## College Algebra 7th Edition

$\left\{-2\sqrt{2}, 2\sqrt{2}\right\}$
Add $64$ to both sides: $8x^2=64$ Divide $8$ on both sides to obtain: $x^2=8$ Factor $8$ so that one factor is a perfect square: $x=\pm \sqrt{4(2)} \\x=\pm \sqrt{2^2(2)} \\x=\pm 2\sqrt{2}$ Thus, the solution set of the given equation is $\left\{-2\sqrt{2}, 2\sqrt{2}\right\}$.