11.1 E x a m p l e 1 714SECTION 11.1 OBJECTIVES 1. Solve quadratic equations by using the square root method 2. Solve quadratic equations by completing the square Here, we factor the quadratic member of the equation as a difference of squares. Solving Quadratic Equations by Completing the Square Recall that a quadratic equation is an equation of the form ax 2 bx c 0 where a is not equal to zero. In Section 6.5, we solved quadratic equations by factoring and using the zeroproduct principle. However, not all equations are solvable by that method. In this section, we will learn another technique that can be used to solve some quadratic equation. This technique is called the square root method. Let s begin by solving a special type of equation by factoring. Solving Equations by Factoring Solve the quadratic equation x 2 16 by factoring. We write the equation in standard form: Factoring, we have Finally, the solutions are x 2 16 0 (x 4)(x 4) 0 x 4 or x 4 or {4} CHECK YOURSELF 1 Solve each of the following quadratic equations. (a) x 2 25 (b) 5x 2 180
Section 11.1 Solving Quadratic Equations by Completing the Square 715 The Square Root Method The equation in Example 1 could have been solved in an alternative fashion. We could have used what is called the square root method. Again, given the equation we can write the equivalent statement x 2 16 x 16 or x 16 Note: Be sure to include both the positive and the negative square roots when you use the square root method. This yields the solutions x 4 or x 4 or {4} This discussion leads us to the following general result. Square Root Property If x 2 k, where k is a complex number, then x k or x k Example 2 further illustrates the use of this property. E x a m p l e 2 Using the Square Root Method Solve each equation by using the square root method. (a) x 2 9 By the square root property, x 9 or x 9 3 3 or {3} If a calculator were used, 17 4.123 (rounded to three decimal places). (b) x 2 17 0 Add 17 to both sides of the equation. x 2 17 or {17} or {17, 17}
716 Chapter 11 Quadratic Functions (c) 4x 2 3 0 4x 2 3 x 2 3 4 x 3 4 x 3 or 2 3 2 In Example 2, part d, we see that complex-number solutions may result. (d) x 2 1 0 x 2 1 x 1 x i or {i} CHECK YOURSELF 2 Solve each equation. (a) x 2 5 (b) x 2 2 0 (c) 9x 2 8 0 (d) x 2 9 0 We can also use the approach in Example 2 to solve an equation of the form (x 3) 2 16 As before, by the square root property we have x 3 4 Add 3 to both sides of the equation. Solving for x yields x 3 4 which means that there are two solutions: x 3 4 or x 3 4 1 7 or {1, 7}
Section 11.1 Solving Quadratic Equations by Completing the Square 717 E x a m p l e 3 The two solutions 5 5 and 5 5 are abbreviated as 5 5. Using the Square Root Method Use the square root method to solve each equation. (a) (x 5) 2 5 0 (x 5) 2 5 x 5 5 x 5 5 or {5 5} (b) 9(y 1) 2 2 0 9(y 1) 2 2 (y 1) 2 2 9 y 1 2 9 y 1 2 3 3 2 3 3 3 2 or 3 3 2 3 CHECK YOURSELF 3 Using the square root method, solve each equation. (a) (x 2) 2 3 0 (b) 4(x 1) 2 3 Completing the Square Not all quadratic equations can be solved directly by factoring or using the square root method. We must extend our techniques. The square root method is useful in this process because any quadratic equation can be written in the form If (x h) 2 k, then x h k and x h k which yields the solution (x h) 2 k x h k
718 Chapter 11 Quadratic Functions The process of changing an equation in standard form to the form ax 2 bx c 0 (x h) 2 k is called the method of completing the square, and it is based on the relationship between the middle term and the last term of any perfect-square trinomial. Let s look at three perfect-square trinomials to see whether we can detect a pattern: Note that this relationship is true only if the leading, or x 2 coefficient is 1. That will be important later. x 2 4x 4 (x 2) 2 (1) x 2 6x 9 (x 3) 2 (2) x 2 8x 16 (x 4) 2 (3) Note that in each case the last (or constant) term is the square of one-half of the coefficient of x in the middle (or linear) term. For example, in equation (2), x 2 6x 9 (x 3) 2 1 2 of this coefficient is 3, and (3) 2 9, the constant. Verify this relationship for yourself in equation (3). To summarize, in perfect-square trinomials, the constant is always the square of one-half the coefficient of x. We are now ready to use the above observation in the solution of quadratic equations by completing the square. Consider Example 4. E x a m p l e 4 When you graph the related function, y x 2 8x 7, you will note that the x values for the x intercepts are just below 1 and just above 9. Be certain that you see how these points relate to the exact solutions, 4 23 and 4 23. Completing the Square to Solve an Equation Solve x 2 8x 7 0 by completing the square. First, we rewrite the equation with the constant on the right-hand side: x 2 8x 7 Our objective is to have a perfect-square trinomial on the left-hand side. We know that we must add the square of one-half of the x coefficient to complete the square. In this case, that value is 16, so now we add 16 to each side of the equation. x 2 8x 16 7 16
Section 11.1 Solving Quadratic Equations by Completing the Square 719 1 2 8 4 and 42 16 Factor the perfect-square trinomial on the left, and combine like terms on the right to yield Now the square root property yields (x 4) 2 23 Remember that if (x h) 2 k, then x h k. x 4 23 Subtracting 4 from both sides of the equation gives x 4 23 or {4 23} CHECK YOURSELF 4 Solve x 2 6x 2 0 by completing the square. E x a m p l e 5 Completing the Square to Solve an Equation Solve x 2 5x 3 0 by completing the square. x 2 5x 3 0 x 2 5x 3 Add 3 to both sides. Make the left-hand side a perfect square. Add the square of one-half of the x coefficient to both sides of the equation. Note that 1 2 5 5 2 x 2 5x 5 2 2 3 5 2 2 x 5 2 2 3 7 4 x 5 2 37 Solve for x. 2 Take the square root of both sides. x 5 37 or 2 5 37 2 CHECK YOURSELF 5 Solve x 2 3x 7 0 by completing the square. Some equations have nonreal complex solutions, as Example 6 illustrates. E x a m p l e 6 Completing the Square to Solve an Equation Solve x 2 4x 13 0 by completing the square. x 2 4x 13 0 Subtract 13 from both sides.
720 Chapter 11 Quadratic Functions x 2 4x 13 Add 1 2 (4) 2 to both sides. Note that the graph of y x 2 4x 13 does not intercept the x axis. x 2 4x 4 13 4 Factor the left-hand side. (x 2) 2 9 Take the square root of both sides. x 2 9 Simplify the radical. x 2 i9 CHECK YOURSELF 6 Solve x 2 10x 41 0. x 2 3i x 2 3i or {2 3i} Example 7 illustrates a situation in which the leading coefficient of the quadratic member is not equal to 1. As you will see, an extra step will be required. E x a m p l e 7! Caution Before you can complete the square on the left, the coefficient of x 2 must be equal to 1. Otherwise, we must divide both sides of the equation by that coefficient. Completing the Square to Solve an Equation Solve 4x 2 8x 7 0 by completing the square. 4x 2 8x 7 0 Add 7 to both sides. 4x 2 8x 7 Divide both sides by 4. x 2 2x 7 4 x 2 2x 1 7 1 4 Now, complete the square on the left. The left side is now a perfect square. (x 1) 2 1 1 4 x 1 11 4 1 x 1 1 4 1 11 2 2 11 or 2 2 11 2
Section 11.1 Solving Quadratic Equations by Completing the Square 721 CHECK YOURSELF 7 Solve 4x 2 8x 3 0 by completing the square. The following algorithm summarizes our work in this section with solving quadratic equations by completing the square. Completing the Square Step 1 Step 2 Step 3 Step 4 Step 5 Isolate the constant on the right side of the equation. Divide both sides of the equation by the coefficient of the x 2 term if that coefficient is not equal to 1. Add the square of one-half of the coefficient of the linear term to both sides of the equation. This will give a perfect-square trinomial on the left side of the equation. Write the left side of the equation as the square of a binomial, and simplify on the right side. Use the square root property, and then solve the resulting linear equations. CHECK YOURSELF ANSWERS 1. (a) {5, 5}; (b) {6, 6}. 2. (a) {5, 5}; (b) {2, 2}; 2 (c) 2, 2 2 3 3 ; and (d) {3i, 3i}. 3. (a) {2 3}; (b) 2 3 2. 4. {3 11}. 5. 37 3 2. 6. {5 4i}. 7. 1 2, 3 2.
E x e r c i s e s 11.1 1. {5, 1} 2. {2, 3} 3. {5, 7} 4. {8, 3} 5. 1 2, 3 6. 2 3, 4 7. 1 2, 2 3 8. 1 5, 1 2 In Exercises 1 to 8, solve by factoring or completing the square. 1. x 2 6x 5 0 2. x 2 5x 6 0 3. z 2 2z 35 0 4. q 2 5q 24 0 5. 2x 2 5x 3 0 6. 3x 2 10x 8 0 9. {6} 10. {12, 12} 11. {7, 7} 12. {32, 32} 13. {6, 6} 14. {22, 22} 15. {2i} 16. {3i} 17. {1 23} 18. 3 5 2 19. 3 1 2 20. 3i 4 3 21. 36 22. 49 23. 16 24. 64 25. 9 4 26. 2 5 4 27. 1 4 28. 1 4 1 1 29. 30. 1 6 3 6 1 31. 1 00 32. 1 64 7. 6y 2 y 2 0 8. 10z 2 3z 1 0 In Exercises 9 to 20, use the square root method to find solutions for the equations. 9. x 2 36 10. x 2 144 11. y 2 7 12. p 2 18 13. 2x 2 12 0 14. 3x 2 66 0 15. 2t 2 12 4 16. 3u 2 5 32 17. (x 1) 2 12 18. (2x 3) 2 5 19. (2z 1) 2 3 0 20. (3p 4) 2 9 0 In Exercises 21 to 32, find the constant that must be added to each binomial expression to form a perfect-square trinomial. 21. x 2 12x 22. r 2 14r 23. y 2 8y 24. w 2 16w 25. x 2 3x 26. z 2 5z 27. n 2 n 28. x 2 x 29. x 2 1 2 x 30. x 2 1 3 x 31. x2 1 5 x 32. y2 1 4 y 722
Section 11.1 Solving Quadratic Equations by Completing the Square 723 33. {6 38} 34. {7 214} 35. {2, 4} 36. {2 219} 37. {1 6} 38. {1, 3} 39. {5 23} 40. 3 77 2 41. 53 5 2 42. {4 2i} 43. 1 13 2 44. 21 1 2 45. 17 1 4 46. 73 1 6 47. 3 1 2 48. 7 2 3 49. 22 4 3 50. 5 2 2 51. i 35 1 3 In Exercises 33 to 54, solve each equation by completing the square. 33. x 2 12x 2 0 34. x 2 14x 7 0 35. y 2 2y 8 36. z 2 4z 72 0 37. x 2 2x 5 0 38. x 2 2x 3 39. x 2 10x 13 0 40. x 2 3x 17 0 41. z 2 5z 7 0 42. q 2 8q 20 0 43. m 2 m 3 0 44. y 2 y 5 0 45. x 2 1 2 x 1 46. x2 1 3 x 2 47. 2x2 2x 1 0 48. 3x 2 4x 1 49. 3x 2 8x 2 50. 4x 2 8x 1 0 51. 3x 2 2x 12 0 52. 7y 2 2y 3 0 53. x 2 8x 20 0 54. x 2 2x 10 0 55. Why must the leading coefficient of the quadratic member be set equal to 1 before using the technique of completing the square? 56. What relationship exists between the solution(s) of a quadratic equation and the graph of a quadratic function? In Exercises 57 to 62, find the constant that must be added to each binomial to form a perfect-square trinomial. Let x be the variable; other letters represent constants. 57. x 2 2ax 58. x 2 2abx 59. x 2 3ax 60. x 2 abx 61. a 2 x 2 2ax 62. a 2 x 2 4abx 52. 2 i5 1 7 53. {4 2i} 54. {1 3i} 57. a 2 58. a 2 b 2 59. 9 4 a2 60. a2 b 2 4 61. 1 62. 4b 2
724 Chapter 11 Quadratic Functions 63. {a 4 a 2 } 64. {a a 2 8} 65. In Exercises 63 and 64, solve each equation by completing the square. 63. x 2 2ax 4 64. x 2 2ax 8 0 In Exercises 65 to 68, use your graphing utility to find the graph. Approximate the x intercepts for each graph. (You may have to adjust the viewing window to see both intercepts.) 10 12.2.2 5 10 (6, 38) 66. 0.5 4 8 4 12 10 (14.5) 65. y x 2 12x 2 66. y x 2 14x 7 67. y x 2 2x 8 68. y x 2 4x 72 69. On your graphing calculator, view the graph of f(x) x 2 1. (a) What can you say about the x intercepts of the graph? (b) Determine the zeros of the function, using the square root method. (c) How does your answer to part a relate to your answer to part b? 70. Consider the following representation of completing the square : Suppose we wish to complete the square for x 2 10x. A square with dimensions x by x has area equal to x 2. (7, 56) x x 2 67. 2 2 2 4 x We divide the quantity 10x by 2 and get 5x. If we extend the base x by 5 units, and draw the rectangle attached to the square, the rectangle s dimensions are 5 by x with an area of 5x. (1, 9) x x 2 5x 68. x 5 Now we extend the height by 5 units, and draw another rectangle whose area is 5x. 10.7 10 24 6.7 5 5x 30 50 70 (2, 76) x x 2 5x 69. (a) There are none; (b) i; (c) If the graph of f(x) has no x intercepts, the zeros of the function do not exist. x 5 (a) What is the total area represented in the figure so far? (b) How much area must be added to the figure to complete the square? (c) Write the area of the completed square as a binomial squared. 71. Repeat the process described in Exercise 70 with x 2 16x. 70. (a) x 2 5x 5x; (b) 25; (c) x 2 10x 25 (x 5) 2 71. (a) x 2 8x 8x; (b) 64; (c) x 2 16x 64 (x 8) 2