## Intermediate Algebra (12th Edition)

$r=\left\{ 5-2i,5+2i \right\}$
$\bf{\text{Solution Outline:}}$ To find the non real complex solutions of the given equation, $(r-5)^2=-4 ,$ take the square root of both sides (Square Root Property). Then use the properties of radicals and use $i=\sqrt{-1}.$ Finally, simplify the radical and isolate the variable. $\bf{\text{Solution Details:}}$ Taking the square root of both sides (Square Root Property), the equation above is equivalent to \begin{array}{l}\require{cancel} r-5=\pm\sqrt{-4} .\end{array} Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy},$ the equation above is equivalent to\begin{array}{l}\require{cancel} r-5=\pm\sqrt{-1}\cdot\sqrt{4} .\end{array} Using $i=\sqrt{-1},$ the equation above is equivalent to \begin{array}{l}\require{cancel} r-5=\pm i\sqrt{4} .\end{array} Writing the radicand as an expression that contains a factor that is a perfect power of the given index and then extracting the root of that factor, the equation above is equivalent to \begin{array}{l}\require{cancel} r-5=\pm i\sqrt{(2)^2} \\\\ r-5=\pm i(2) \\\\ r-5=\pm 2i .\end{array} Using the properties of equality to isolate the variable, the equation above is equivalent to \begin{array}{l}\require{cancel} r=5\pm 2i .\end{array} Hence, $r=\left\{ 5-2i,5+2i \right\} .$