Solve 2(3a 1)2 7(3a 1) 4 0. Solution The binomial 3a 1 appears as a squared quantity and as a linear quantity. Begin by using substitution. Let u 3a 1. Then, u2 (3a 1)2. Substitute: 2(3a 1)2 7(3a 1) 4 0 2u2 7u 4 0 Does 2u2 7u 4 0 factor? Yes. Solve by factoring. (2u 1)(u 4) 0 Factor 2u2 7u 4 0. b R 2u 1 0 or u 4 0 Set each factor equal to 0. u 1 2 or u 4 Solve for u. Solve for a using u 3a 1. When u 1 2 : When u 4: u 3a 1 u 3a 1 1 2 3a 1 4 3a 1 3 2 3a 3 3a Subtract 1. Subtract 1. 1 3 1 2 a 1 a Divide by 3. Multiply by . The solution set is e 1 2 , 1 f . Check these values in the original equation. Don’t forget to solve for the variable in the original equation. YOU TRY 6 Solve 3(2p 1)2 11(2p 1) 10 0. ANSWERS TO YOU TRY EXERCISES 1) e2, 4 3 f 2) {4} 3) a) {3, 2, 2, 3} b) {125, 1} 4) a) {3, 2, 2, 3} b) {125, 1} 5) e 29 157 2 , 29 157 2 , 29 157 2 , 29 157 2 f 6) e 4 3 , 3 2 f EXAMPLE 6 642 CHAPTER 10 Quadratic Equations and Functions www.mhhe.com/messersmith
messersmith_power_intermediate_algebra_1e_ch4_7_10
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