## Intermediate Algebra (12th Edition)

$x=\left\{ -2i\sqrt{3},2i\sqrt{3} \right\}$
$\bf{\text{Solution Outline:}}$ To find the nonreal complex solutions of the given equation, $x^2=-12 ,$ 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. $\bf{\text{Solution Details:}}$ Taking the square root of both sides (Square Root Property), the equation above is equivalent to \begin{array}{l}\require{cancel} x=\pm\sqrt{-12} .\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} x=\pm\sqrt{-1}\cdot\sqrt{12} .\end{array} Using $i=\sqrt{-1},$ the equation above is equivalent to \begin{array}{l}\require{cancel} x=\pm i\sqrt{12} .\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} x=\pm i\sqrt{4\cdot3} \\\\ x=\pm i\sqrt{(2)^2\cdot3} \\\\ x=\pm i(2)\sqrt{3} \\\\ x=\pm 2i\sqrt{3} .\end{array} Hence, $x=\left\{ -2i\sqrt{3},2i\sqrt{3} \right\} .$