Answer
The solutions are $x=\pm1$ and $x=\pm2$
Work Step by Step
$x^{4}-5x^{2}+4=0$
Let $u$ be equal to $x^{2}$
If $u=x^{2}$, then $u^{2}=x^{4}$
Rewrite the original equation using the new variable $u$:
$u^{2}-5u+4=0$
Solve by factoring:
$(u-4)(u-1)=0$
Set both factors equal to $0$ and solve each individual equation for $u$:
$u-4=0$
$u=4$
$u-1=0$
$u=1$
Substitute $u$ back to $x^{2}$ and solve for $x$:
$x^{2}=4$
$\sqrt{x^{2}}=\pm\sqrt{4}$
$x=\pm2$
$x^{2}=1$
$\sqrt{x^{2}}=\pm\sqrt{1}$
$x=\pm1$
The solutions are $x=\pm1$ and $x=\pm2$