Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.6 - Complex Numbers - 1.6 Exercises: 60

Answer

$\dfrac{\sqrt{-36}}{\sqrt{-2}\sqrt{-9}}=0-\sqrt{2}i$

Work Step by Step

$\dfrac{\sqrt{-36}}{\sqrt{-2}\sqrt{-9}}$ Rewrite $\sqrt{-36}$ as $(\sqrt{-1})(\sqrt{36})$, $\sqrt{-2}$ as $(\sqrt{-1})(\sqrt{2})$ and $\sqrt{-9}$ as $(\sqrt{-1})(\sqrt{9})$: $\dfrac{\sqrt{-36}}{\sqrt{-2}\sqrt{-9}}=\dfrac{(\sqrt{-1})(\sqrt{36})}{(\sqrt{-1})(\sqrt{2})(\sqrt{-1})(\sqrt{9})}=...$ Evaluate $\sqrt{9}$ and $\sqrt{36}$: $...=\dfrac{(\sqrt{-1})(6)}{(\sqrt{-1})(\sqrt{2})(\sqrt{-1})(3)}=...$ Since $\sqrt{-1}=i$, this expression becomes: $...=\dfrac{6i}{(i)(i)3\sqrt{2}}=\dfrac{6i}{(3\sqrt{2})i^{2}}=...$ Simplify: $...=\dfrac{6i}{(3\sqrt{2})(-1)}=-\dfrac{6i}{3\sqrt{2}}=-\dfrac{2i}{\sqrt{2}}=...$ Rationalizing the denominator: $...=-\dfrac{2}{\sqrt{2}}i\cdot\dfrac{\sqrt{2}}{\sqrt{2}}=-\dfrac{2\sqrt{2}}{2}i=-\sqrt{2}i$ In $a+bi$ form: $\dfrac{\sqrt{-36}}{\sqrt{-2}\sqrt{-9}}=0-\sqrt{2}i$
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