Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 9 - Roots and Radicals - Chapters 1-9 Cumulative Review Problem Set - Page 433: 78



Work Step by Step

Rearranging the equation and taking the square root on both sides, we obtain: $2x^{2}-8=0$ $2x^{2}=8$ $x^{2}=\frac{8}{2}$ $x^{2}=4$ $\sqrt {x^{2}}=\pm\sqrt 4$ $x=\pm2$ Therefore, the solution set is {$-2,2$}.
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