## Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole

# Chapter 1 - Section 1.6 - Complex Numbers - 1.6 Exercises: 58

#### Answer

$(\sqrt{3}-\sqrt{-4})(\sqrt{6}-\sqrt{-8})=-\sqrt{2}-4\sqrt{6}i$

#### Work Step by Step

$(\sqrt{3}-\sqrt{-4})(\sqrt{6}-\sqrt{-8})$ Rewrite $\sqrt{-4}$ as $(\sqrt{-1})(\sqrt{4})$ and $\sqrt{-8}$ as $(\sqrt{-1})(\sqrt{8})$: $[\sqrt{3}-(\sqrt{-1})(\sqrt{4})][\sqrt{6}-(\sqrt{-1})(\sqrt{8})]=...$ Evaluate $\sqrt{4}$ and $\sqrt{8}$: $...=[\sqrt{3}-(\sqrt{-1})(2)][\sqrt{6}-(\sqrt{-1})(2\sqrt{2})]=...$ Since $\sqrt{-1}=i$, this expression becomes: $...=(\sqrt{3}-2i)(\sqrt{6}-2\sqrt{2}i)=...$ Evaluate the product and simplify: $...=(\sqrt{3})(\sqrt{6})-2(\sqrt{3})(\sqrt{2})i-(2)(\sqrt{6})i+(2i)(2\sqrt{2}i)=...$ $...=\sqrt{18}-2\sqrt{6}i-2\sqrt{6}i+4\sqrt{2}(i^{2})=...$ $...=3\sqrt{2}-4\sqrt{6}i+4\sqrt{2}(-1)=...$ $...=3\sqrt{2}-4\sqrt{2}-4\sqrt{6}i=...$ $...=-\sqrt{2}-4\sqrt{6}i$

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