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