Answer
The solutions are $x=\pm3$ and $x=\pm2$
Work Step by Step
$x^{4}-13x^{2}+36=0$
Let $u$ be equal to $x^{2}$
If $u=x^{2}$, then $u^{2}=x^{4}$
Rewrite the original equation using the new variable $u$:
$u^{2}-13u+36=0$
Solve by factoring:
$(u-9)(u-4)=0$
Set both factors equal to $0$ and solve each individual equation for $u$:
$u-9=0$
$u=9$
$u-4=0$
$u=4$
Substitute $u$ back to $x^{2}$ and solve for $x$:
$x^{2}=9$
$\sqrt{x^{2}}=\pm\sqrt{9}$
$x=\pm3$
$x^{2}=4$
$\sqrt{x^{2}}=\pm\sqrt{4}$
$x=\pm2$
The solutions are $x=\pm3$ and $x=\pm2$
