## Precalculus (6th Edition)

$\left\{-2, 2, -i, i\right\}$
Let $u=x^2 \\u^2=x^4$ The given equation becomes: $u^2-3u-4=0$ Factor the trinomial to obtain: $(u-4)(u+1)=0$ Use the Zero-Factor Property by equating each factor to zero to obtain: $u-4=0$ or $u+1=0$ Solve each equation: $u=4$ or $u=-1$ Replace $u$ with $x^2$ to obtain: $x^2=4$ or $x^2=-1$ Take the square root of both sides to obtain (note that $\sqrt{-1}=i$): $x=\pm \sqrt{4}$ or $x=\pm \sqrt{-1}$ $x = \pm 2$ or $x = \pm i$ Therefore, the solution set is $\left\{-2, 2, -i, i\right\}$.