## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$\left\{\pm3, \pm 3i\right\}$
Recall the definition of absolute value: $|x|=a \longrightarrow x=a \text{ or } x=-a$ Thus, $|x^2|=9\longrightarrow x^2=9 \text{ or } x^2 = -9$ Solving each equation gives: \begin{align*} x^2&=9 &\text{ or}& &x^2=-9\\ \sqrt{x^2}&=\pm\sqrt9 &\text{or}& &\sqrt{x^2}=\pm\sqrt{-9}\\ x&=\pm3 &\text{or}& &x=\pm3i \end{align*} So, our answers are: $x=\pm3$ or $x=\pm3i$