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messersmith_power_introductory_algebra_1e_ch4_7_10

Compare the signs of the constant on the right side of the equal sign in Step 4 of a) and b). What do you notice? Note Notice that we would have obtained the same result in Example 2a) if we had solved the equation by factoring. b R b) k2 2k 5 0 x2 12x 27 0 (x 9)(x 3) 0 x 9 0 or x 3 0 x 9 or x 3 Step 1: The coeffi cient of k2 is already 1. Step 2: Get the variables on one side of the equal sign and the constant on the other side: k2 2k 5 Step 3: Complete the square: 1 2 (2) 1 (1)2 1 Add 1 to both sides of the equation: k2 2k 1 5 1 k2 2k 1 4 Step 4: Factor: (k 1)2 4 Step 5: Solve using the square root property. (k 1)2 4 k 1 14 Since 14 is not a real number, there is no real number solution to k2 2k 5 0. The solution set is . YOU TRY 2 Solve by completing the square. a) w2 4w 21 0 b) a2 6a 34 0 Remember that in order to complete the square, the coeffi cient of the quadratic term must be 1. Solve by completing the square. a) 4n2 16n 15 0 b) 10y 2y2 3 Solution a) 4n2 16n 15 0 Step 1: Since the coeffi cient of n2 is not 1, divide the whole equation by 4. n2 4 4 16n 4 15 4 0 4 Divide by 4. n2 4n 15 4 0 Simplify. EXAMPLE 3 In-Class Example 3 Solve by completing the square. a) 4x2 24x 35 0 b) 6h 2h2 7 Answer: a) e 5 2 , 7 2 f b) e 3 2 123 2 , 3 2 123 2 f 616 CHAPTER 10 Quadratic Equations www.mhhe.com/messersmith


messersmith_power_introductory_algebra_1e_ch4_7_10
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