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: 47

Answer

-2

Work Step by Step

$\frac{\sqrt -6\times\sqrt -2}{\sqrt 3}=\frac{i\sqrt 6\times i\sqrt 2}{\sqrt 3}=\frac{i^{2}\times\sqrt 6\times\sqrt 2}{\sqrt 3}=\frac{-1\times\sqrt (6\times 2)}{\sqrt 3}=\frac{-1\times\sqrt 12}{\sqrt 3}=-\sqrt\frac{12}{3}=-\sqrt4=-2$ We know that $\sqrt-1 =i$, because $i^{2}=-1$.
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