## Precalculus (10th Edition)

$x=4$. Please refer to the step-by-step part below for the explanation.
$(x-4)^2\leq0$, since the square of a real number can only be non-negative, the only solution is when $(x-4)^2=0$. Solve this equation to obtain: \begin{align*}(x-4)^2&=0\\ \sqrt{(x-4)^2}&=\pm \sqrt{0}\\ x-4&=0\\ x&=4\\ \end{align*} Thus, the only solution to the given inequality is $x=4$.