Note After multiplying the original equation in Example 3 by 18, we got the equation 4k2 12k 9 0. Notice that we could have solved this equation by factoring: 4k2 12k 9 0 (2k 3)2 0 Factor. 2k 3 0 Use the square root property. 2k 3 Add 3. k 3 2 Divide by 2. The solution set is e 3 2 f . This is the same as the result we obtained using the quadratic formula. ANSWERS TO YOU TRY EXERCISES 1) a) e 1 161 10 , 1 161 10 f b) e2, 5 3 f 2) a) {3 15, 3 15} b) 3) e 1 3 f 10.3 Exercises Do the exercises, and check your work. *Additional answers can be found in the Answers to Exercises appendix. Objective 2: Solve a Quadratic Equation Using the Quadratic Formula 1) To solve a quadratic equation, ax2 bx c 0 (a 0), for x, we can use the quadratic formula. Write the quadratic formula. x b 2b2 4ac 2a Find the error in each, and correct the mistake. 2) The solution to ax2 bx c 0 (a 0) can be found using the quadratic formula x b 2b2 4ac 2a The fraction bar should also be under b: b 2b2 4ac x 2a 3) In order to solve 3x2 5x 4 using the quadratic formula, a student substitutes a, b, and c into the formula in this way: a 3, b 5, c 4. x (5) 2(5)2 4(3)(4) 2(3) 4) 3 1215 3 1 1215 Solve using the quadratic formula. 5) x2 2x 8 0 {4, 2} 6) p2 8p 12 0 7) 6z2 7z 2 0 8) 2h2 h 15 0 {6, 2} e 1 2 , 2 3 f e3, 5 2 f 9) k2 2 5k 10) d2 5 3d 11) 3w2 2w 4 12) 8r 2 5r2 13) y 2y2 6 14) 3v 4v2 3 15) m2 11m 0 {11, 0} 16) w2 8w 0 {0, 8} 17) 2p(p 3) 3 18) 3q(q 3) 7q 4 19) (2s 3)(s 1) s2 s 6 {1 110, 1 110} 20) 2n2 2n 4 (3n 4)(n 2) {2 212, 2 212} 21) k(k 2) 5 22) 3r 2(r2 4) 23) (x 8)(x 3) 3(3 x) {3, 5} 24) 2(k 10) (k 10)(k 2) {10, 4} 25) 8t 1 16t2 e 1 4 f 26) 9u2 12u 4 0 27) 1 8 z2 3 4 z 1 2 0 28) 1 6 p2 4 3 p 5 2 29) 1 6 k 1 2 3 4 k2 30) 1 5 1 2 w 1 5 w2 e 2 3 f {3 15, 3 15} {4 131, 4 131} e 1 155 9 , 1 155 9 f e 5 141 4 , 5 141 4 f 626 CHAPTER 10 Quadratic Equations www.mhhe.com/messersmith
messersmith_power_introductory_algebra_1e_ch4_7_10
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