Answer
The solution set is $\left( -\infty ,2 \right)\cup \left( 2,\infty \right)$
Work Step by Step
Consider the inequality ${{\left( x-2 \right)}^{2}}>0$.
The function $f\left( x \right)>0$.
Express the function as:
$f\left( x \right)={{\left( x-2 \right)}^{2}}$
The function has an $x$ -intercept at $2$ and the graph of the function passes through point $\left( 2,0 \right)$.
Thus, the boundary of the function is $\left( -\infty ,2 \right)$ and $\left( 2,\infty \right)$.
Now, check the interval.
Substitute $x=-2$
$\begin{align}
& f\left( x \right)={{\left( x-2 \right)}^{2}} \\
& f\left( -2 \right)={{\left( -2-2 \right)}^{2}} \\
& =16 \\
& >0
\end{align}$
Thus, the function is positive for all $x$ in $\left( -\infty ,2 \right)$.
Substitute $x=3$
$\begin{align}
& f\left( x \right)={{\left( x-2 \right)}^{2}} \\
& f\left( 3 \right)={{\left( 3-2 \right)}^{2}} \\
& =1 \\
& >0
\end{align}$
Thus, the function is positive for all $x$ in $\left( 2,\infty \right)$.
Therefore, the solution set of the inequality is given as $\left( -\infty ,2 \right)\cup \left( 2,\infty \right)$
