## Precalculus (6th Edition)

$\color{blue}{\left\{-i, i, -\frac{1}{2}, \frac{1}{2}\right\}}$
Let $u=x^2 \\u^2=x^4$ The given equation can be written as: $$4u^2+3u-1=0$$ Factor the trinomial to obtain: $(4u-1)(u+1)=0$ Use the Zero-Factor Property by equating each factor to zero, then solve each equation to obtain: \begin{array}{ccc} &4u-1=0 &\text{or} &u+1=0 \\&4u=1 &\text{or} &u=-1 \\&u=\frac{1}{4} &\text{or} &u=-1 \end{array} Replace $u$ with $x62$ to obtain: $x^2=\frac{1}{4}$ or $x^2=-1$ Solve each equation by taking the square root of both sides to obtain: $x=\pm\sqrt{\frac{1}{4}}$ or $x=\pm \sqrt{-1}$ $x=\pm \frac{1}{2}$ or $x = \pm i$ Thus, the solution set is $\color{blue}{\left\{-i, i, -\frac{1}{2}, \frac{1}{2}\right\}}$.