## Precalculus: Mathematics for Calculus, 7th Edition

$\dfrac{\sqrt{-36}}{\sqrt{-2}\sqrt{-9}}=0-\sqrt{2}i$
$\dfrac{\sqrt{-36}}{\sqrt{-2}\sqrt{-9}}$ Rewrite $\sqrt{-36}$ as $(\sqrt{-1})(\sqrt{36})$, $\sqrt{-2}$ as $(\sqrt{-1})(\sqrt{2})$ and $\sqrt{-9}$ as $(\sqrt{-1})(\sqrt{9})$: $\dfrac{\sqrt{-36}}{\sqrt{-2}\sqrt{-9}}=\dfrac{(\sqrt{-1})(\sqrt{36})}{(\sqrt{-1})(\sqrt{2})(\sqrt{-1})(\sqrt{9})}=...$ Evaluate $\sqrt{9}$ and $\sqrt{36}$: $...=\dfrac{(\sqrt{-1})(6)}{(\sqrt{-1})(\sqrt{2})(\sqrt{-1})(3)}=...$ Since $\sqrt{-1}=i$, this expression becomes: $...=\dfrac{6i}{(i)(i)3\sqrt{2}}=\dfrac{6i}{(3\sqrt{2})i^{2}}=...$ Simplify: $...=\dfrac{6i}{(3\sqrt{2})(-1)}=-\dfrac{6i}{3\sqrt{2}}=-\dfrac{2i}{\sqrt{2}}=...$ Rationalizing the denominator: $...=-\dfrac{2}{\sqrt{2}}i\cdot\dfrac{\sqrt{2}}{\sqrt{2}}=-\dfrac{2\sqrt{2}}{2}i=-\sqrt{2}i$ In $a+bi$ form: $\dfrac{\sqrt{-36}}{\sqrt{-2}\sqrt{-9}}=0-\sqrt{2}i$