Answer
$x=\pm i$ or $x=\pm 2i$
Work Step by Step
$ x^{4}+5x^{2}+4=0\qquad$...substitute $x^{2}$ for $u$
$ u^{2}+5u+4=0\qquad$... solve with the Quadractic formula.
$u=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$u=\displaystyle \frac{-5\pm\sqrt{25-16}}{2}$
$u=\displaystyle \frac{-5\pm\sqrt{9}}{2}$
$u=\displaystyle \frac{-5\pm 3}{2}$
$u=\displaystyle \frac{-5+3}{2}=\frac{-2}{2}=-1$ or $u=\displaystyle \frac{-5-3}{2}=\frac{-8}{2}=-4$
Bring back $x^{2}=u$.
$x^{2}=-1$ or $x^{2}=-4$
$x=\pm i$ or $x=\pm 2i$
