Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 7 - Section 7.7 - Complex Numbers - Exercise Set: 60

Answer

$-6\sqrt{3}$

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: $=\sqrt{3(-1)} \cdot \sqrt{36(-1)} \\=(\sqrt{3} \cdot \sqrt{-1}) \cdot \sqrt{6^2(-1)} \\=(\sqrt{3} \cdot i) \cdot 6\sqrt{-1} \\=i\sqrt{3} \cdot 6i \\=6i^2\sqrt{3}$ Use rule (2) above to obtain: $=6(-1)\sqrt{3} \\=-6\sqrt{3}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.