SECTION.4 Equations Quadratic in Form 785 In Eercises, without solving the equation, determine the number and type of solutions... In Eercises 3 4, write a quadratic equation in standard form with the given solution set. 3. + 5 + 4 = 0 0 + 4 = 5-5 e -, 3 4 f 4. E -3, 3F 5. A company manufactures and sells bath cabinets. The function P = - + 50-445 models the company s daily profit, P, when cabinets are manufactured and sold per day. How many cabinets should be manufactured and sold per day to maimize the company s profit? What is the maimum daily profit? 6. Among all pairs of numbers whose sum is -8, find a pair whose product is as large as possible. What is the maimum product? 7. The base of a triangle measures 40 inches minus twice the measure of its height. For what measure of the height does the triangle have a maimum area? What is the maimum area? SECTION.4 Equations Quadratic in Form Objective Solve equations that are quadratic in form. NOT AVAILABLE FOR ELECTRONIC VIEWING How important is it for you to have a clean house? The percentage of people who find this to be quite important varies by age. In the eercise set, you will work with a function that models this phenomenon. Your work will be based on equations that are not quadratic, but that can be written as quadratic equations using an appropriate substitution. Here are some eamples: Given Equation Substitution New Equation 4-0 + 9 = 0 or - 0 + 9 = 0 5 5 A 3 + 3 + = 0 or 3 B + 3 + = 0 u = u = 3 u - 0u + 9 = 0 5u + u + = 0 ISBN 0-558-957-4 Solve equations that are quadratic in form. An equation that is quadratic in form is one that can be epressed as a quadratic equation using an appropriate substitution. Both of the preceding given equations are quadratic in form. In an equation that is quadratic in form, the variable factor in one term is the square of the variable factor in the other variable term. The third term is a constant. By letting u equal the variable factor that reappears squared, a quadratic equation in u will result. Now it s easy. Solve this quadratic equation for u. Finally, use your substitution to find the values for the variable in the given equation. Eample shows how this is done. Introductory & Intermediate Algebra for College Students, Third Edition, by Robert Blitzer. Published by Prentice Hall. Copyright 009 by Pearson Education, Inc.
786 CHAPTER Quadratic Equations and Functions EXAMPLE Solve: 4-8 - 9 = 0. Solution Can you see that the variable factor in one term is the square of the variable factor in the other variable term? 4-8 -9=0 4 is the square of : ( ) = 4. We will let u equal the variable factor that reappears squared. Thus, Using Technology Graphic Connections The graph of y = 4-8 - 9 has -intercepts at -3 and 3. This verifies that the real solutions of 4-8 - 9 = 0 are -3 and 3. The imaginary solutions, -i and i, are not shown as intercepts. -intercept: 3 -intercept: 3 [ 5, 5, ] by [ 5, 0, 5] let u =. Now we write the given equation as a quadratic equation in u and solve for u. 4-8 - 9 = 0-8 - 9 = 0 The given equation contains and squared. u - 8u - 9 = 0 u - 9u + = 0 u - 9 = 0 or u + = 0 Let u =. Replace with u. Apply the zero-product principle. u = 9 u = - Solve for u. We re not done! Why not? We were asked to solve for and we have values for u. We use the original substitution, u =, to solve for. Replace u with in each equation shown, namely u = 9 and u = -. = 9 = - = ;9 = ;- Apply the square root property. = ;3 = ;i Substitute these values into the given equation and verify that the solutions are -3, 3, -i, and i. The solution set is 5-3, 3, -i, i6. The graph in the Using Technology bo shows that only the real solutions, -3 and 3, appear as -intercepts. CHECK POINT Solve: 4-5 + 6 = 0. If checking proposed solutions is not overly cumbersome, you should do so either algebraically or with a graphing utility. The Using Technology bo shows a check of the two real solutions in Eample. Are there situations when solving equations quadratic in form where a check is essential? Yes. If at any point in the solution process both sides of an equation are raised to an even power, a check is required. Etraneous solutions that are not solutions of the given equation may have been introduced. EXAMPLE Solve: - - 0 = 0. Solution To identify eponents on the terms, let s rewrite as. The equation can be epressed as - - 0 = 0. ISBN 0-558-957-4 Introductory & Intermediate Algebra for College Students, Third Edition, by Robert Blitzer. Published by Prentice Hall. Copyright 009 by Pearson Education, Inc.
SECTION.4 Equations Quadratic in Form 787 By epressing the equation as - - 0 = 0, we can see that the variable factor in one term is the square of the variable factor in the other variable term. - -0=0 is the square We will let u equal the variable factor that reappears squared. Thus, let u =. Now we write the given equation as a quadratic equation in u and solve for u. of : ( ) =. - - 0 = 0 - - 0 = 0 - - 0 = 0 u - u - 0 = 0 u - 5u + = 0 u - 5 = 0 or u + = 0 This is the given equation in eponential form. The equation contains and squared. Let u =. Replace with u. Set each factor equal to zero. u = 5 u = - Solve for u. Use the original substitution, u =, to solve for. Replace u with in each of the preceding equations. = 5 or = - Replace u with. 5 Q R = a Both sides are raised to even powers. We must check. = 5 4 b Q R =( ) = 4 Solve for by squaring both sides of each equation. Square 5 and -. It is essential to check both proposed solutions in the original equation. Check 5 4 : - - 0 = 0 # 5 4-5 A 4-0 0 5-5 - 0 0 0-0 0 0 = 0, true Check 4: - - 0 = 0 # 4-4 - 0 0 8 - - 0 0 6-0 0-4 = 0, false ISBN 0-558-957-4 The check indicates that 4 is not a solution. It is an etraneous solution brought about 5 by squaring each side of one of the equations. The only solution is and the solution set is e 5 4 4 f. CHECK POINT Solve: - - 8 = 0. Introductory & Intermediate Algebra for College Students, Third Edition, by Robert Blitzer. Published by Prentice Hall. Copyright 009 by Pearson Education, Inc.
788 CHAPTER Quadratic Equations and Functions The equations in Eamples and can be solved by methods other than using substitutions. 4-8 -9=0 - -0=0 This equation can be solved directly by factoring: ( 9)( + ) = 0. This equation can be solved by isolating the radical term: 0 =. Then square both sides. In the eamples that follow, solving the equations by methods other than first introducing a substitution becomes increasingly difficult. EXAMPLE 3 Solve: - 5 + 3-5 - 0 = 0. Solution This equation contains - 5 and - 5 squared. We let u = - 5. - 5 + 3-5 - 0 = 0 u + 3u - 0 = 0 u + 5u - = 0 u + 5 = 0 or u - = 0 u = -5 u = Let u = - 5. Set each factor equal to zero. Solve for u. Study Tip Solve Eample 3 by first simplifying the given equation s left side. Then factor out and solve the resulting equation. Do you get the same solutions? Which method, substitution or first simplifying, is faster? Use the original substitution, u = - 5, to solve for. Replace u with - 5 in each of the preceding equations. - 5 = -5 or - 5 = Replace u with - 5. = 0 = 0 = 7 = ;7 Solve for by isolating. Apply the square root property. Although we did not raise both sides of an equation to an even power, checking the three proposed solutions in the original equation is a good idea. Do this now and verify that the solutions are - 7, 0, and 7, and the solution set is E - 7, 0, 7F. CHECK POINT 3 Solve: - 4 + - 4-6 = 0. EXAMPLE 4 Solve: 0 - + 7 - + = 0. Solution The variable factor in one term is the square of the variable factor in the other variable term. 0 +7 +=0 is the square of : ( ) =. We will let u equal the variable factor that reappears squared. Thus, let u = -. ISBN 0-558-957-4 Introductory & Intermediate Algebra for College Students, Third Edition, by Robert Blitzer. Published by Prentice Hall. Copyright 009 by Pearson Education, Inc.
SECTION.4 Equations Quadratic in Form 789 Now we write the given equation as a quadratic equation in u and solve for u. 0 - + 7 - + = 0 0 - + 7 - + = 0 0u + 7u + = 0 5u + u + = 0 5u + = 0 or u + = 0 5u = - u = - The equation contains - and - squared. Let u = -. Set each factor equal to zero. Solve each equation for u. u = - 5 u = - Use the original substitution, u = -, to solve for. Replace u with in each of the preceding equations. - - = - 5 ( ) =a b 5 = 5 or - = - ( ) =a b = Replace u with -. Solve for by raising both sides of each equation to the - power. A 5B = 5 = 5 A B = = We did not raise both sides of an equation to an even power.a check will show that both -5 and - are solutions of the original equation. The solution set is 5-5, -6. CHECK POINT 4 Solve: - + - - = 0. EXAMPLE 5 Solve: 5 3 + 3 + = 0. Solution The variable factor in one term is the square of the variable factor in the other variable term. 3 3 5 + +=0 We will let u equal the variable factor that reappears squared. Thus, let u = 3. Now we write the given equation as a quadratic equation in u and solve for u. 3 is the square of : ( ) 3 3 = 3. ISBN 0-558-957-4 5 3 + 3 + = 0 5 A 3 B + A 3 B + = 0 5 u + u + = 0 5u + u + = 0 5u + = 0 or u + = 0 u = - 5 u = - The given equation contains 3 and 3 squared. Let u = 3. Set each factor equal to 0. Solve for u. Use the original substitution, u = 3, to solve for. Replace u with 3 in each of the preceding equations. Introductory & Intermediate Algebra for College Students, Third Edition, by Robert Blitzer. Published by Prentice Hall. Copyright 009 by Pearson Education, Inc.
790 CHAPTER Quadratic Equations and Functions 3 = - 5 A 3 B 3 = a - 3 5 b or 3 = - A 3 B 3 = - 3 Replace u with in u = - 3 and u = -. 5 Solve for by cubing both sides of each equation. = - 5 = -8 We did not raise both sides of an equation to an even power. A check will show that both -8 and - are solutions of the original equation, 5 3 + 3 5 + = 0. The solution set is E -8, - 5F. CHECK POINT 5 Solve: 3 3-3 - 4 = 0..4 EXERCISE SET Practice Eercises In Eercises 3, solve each equation by making an appropriate substitution. If at any point in the solution process both sides of an equation are raised to an even power, a check is required.. 4-5 + 4 = 0. 4-3 + 36 = 0 3. 4 - + 8 = 0 4. 4-9 + 0 = 0 5. 4 + = 8 6. 4 + 4 = 5 7. + - = 0 8. + - 6 = 0 9. - 4 - = 0 0. - 6 + 8 = 0. - 3 + 40 = 0. - 7-30 = 0 3. - 5-4 - 5 - = 0 4. + 3 + 7 + 3-8 = 0 5. - - - = 6. - - - = 6 7. + 3-8 + 3-0 = 0 8. - - - + 4 = 0 9. - - - - 0 = 0 0. - - - - 6 = 0. - - 7 - + 3 = 0. 0 - + 9 - + = 0 3. - - 4 - = 3 4. - - 6 - = -4 5. 6. 7. 8. 3-3 - 6 = 0 3 + 3-3 = 0 5 + 5-6 = 0 5 + 5 - = 0 9. - 4 = 30. - 5 4 = 3 3. a - 8 b + 5a - 8 b - 4 = 0 3. a - 0 b + 6a - 0 b - 7 = 0 ISBN 0-558-957-4 Introductory & Intermediate Algebra for College Students, Third Edition, by Robert Blitzer. Published by Prentice Hall. Copyright 009 by Pearson Education, Inc.
SECTION.4 Equations Quadratic in Form 79 In Eercises 33 38, find the -intercepts of the given function, f. Then use the -intercepts to match each function with its graph. [The graphs are labeled a through f.] 33. f = 4-5 + 4 34. f = 4-3 + 36 35. f = 3 + 6-3 36. f = - - - - 6 37. f = + - 9 + + 0 38. f = + + 5 + - 3 a. c. e. [ 5, 5, ] by [ 0, 40, 5] [ 5, 5, ] by [ 4, 0, ] [, 0, ] by [ 3, 3, ] b. d. f. [ 3, 3, ] by [ 0, 0, ] [ 6, 3, ] by [ 0, 0, ] [, 6, ] by [, 0, ] ISBN 0-558-957-4 Practice PLUS 39. Let f = + 3 - - 0 + 3 -. Find all such that f = -6. 40. Let f = + - - 7 + -. Find all such that f = -6. 4. Let f = 3a Find all such that + b + 5a + b. f =. 4. Let f = 3 + 3 3. Find all such that f =. 43. Let f = and g = 3 Find all - 4 A - 4-36. such that f = g. 44. Let f = and g = - Find all - + 0 A -. such that f = g. 45. Let f = 3-4 - and g = 6-4 -. Find all such that f eceeds g by. 46. Let f = 6a and g = 5a Find all - 3 b - 3 b. such that f eceeds g by 6. Application Eercises How important is it for you to have a clean house? The bar graph indicates that the percentage of people who find this to be quite important varies by age. The percentage, P, who find having a clean house very important can be modeled by the function where is the number of years a person s age is above or below 40. Thus, is positive for people over 40 and negative for people under 40. Use the function to solve Eercises 47 48. Percentage for Whom Having a Clean House Is Very Important P = 0.04 + 40-3 + 40 + 04, 00 75 50 5 58% 0 9 Source: Soap and Detergent Association The Importance of Having a Clean House, by Age 49% 47% 5% 30 39 40 49 50 59 Age Group 67% 60 69 47. According to the model, at which ages do 60% of us feel that having a clean house is very important? Substitute 60 for P and solve the quadratic-in-form equation. How well does the function model the data shown in the bar graph? 48. According to the model, at which ages do 50% of us feel that having a clean house is very important? Substitute 50 for P and solve the quadratic-in-form equation. How well does the function model the data shown in the bar graph? Introductory & Intermediate Algebra for College Students, Third Edition, by Robert Blitzer. Published by Prentice Hall. Copyright 009 by Pearson Education, Inc.
79 CHAPTER Quadratic Equations and Functions Writing in Mathematics 49. Eplain how to recognize an equation that is quadratic in form. Provide two original eamples with your eplanation. 50. Describe two methods for solving this equation: Technology Eercises - 5 + 4 = 0. 5. Use a graphing utility to verify the solutions of any five equations in Eercises 3 that you solved algebraically. The real solutions should appear as -intercepts on the graph of the function related to the given equation. Use a graphing utility to solve the equations in Eercises 5 59. Check by direct substitution. 5. 6-7 3-8 = 0 53. 3 - - - 4 - - + = 0 54. 4-0 + 9 = 0 55. + 6 = 8 56. + = 5 + + 3 57. - 3 + - 3-4 = 0 58. + 4 4 = 5 59. 3-3 3 + = 0 Critical Thinking Eercises Make Sense? In Eercises 60 63, determine whether each statement makes sense or does not make sense and eplain your reasoning. 60. When I solve an equation that is quadratic in form, it s important to write down the substitution that I am making. 6. Although I ve rewritten an equation that is quadratic in form as au + bu + c = 0 and solved for u, I m not finished. 6. Checking is always a good idea, but it s never necessary when solving an equation that is quadratic in form. 63. The equation 5 3 + 3 + = 0 is quadratic in form, but when I reverse the variable terms and obtain 3 + 5 3 + = 0, the resulting equation is no longer quadratic in form. In Eercises 64 67, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. 64. If an equation is quadratic in form, there is only one method that can be used to obtain its solution set. 65. An equation with three terms that is quadratic in form has a variable factor in one term that is the square of the variable factor in another term. 66. Because 6 is the square of 3, the equation 6-5 3 + 6 = 0 is quadratic in form. 67. To solve - 9 + 4 = 0, we let u =. In Eercises 68 70, use a substitution to solve each equation. 68. 4-5 - = 0 69. 5 6 + 3 = 8 + 4 70. A - + - A + 4 = 5 alet u = + 4 A -.b Review Eercises 7. Simplify: (Section 7., Eample 3) + i 7. Divide: (Section 0.7, Eample 5) - i. 73. If f() = +, find f(3) - f(4). (Section 8., Eample 3) Preview Eercises Eercises 74 76 will help you prepare for the material covered in the net section. 74. Solve: 75. Factor: 76. Simplify: + = 5. 3 + - 4-4. + + 3 -. 0 3 -. ISBN 0-558-957-4 Introductory & Intermediate Algebra for College Students, Third Edition, by Robert Blitzer. Published by Prentice Hall. Copyright 009 by Pearson Education, Inc.