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messersmith_power_intermediate_algebra_1e_ch4_7_10

Remember: i 2 1. YOU TRY 1 c) 2a2 21 3 2a2 18 Subtract 21. a2 9 Divide by 2. a 19 Square root property a 3i Check: a 3i: 2a2 21 3 a 3i: 2a2 21 3 2(3i)2 21 3 2(3i)2 21 3 2(9i2) 21 3 2(9i2) 21 3 2(9)(1) 21 3 2(9)(1) 21 3 18 21 3 ✓ 18 21 3 ✓ The solution set is {3i, 3i}. Solve using the square root property. a) p2 100 b) w2 32 0 c) 3m2 19 7 Can we solve (w 4)2 25 using the square root property? Yes. The equation has a squared quantity and a constant. 2 Solve an Equation of the Form (ax b)2 k Solve x2 25 and (w 4)2 25 using the square root property. Solution While the equation (w 4)2 25 has a binomial that is being squared, the two equations are actually in the same form. x2 25 (w 4)2 25 c c c c x squared constant (w 4) squared constant Solve x2 25: x2 25 x 125 Square root property x 5 The solution set is {5, 5}. We solve (w 4)2 25 in the same way with some additional steps. (w 4)2 25 w 4 125 Square root property w 4 5 This means w 4 5 or w 4 5. Solve both equations. w 4 5 or w 4 5 w 9  or w 1 Add 4 to each side. EXAMPLE 2 The squared quantity must be isolated before taking the square root of each side. www.mhhe.com/messersmith SECTION 10.1 The Square Root Property and Completing the Square 613


messersmith_power_intermediate_algebra_1e_ch4_7_10
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