Answer
The solutions are $x=\pm1$ and $x=\pm2$
Work Step by Step
$x^{4}-5x^{2}+4=0$
To solve this equation, let $x^{2}$ be equal to $t$.
$t=x^{2}$ so $t^{2}=x^{4}$
Substitute $x^{4}$ by $t^{2}$ and $x^{2}$ by $t$ in the original equation:
$t^{2}-5t+4=0$
Solve this equation by factoring:
$(t-1)(t-4)=0$
Set both factors equal to $0$ and solve each individual equation for $t$:
$t-1=0$
$t=1$
$t-4=0$
$t=4$
Substitute $t$ back to $x^{2}$ and solve for $x$:
$t=1$
$x^{2}=1$
$x=\sqrt{1}$
$x=\pm1$
$t=4$
$x^{2}=4$
$x=\sqrt{4}$
$x=\pm2$
The solutions are $x=\pm1$ and $x=\pm2$