Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 1 - Equations and Inequalities - 1.3 Complex Numbers - 1.3 Exercises - Page 112: 40

Answer

$$-3$$

Work Step by Step

Recall that the definition $\sqrt{-a} = i\sqrt{a}$ must be applied BEFORE the other rules regarding radicals, therefore, $\frac{\sqrt{-12}*\sqrt{-6}}{\sqrt 8}$ $\frac{i\sqrt{12}*i\sqrt{6}}{\sqrt 8}$ $\frac{i^2\sqrt{72}}{\sqrt 8}$ $(i^2)(\sqrt{\frac{72}{8}})$ $(-1)(\sqrt 9)$ $$-3$$
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