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messersmith_power_intermediate_algebra_1e_ch4_7_10

Check: w 9:  (w 4)2 25 w 1:  (w 4)2 25 (9 4)2 ? 25 (1 4)2 ? 25 52 25 ✓ (5)2 25 ✓ The solution set is {1, 9}. YOU TRY 2 Solve (c 6)2 81 using the square root property. EXAMPLE 3 Solve. a) (3t 4)2 9 b) (2m 5)2 12 c) (z 8)2 11 7 d) (6k 5)2 20 0 Solution a) (3t 4)2 9 3t 4 19 Square root property 3t 4 3 This means 3t 4 3 or 3t 4 3. Solve both equations. 3t 4 3 or 3t 4 3 3t 1            3t 7 Subtract 4 from each side. t 1 3  or        t 7 3 Divide by 3. The solution set is e 7 3 , 1 3 f . b) (2m 5)2 12 2m 5 112 Square root property 2m 5 213 Simplify 112. 2m 5 213 Add 5 to each side. 5 213 m 2 Divide by 2. One solution is 5 213 2 , and the other is 5 213 2 . The solution set, e 5 213 2 , 5 213 2 f , can also be written as e 5 213 2 f . c) (z 8)2 11 7 (z 8)2 4 Subtract 11 from each side. z 8 14 Square root property z 8 2i Simplify 14. z 8 2i Subtract 8 from each side. The check is left to the student. The solution set is {8 2i, 8 2i}. Do you need to review how to simplify square roots? 614 CHAPTER 10 Quadratic Equations and Functions www.mhhe.com/messersmith


messersmith_power_intermediate_algebra_1e_ch4_7_10
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