## Precalculus (6th Edition)

$\color{blue}{\left\{4-i\sqrt5, 4+i\sqrt5\right\}}$.
RECALL: If $x^2=a$, then taking the square root of both sides gives $x = \pm \sqrt{a}$. Take the square root of both sides of the given equation to obtain: $\sqrt{(x-4)^2}=\pm \sqrt{-5} \\x-4 =\pm \sqrt{-1(5)}$ Since $\sqrt{-1}=i$, then the expression above is equivalent to: $x-4=\pm i\sqrt{5}$ Add $4$ to both sides: $x =4 \pm i\sqrt{5}$ Thus, the solution set is $\color{blue}{\left\{4-i\sqrt5, 4+i\sqrt5\right\}}$.