Answer
{$-2,2$}
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 {$-2,2$}.
Elementary Algebra
by Kaufmann, Jerome; Schwitters, Karen;
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
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