Answer
$\left\{-5,-2,5,2\right\}$
Work Step by Step
Using factoring of trinomials, the given equation, $
x^4-29x^2+100=0
,$ is equivalent to
\begin{align*}
(x^2-25)(x^2-4)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
x^2-25=0 & x^2-4=0
\\
x^2=25 & x^2=4
\\
x=\pm\sqrt{25} & x=\pm\sqrt{4}
\\
x=\pm5 & x=\pm2
.\end{array}
Checking the solutions by substitution in the given equation results to
\begin{array}{l|r}
\text{If }x=\pm5: & \text{If }x=\pm2
\\\\
(\pm5)^4-29(\pm5)^2+100\overset{?}=0 & (\pm2)^4-29(\pm2)^2+100\overset{?}=0
\\
625-29(25)+100\overset{?}=0 & 16-29(4)+100\overset{?}=0
\\
625-725+100\overset{?}=0 & 16-116+100\overset{?}=0
\\
0\overset{\checkmark}=0 & 0\overset{\checkmark}=0
.\end{array}
Since both solutions satisfy the original equation, then the solution set of the equation $
x^4-29x^2+100=0
$ is $\left\{-5,-2,5,2\right\}$.
