Answer
The solution set for the inequality is $\left\{ 2 \right\}$.
Work Step by Step
Consider the inequality ${{\left( x-2 \right)}^{2}}\le 0$.
The function $f\left( x \right)\le 0$.
Express the inequality as:
$f\left( x \right)={{\left( x-2 \right)}^{2}}$
Now, check the solution set.
For $\left( -\infty ,2 \right)$
Substitute $x=-1$
$\begin{align}
& f\left( x \right)={{\left( x-2 \right)}^{2}} \\
& f\left( -1 \right)={{\left( -1-2 \right)}^{2}} \\
& =9 \\
& >0
\end{align}$
And,
For $\left( 2,\infty \right)$
Substitute $x=3$
$\begin{align}
& f\left( x \right)={{\left( x-2 \right)}^{2}} \\
& f\left( 3 \right)={{\left( 3-2 \right)}^{2}} \\
& =1 \\
& >0
\end{align}$
The $x$ intercept is at $2$ and it is negative so there is only one real solution, and there are no boundary points due to the negative sign.
Thus, the provided inequality has only one solution $\left\{ 2 \right\}$.
