Answer
The values of x in the equation ${{x}^{4}}-13{{x}^{2}}+36=0$ are –2, –3, 2 and 3.
Work Step by Step
Consider the equation.
${{x}^{4}}-13{{x}^{2}}+36=0$ …… (1)
The value of x in the equation ${{x}^{4}}-13{{x}^{2}}+36=0$ is calculated as follows.
Consider $u={{x}^{2}}$; then ${{u}^{2}}={{x}^{4}}$.
Substitute u for ${{x}^{2}}$ in the equation (1).
$\begin{align}
& {{x}^{4}}-13{{x}^{2}}+36=0 \\
& {{u}^{2}}-13u+36=0 \\
& {{u}^{2}}-4u-9u+36=0 \\
& u\left( u-4 \right)-9\left( u-4 \right)=0
\end{align}$
Simplify further as follows.
$\left( u-4 \right)\left( u-9 \right)=0$
$\begin{align}
& u-4=0 \\
& u=4
\end{align}$
Or,
$\begin{align}
& u-9=0 \\
& u=9
\end{align}$
Substitute $u={{x}^{2}}$ and apply the square root property.
$\begin{align}
& {{x}^{2}}=4 \\
& x=\sqrt{4} \\
& x=\pm 2 \\
\end{align}$
Or,
$\begin{align}
& {{x}^{2}}=9 \\
& x=\sqrt{9} \\
& x=\pm 3 \\
\end{align}$
Therefore, the values of x in the equation ${{x}^{4}}-13{{x}^{2}}+36=0$ are $-2,2,3\text{ and}-3$.
Check,
Substitute $\pm 2$ for $x$ in equation (1).
$\begin{align}
{{\left( \pm 2 \right)}^{4}}-13{{\left( \pm 2 \right)}^{2}}+36\overset{?}{\mathop{=}}\,0 & \\
16-52+36\overset{?}{\mathop{=}}\,0 & \\
52-52\overset{?}{\mathop{=}}\,0 & \\
0=0 & \\
\end{align}$
It is true.
Substitute $\pm 3$ for $x$ in equation (1).
$\begin{align}
{{\left( \pm 3 \right)}^{4}}-13{{\left( \pm 3 \right)}^{2}}+36\overset{?}{\mathop{=}}\,0 & \\
81-117+36\overset{?}{\mathop{=}}\,0 & \\
117-117\overset{?}{\mathop{=}}\,0 & \\
0=0 & \\
\end{align}$
It is also true.
Thus, the values of x are $-2,2, 3 $ and $-3$.
