Answer
$\left\{\pm1, \pm2\right\}$
Work Step by Step
Using the factoring of trinomials, the given equation, $
x^4-5x^2+4=0
$, is equivalent to
\begin{align*}
(x^2-1)(x^2-4)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property), then
\begin{array}{l|r}
x^2-1=0 & x^2-4=0
\\
x^2=1 & x^2=4
.\end{array}
Taking the square root of both sides (Square Root Property), the equations above are equivalent to
\begin{array}{l|r}
x=\pm\sqrt{1} & x=\pm\sqrt{4}
\\
x=\pm1 & x=\pm2
.\end{array}
Hence, the solution set of the equation $
x^4-5x^2+4=0
$ is $
\left\{\pm1, \pm2\right\}
.$
