## Precalculus (6th Edition)

$\color{blue}{\left\{-20i, 20i\right\}}$
RECALL: If $x^2=a$, then taking the square root of both sides gives $x = \pm \sqrt{a}$. Take the square root of both sides of the given equation to obtain: $\sqrt{x^2}=\pm \sqrt{-400} \\x =\pm \sqrt{400(-1)} \\x=\pm\sqrt{20^2(-1)} \\x =\pm 20\sqrt{-1}$ Since $\sqrt{-1}=i$, then the solution is: $x=\pm 20i$ Thus, the solution set is $\color{blue}{\left\{-20i, 20i\right\}}$.