Note We would have obtained the same result if we had solved the equation by factoring. x2 6x 8 0 (x 4)(x 2) 0 b R x 4 or x 2 b) 12h 4h2 24 x 4 0 or x 2 0 Step 1: Because the coeffi cient of h2 is not 1, divide the whole equation by 4. 12h 4 4h2 4 24 4 3h h2 6 Step 2: The constant is on a side by itself. Rewrite the left side of the equation. h2 3h 6 Step 3: Complete the square: 1 2 (3) 3 2 a3 2 b 2 9 4 Add 9 4 to both sides of the equation. h2 3h 9 4 6 9 4 h2 3h 9 4 24 4 9 4 Get a common denominator. h2 3h 9 4 15 4 Step 4: Factor. ah 3 2 b 2 15 4 c 3 2 is 1 2 (3), the coeffi cient of h. Step 5: Solve using the square root property. ah 3 2 b 2 15 4 h 3 2 A 15 4 h 3 2 115 2 i Simplify the radical. h 3 2 115 2 i Subtract 3 2 . The check is left to the student. The solution set is e 3 2 115 2 i, 3 2 115 2 i f . This quadratic equation produced nonreal, complex solutions. 620 CHAPTER 10 Quadratic Equations and Functions www.mhhe.com/messersmith
messersmith_power_intermediate_algebra_1e_ch4_7_10
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