## Precalculus: Mathematics for Calculus, 7th Edition

$\dfrac{2+\sqrt{-8}}{1+\sqrt{-2}}=2+0i$
$\dfrac{2+\sqrt{-8}}{1+\sqrt{-2}}$ Rewrite $\sqrt{-8}$ as $(\sqrt{-1})(\sqrt{8})$ and $\sqrt{-2}$ as $(\sqrt{-1})(\sqrt{2})$: $\dfrac{2+\sqrt{-8}}{1+\sqrt{-2}}=\dfrac{2+(\sqrt{-1})(\sqrt{8})}{1+(\sqrt{-1})(\sqrt{2})}=...$ Evaluate $\sqrt{8}$: $...=\dfrac{2+(\sqrt{-1})(2\sqrt{2})}{1+(\sqrt{-1})(\sqrt{2})}=...$ Since $\sqrt{-1}=i$, this expression becomes: $...=\dfrac{2+2\sqrt{2}i}{1+\sqrt{2}i}=...$ Evaluate the division and simplify: $...=\dfrac{2+2\sqrt{2}i}{1+\sqrt{2}i}\cdot\dfrac{1-\sqrt{2}i}{1-\sqrt{2}i}=\dfrac{(2+2\sqrt{2}i)(1-\sqrt{2}i)}{1^{2}-(\sqrt{2}i)^{2}}=...$ $...=\dfrac{2-2\sqrt{2}i+2\sqrt{2}i-2\sqrt{4}i^{2}}{1-2i^{2}}=\dfrac{2-2(2)(-1)}{1-2(-1)}=...$ $...=\dfrac{2+4}{1+2}=\dfrac{6}{3}=2$ In $a+bi$ form: $\dfrac{2+\sqrt{-8}}{1+\sqrt{-2}}=2+0i$