Answer
$\left\{-6,-1,1,6\right\}$
Work Step by Step
Using factoring of trinomials, the given equation, $
x^4-37x^2+36=0
,$ is equivalent to
\begin{align*}
(x^2-36)(x^2-1)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
x^2-36=0 & x^2-1=0
\\
x^2=36 & x^2=1
\\
x=\pm\sqrt{36} & x=\pm\sqrt{1}
\\
x=\pm6 & x=\pm1
.\end{array}
Checking the solutions by substitution in the given equation results to
\begin{array}{l|r}
\text{If }x=\pm6: & \text{If }x=\pm1
\\\\
(\pm6)^4-37(\pm6)^2+36\overset{?}=0 &
(\pm1)^4-37(\pm1)^2+36\overset{?}=0
\\
1296-37(36)+36\overset{?}=0 &
1-37(1)+36\overset{?}=0
\\
1296-1332+36\overset{?}=0 &
1-37+36\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^4-37x^2+36=0
$ is $\left\{-6,-1,1,6\right\}$.
