## College Algebra 7th Edition

$x=\left\{-3, 3\right\}$
Use the property $a^m = a^n \longrightarrow m=n$ to obtain: $\begin{array}{ccc} &x^2&=&9 \\&\sqrt{x}&=&\pm\sqrt{9} \\&x&=&\pm3 \end{array}$ Thus, the solutions are:$x=\left\{-3, 3\right\}$.