## Intermediate Algebra for College Students (7th Edition)

Published by Pearson

# Chapter 7 - Section 7.7 - Complex Numbers - Exercise Set - Page 570: 61

#### Answer

$-2\sqrt{6}$

#### Work Step by Step

RECALL: (1) $\sqrt{-1}=i$ (2) $i^2=-1$ (3) $\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}$ Simplify each radical by using rules (1) and (3) above: $=\sqrt{8(-1)} \cdot \sqrt{3(-1)} \\=(\sqrt{8} \cdot \sqrt{-1}) \cdot (\sqrt{3} \cdot \sqrt{-1}) \\=(\sqrt{8} \cdot i) \cdot (\sqrt{3} \cdot i) \\=i^2\sqrt{8(3)} \\=i^2\sqrt{24} \\=i^2\sqrt{4(6)} \\=i^2\sqrt{2^2(6)} \\=i^2\cdot 2\sqrt{6} \\=2i^2\sqrt{6}$ Use rule (2) above to obtain: $=2(-1)\sqrt{6} \\=-2\sqrt{6}$

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