Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 8 - Complex Numbers, Polar Equations, and Parametric Equations - Section 8.1 Complex Numbers - 8.1 Exercises - Page 357: 48

Answer

-3

Work Step by Step

$\frac{\sqrt -12\times\sqrt -6}{\sqrt 8}=\frac{i\sqrt 12\times i\sqrt 6}{\sqrt 8}=\frac{i^{2}\times\sqrt 12\times\sqrt 6}{\sqrt 8}=\frac{-1\times\sqrt (12\times 6)}{\sqrt 8}=\frac{-1\times\sqrt 72}{\sqrt 8}=-\sqrt\frac{72}{8}=-\sqrt9=-3$ We know that $\sqrt -1=i$, because $i^{2}=-1$.
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