 h(x) = 2x + 5 Find g( − − 16) g(n) = n2 h(n) = 2n Find (g − 18) g(x) = 2x h(x) = x3 Find (g − 6) + h( 6) − 3 − 3 − h)(n) 3 2x2 + 2x − − h)(x) 4 20) g(t) = t h(t) = 2t Find (g − h)(t) · 22) g(t) = 2t2 h(t) = t + 5 Find g(t) − 5t − h(t) · 398 23) f (x) = x2 5x − g(x) = x + 5 Find (f + g)(x) 25) g(n) = n2 + 5 f (n) = 3n + 5 Find g(n) 27) g(a) = − f (a) = 3a + 5 Find ( g f )(a) f (n) ÷ 2a + 5 29) h(n) = n3 + 4n g(n) = 4n + 5 Find h(n) + g(n) 31) g(n) = n2 4n − h(n) = n 5 − h(n2) Find g(n2) · 33) f (x) = 2x g(x) = − − Find (f + g)( 3x 1 − x) 4 − 35) f (t) = t2 + 4t g(t) = 4t + 2 Find f (t2) + g(t2) 37) g(a) = a3 + 2a h(a) = 3a + 4 Find ( g x) h )( 39) f (n) = − g(n) = 2n + 1 Find (f − 3n2 + 1 g)( n 3 ) − 4x + 1 41) f (x) = g(x) = 4x + 3 Find (f g)(9) − ◦ 43) h(a) = 3a + 3 g(a) = a + 1 Find (h g)(5) ◦ 45) g(x) = x + 4 h(x) = x2 1 − h)(10) Find (g ◦ 24) f (x) = 4x g(x) = 3x2 Find (f + g)(x) − − 4 5 26) f (x) = 2x + 4 g(x) = 4x 5 g(x) Find f (x) − − 28) g(t) = t3 + 3t2 h(t) = 3t Find g(t) 5 h(t) − − 30) f (x) = 4x + 2 g(x) = x2 + 2x Find f (x) ÷ 32) g(n) = n + 5 h(n) = 2n Find (g − h)( · g(x) 5 − 3n) 34) g(a) = 2a − h(a) = 3a Find g(4n) h(4n) ÷ 36) h(n) = 3n 2 − 3n2 g(n) = − Find h( n 3 ) 4n − g( n 3 ) ÷ 4x + 2 5 38) g(x) = − h(x) = x2 − Find g(x2) + h(x2) 40) f (n) = 3n + 4 g(n) = n3 5n − g( n Find f ( n 2 ) 2 ) − 42) g(x) = x Find (g 1 − g)(7) ◦ 44) g(t) = t + 3 h(t) = 2t 5 − h)(3) Find (g ◦ 46) f (a) = 2a 4 g(a) = a2 + 2a g)( Find (f − ◦ 4) − 399 47) f (n) = 4n + 2 − g(n) = n + 4 Find (f g)(9) ◦ 49) g(x) = 2x 4 − h(x) = 2x3 + 4x2 Find (g h)(3) ◦ 51) g(x) = x2 5x h(x) = 4x + 4 Find (g h)(x) − 53) f (a) = − g(a) = 4a Find (f 2a + 2 g)(a) ◦ ◦ 55) g(x) = 4x + 4 f (x) = x3 1 − f )(x) Find (g ◦ 57) g(x) = x + 5 3 − f )(x) − f (x) = 2x Find (g ◦ 59) f (t) = 4t + 3 g(t) = − Find (f ◦ 4t 2 − g)(t) 48) g(x) = 3x + 4 h(x) = x3 + 3x h)(3) Find (g ◦ 50) g(a) = a2 + 3 g)( Find (g ◦ 3) − 52) g(a) = 2a + 4 4a + 5 h)(a) h(a) = − Find (g ◦ t 4 − g)(t) − Find (g ◦ 54) g(t) = 4n − 56) f (n) = 2n2 g(n) = n + 2 Find (f − g)(n) ◦ 58) g(t) = t3 f (t) = 3t Find (g t − 4 − f )(t) ◦ 60) f (x) = 3x 4 − g(x) = x3 + 2x2 g)(x) Find (f ◦ 400 10.3 Functions - Inverse Functions Objective: Identify and ﬁnd inverse functions. When a value goes into a function it is called the input. The result that we get when we evaluate the function is called the output. When working with functions sometimes we will know the output and be interested in what input gave us the output. To ﬁnd this we use an inverse function. As the name suggests an inverse function undoes whatever the function did. If a function is named f (x), the 1(x) (read “f inverse of x”). The negative one is inverse function will be named f − not an exponent, but mearly a symbol to let us know that this function is the inverse of f . World View Note: The notation used for functions was ﬁrst introduced by the great Swiss mathematician, Leonhard Euler in the 18th century. 1(x) = x For example, if f (x) = x + 5, we could deduce that the inverse function would be 5. If we had an input of 3, we could calculate f (3) = (3) + 5 = 8. Our f − output is 8. If we plug this output into the inverse function we get f − 5 = 3, which is the original input. 1(8) = (8) − − Often the functions are much more involved than those described above. It may be diﬃcult to determine just by looking at the functions if they are inverses. In order to test if two functions, f (x) and g(x) are inverses we will calculate the composition of the two functions at x. If f changes the variable x in some way, then g undoes whatever f did, then we will be back at x again for our ﬁnal solution. In otherwords, if we simplify (f g)(x) the solution will be x. If it is anything but x the functions are not inverses. ◦ 401 Example 524. Are f (x) = 3x + 4 3√ and g(x) = x3 4 − 3 f inverses? Caculate composition x3 f (g(x)) Replace g(x) with 4 − 3 3 s 3 x3 − 3 3√ x3 − x3 4 − 3 Substitute x3 4 − 3 4 + 4 Divide out the 3′s 4 + 4 3√ x3 x Combine like terms Take cubed root Simpliﬁed to x! Yes, they are inverses! Our Solution for variable in f Example 525. Are h(x) = 2x + 5 and g(x) = x 2 − 2 5 inverses? Calculate composition x 2 − for variable in h h(g(x)) Replace g(x) with x 2 − 5 + 5 Distrubte 2 Substitute 10 + 5 Combine like terms 5 Did not simplify to x x No, they are not inverses Our Solution − − Example 526. Are f (x) = 3x 2 − 4x + 1 and g(x) = x + 2 4x 3 − inverses? Calculate composition f (g(x)) Replace g(x) with x + 2 4x 3 − f Substitute x + 2 4x 3 − x + 2 4x 3 − for variable in f x + 2 4x 3 − x + 2 4x 3 − 3 4 Distribute 3 and 4 into numerators 2 − + 1 402 2 3x + 6 3 − 4x + 8 3 4x − 4x + 1 − (3x + 6)(3 4x 3 (4x + 8)(3 4x 3 − − − − 4x) 4x) 2(3 − + 1(3 3x + 6 2(3 4x + 8 + 1(3 − 4x) 4x) 4x) 4x) − − − − 3x + 6 4x + 8 + 3 6 + 8x 4x − − 11x 11 Multiply each term by LCD: 3 4x − Reduce fractions Distribute Combine like terms Divide out 11 x Simpliﬁed to x! Yes, they are inverses Our Solution While the composition is useful to show two functions are inverses, a more common problem is to ﬁnd the inverse of a function. If we think of x as our input and y as our output from a function, then the inverse will take y as an input and give x as the output. This means if we switch x and y in our function we will ﬁnd the inverse! This process is called the switch and solve strategy. Switch and solve strategy to ﬁnd an inverse: 1. Replace f (x) with y 2. Switch x and y’s 3. Solve for y 4. Replace y with f − 1(x) Example 527. Find the inverse of f (x) = (x + 4)3 y = (x + 4)3 x = (y + 4)3 + 2 2 Replace f (x) with y − Switch x and y 2 − Solve for y 2 − + 2 Add 2 to both sides 403 x + 2 = (y + 4)3 Cube root both sides 3√ Subtract 4 from both sides = √ 1(x) = x + 2 3√ − − − 4 = y Replace y with f − 4 Our Solution 1(x) f − Example 528. Find the inverse of g(x) = 2x 3 − 4x + 2 Replace g(x) with y y = 2x 3 − 4x + 2 x = 2y 3 − 4y + 2 Switch x and y Multiply by (4y + 2) x(4y + 2) = 2y 4xy + 2x = 2y 4xy + 3 2x + 3 = 2y − 4xy + 3 4xy − − 3 Distribute 3 Move all y ′s to one side, rest to other side Subtract 4xy and add 3 to both sides Factor out y − − 2x + 3 = y(2 4x 2 2 4x) Divide by 2 4x 4x − − − − 2x + 3 4x 2 − = y Replace y with g− 1(x) g− 1(x) = 2x + 3 4x 2 − Our Solution In this lesson we looked at two diﬀerent things, ﬁrst showing functions are inverses by calculating the composition, and second ﬁnding an inverse when we only have one function. Be careful not to get them backwards. When we already have two functions and are asked to show they are inverses, we do not want to use the switch and solve strategy, what we want to do is calculate the inverse. There may be several ways to represent the same function so the switch and solve strategy may not look the way we expect and can lead us to conclude two functions are not inverses when they are in fact inverses. 404 10.3 Practice - Inverse Functions State if the given functions are inverses. − 5√ 3 3 1) g(x) = f (x) = x5 − x − − x 1 3) f (x) = − − x 2 − 2x + 1 g(x) = − x 1 − 10x + 5 5 − 10 − f (x) = x 5) g(x) = − 7) f (x) = 2 x + 3 − g(x) = 3x + 2 x + 2 5 1 9) g(x) = x f (x) = 2x5 + 1 q − 2 Find the inverse of each functions. x 2) g(x) = 4 − x f (x) = 4 x 2 2x − x 4) h(x) = − f (x) = − 2 x + 2 6) f (x) = x 5 − 10 h(x) = 10x + 5 5 8) f (x) = x + 1 2 g(x) = 2x5 q 1 − 10) g(x) = 8 + 9x f (x) = 5x 9 2 − 2 11) f (x) = (x 2)5 + 3 12) g(x) = x + 1 3√ + 2 − x − 5 (x − 1 13) g(x) = 4 x + 2 15) f (x) = − 2x − x + 2 2 17) f (x) = 10 1)3 − 3)3 19) g(x) = − 21) f (x) = (x 23) g(x) = x x − 25) f (x) = x 1 − x + 1 27) g(x) = 8 29) g(x) = 5x − 4 5x + 1 − 31) g(x) = − 33) h(x) = 4 − 1 + x3 4x3√ 2 35) f
(x) = x + 1 x + 2 3 14) f (x) = − x 3 − 16) g(x) = 9 + x 3 18) f (x) = 5x 20) f (x) = 12 − 2 − 4 15 3x 22) g(x) = − 5 x + 2 2 2x q 3 24) f (x) = − − x + 3 26) h(x) = x x + 2 28) g(x) = − 30) f (x) = 5x x + 2 3 5 − 4 32) f (x) = 3 34) g(x) = (x − 2x5 1)3 + 2 36) f (x) = − − 1 x + 1 37) f (x) = 7 − x − x 39) g(x) = 3x 2 − 38) f (x) = − 40) g(x) = − 3x 4 2x + 1 3 405 10.4 Functions - Exponential Functions Objective: Solve exponential equations by ﬁnding a common base. As our study of algebra gets more advanced we begin to study more involved functions. One pair of inverse functions we will look at are exponential functions and logarithmic functions. Here we will look at exponential functions and then we will consider logarithmic functions in another lesson. Exponential functions are functions where the variable is in the exponent such as f (x) = ax. (It is important not to confuse exponential functions with polynomial functions where the variable is in the base such as f (x) = x2). World View Note One common application of exponential functions is population growth. According to the 2009 CIA World Factbook, the country with the highest population growth rate is a tie between the United Arab Emirates (north of Saudi Arabia) and Burundi (central Africa) at 3.69%. There are 32 countries with negative growth rates, the lowest being the Northern Mariana Islands (north of Australia) at 7.08%. − Solving exponetial equations cannot be done using the skill set we have seen in the past. For example, if 3x = 9, we cannot take the x root of 9 because we do not know what the index is and this doesn’t get us any closer to ﬁnding x. However, we may notice that 9 is 32. We can then conclude that if 3x = 32 then x = 2. This is the process we will use to solve exponential functions. If we can re-write a problem so the bases match, then the exponents must also match. − Example 529. 52x+1 = 125 Rewrite 125 as 53 52x+1 = 53 2x + 1 = 3 1 1 Same base, set exponents equal Solve Subtract 1 from both sides − − 2x = 2 Divide both sides by 2 2 x = 1 Our Solution 2 Sometimes we may have to do work on both sides of the equation to get a common base. As we do so, we will use various exponent properties to help. First we will use the exponent property that states (ax)y = axy. 406 Example 530. 83x = 32 Rewrite 8 as 23 and 32 as 25 (23)3x = 25 Multiply exponents 3 and 3x 29x = 25 9x = 5 9 9 5 9 x = Same base, set exponents equal Solve Divide both sides by 9 Our Solution As we multiply exponents we may need to distribute if there are several terms involved. Example 531. 273x+5 = 814x+1 Rewrite 27 as 33 and 81 as 34 (92 would not be same base) (33)3x+5 = (34)4x+1 Multiply exponents 3(3x + 5) and 4(4x + 1) 39x+15 = 316x+4 Same base, set exponents equal 9x + 15 = 16x + 4 Move variables to one side 9x Subtract 9x from both sides Subtract 4 from both sides − − 9x 15 = 7x + 4 4 4 − − 7 11 = 7x Divide both sides by 7 7 11 7 = x Our Solution Another useful exponent property is that negative exponents will give us a reciprocal, 1 n an = a− Example 532. 2x = 37x − 1 Rewrite 1 9 as 3− 2 (negative exponet to ﬂip) − 1 Multiply exponents 1 2 and 2x Same base, set exponets equal Subtract 7x from both sides − − 1 1 9 (3− 2)2x = 37x 4x = 37x 3− 4x = 7x 7x 7x − 11x = 11 − − − − − 11 − 1 Divide by − 11 − 1 x = 11 Our Solution If we have several factors with the same base on one side of the equation we can add the exponents using the property that states axay = ax+ y. 407 Example 533. 54x · 52x 56x − 1 = 53x+11 Add exponents on left, combing like terms 1 = 53x+11 Same base, set exponents equal − 1 = 3x + 11 Move variables to one sides Subtract 3x from both sides 3x 6x 3x − − − 3x 1 = 11 Add 1 to both sides − + 1 + 1 3x = 12 Divide both sides by 3 3 3 x = 4 Our Solution It may take a bit of practice to get use to knowing which base to use, but as we practice we will get much quicker at knowing which base to use. As we do so, we will use our exponent properties to help us simplify. Again, below are the properties we used to simplify. (ax)y = axy and 1 an = a−n and axay = ax+ y We could see all three properties used in the same problem as we get a common base. This is shown in the next example. Example 534. 162x − 5 · (24)2x − 5 3x+1 = 32 1 4 2)3x+1 = 25 (2− 6x 20 · 28x − · 2 = 25 2− · x+3 x+3 1 2 1 · (2− Write with a common base of 2 Multiply exponents, distributing as needed − 22x x 2− − · x+2 22 = 2− 3 Add exponents, combining like terms Same base, set exponents equal − 22 = 2x + x − − + x x + 2 Move variables to one side Add x to both sides 22 = 2 Add 22 to both sides 3x − + 22 + 22 3x = 24 Divide both sides by 3 3 3 x = 8 Our Solution All the problems we have solved here we were able to write with a common base. However, not all problems can be written with a common base, for example, 2 = 10x, we cannot write this problem with a common base. To solve problems like this we will need to use the inverse of an exponential function. The inverse is called a logarithmic function, which we will discuss in another secion. 408 10.4 Practice - Exponential Functions Solve each equation. 1) 31 − 2n = 31 − 3n 3) 42a = 1 5) ( 1 25)− k = 125− 2k − 2 7) 62m+1 = 1 36 9) 6− 3x = 36 11) 64b = 25 13) ( 1 4)x = 16 15) 43a = 43 17) 363x = 2162x+1 19) 92n+3 = 243 21) 33x 2 = 33x+1 − 23) 3− 2x = 33 25) 5m+2 = 5− m 27) ( 1 36)b 1 = 216 − 29) 62 − 2x = 62 31) 4 · 3n 2− − 1 = 1 4 33) 43k − 3 35) 9− 2x · 42 2k = 16− k − · ( 1 243)3x = 243− x 37) 64n − 2 39) 5− 3n − · 3 16n+2 = ( 1 4 )3n 1 − 52n = 1 · 2) 42x = 1 16 4) 16− 3p = 64− 3p 6) 625− n 2 = 1 125 − 8) 62r 3 = 6r − − 3 10) 52n = 5− n 12) 216− 3v = 363v 14) 27− 2n 1 = 9 − 16) 4− 3v = 64 18) 64x+2 = 16 20) 162k = 1 64 22) 243p = 27− 3p 24) 42n = 42 − 3n 26) 6252x = 25 28) 2162n = 36 30) ( 1 4)3v 2 = 641 − − v 32) 216 6−2a = 63a 34) 322p − 2 8p = ( 1 2 )2p · 36) 32m 38) 32 − · x 33m = 1 33m = 1 · 40) 43r 4− 3r = 1 64 · 409 10.5 Functions - Logarithmic Functions Objective: Convert between logarithms and exponents and use that relationship to solve basic logarithmic equations. n The inverse of an exponential function is a new function known as a logarithm. Lograithms are studied in detail in advanced algebra, here we will take an introductory look at how logarithms works. When working with radicals we found that anm√ could be written as there were two ways to write radicals. The expression a m. Each form has its advantages, thus we need to be comfortable using both the radical form and the rational exponent form. Similarly an exponent can be written in two forms, each with its own advantages. The ﬁrst form we are very familiar with, bx = a, where b is the base, a can be thought of as our answer, and x is the exponent. The second way to write this is with a logarithm, logba = x. The word “log” tells us that we are in this new form. The variables all still mean the same thing. b is still the base, a can still be thought of as our answer. Using this idea the problem 52 = 25 could also be written as log525 = 2. Both mean the same thing, both are still the same exponent problem, but just as roots can be written in radical form or rational exponent form, both our forms have their own advantages. The most important thing to be comfortable doing with logarithms and exponents is to be able to switch back and forth between the two forms. This is what is shown in the next few examples. Example 535. Write each exponential equation in logarithmic form m3 = 5 Identify base, m, answer, 5, and exponent 3 logm5 = 3 Our Solution 72 = b Identify base, 7, answer, b, and exponent, 2 log7b = 2 Our Solution 4 = 2 3 log 2 3 16 81 16 81 Identify base, 2 3 , answer, 16 81 , and exponent 4 = 4 Our Solution Example 536. Write each logarithmic equation in exponential form log416 = 2 Identify base, 4, answer, 16, and exponent, 2 42 = 16 Our Solution 410 log3x = 7 Identify base, 3, answer, x, and exponent, 7 37 = x Our Solution log93 = 1 1 2 Identify base, 9, answer, 3, and exponent, 1 2 9 2 = 3 Our Solution Once we are comfortable switching between logarithmic and exponential form we are able to evaluate and solve logarithmic expressions and equations. We will ﬁrst evaluate logarithmic expressions. An easy way to evaluate a logarithm is to set the logarithm equal to x and change it into an exponential equation. Example 537. Example 538. Evaluate log264 Set logarithm equal to x log264 = x Change to exponent form 2x = 64 Write as common base, 64 = 26 2x = 26 Same base, set exponents equal x = 6 Our Solution Evaluate log1255 Set logarithm equal to x log1255 = x Change to exponent form 125x = 5 Write as common base, 125 = 53 (53)x = 5 Multiply exponents 53x = 5 3x = 1 3 3 1 3 x = Same base, set exponents equal (5 = 51) Solve Divide both sides by 3 Our Solution Example 539. Evaluate log3 Set logarithm equal to x 1 27 = x Change to exponent form 1 27 Write as common base, = 3− 3 Same base, set exponents equal 3 Our Solution log3 1 27 1 3x = 27 3x = 3− 3 x = − World View Note: Dutch mathematician Adriaan Vlacq published a text in 1628 which listed logarithms calculated out from 1 to 100,000! 411 Solve equations with logarithms is done in a very similar way, we simply will change the equation into exponential form and try to solve the resulting equation. Example 540. Example 541. Example 542. log5x = 2 Change to exponential form 52 = x Evaluate exponent 25 = x Our Solution log2(3x + 5) = 4 Change to exponential form 24 = 3x + 5 Evaluate exponent 16 = 3x + 5 Solve 5 5 Subtract 5 from both sides − − 3 11 = 3x Divide both sides by 3 3 11 3 = x Our Solution logx8 = 3 Change to exponential form x3 = 8 Cube root of both sides x = 2 Our Solution There is one base on a logarithm that gets used more often than any other base, base 10. Similar to square roots not writting the common index of 2 in the radical, we don’t write the common base of 10 in the logarithm. So if we are working on a problem with no base written we will always assume that base is base 10. Example 543. 2 Rewrite as exp

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