Answer
$\color{blue}{\left\{-i, i, -\frac{1}{2}, \frac{1}{2}\right\}}$
Work Step by Step
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 $\color{blue}{\left\{-i, i, -\frac{1}{2}, \frac{1}{2}\right\}}$.
