Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 2 - Section 2.1 - Complex Numbers - Exercise Set: 37

Answer

$\dfrac{-2+i\sqrt{2}}{6}$

Work Step by Step

RECALL: (1) $(a-b)^2 = a^2-2ab+b^2$ (2) $\sqrt{-1}=i$ (3) $i^2=-1$ (4) For any real number $a \gt 0$, $\sqrt{-a} = i\sqrt{a}$. Use rule (4) above to obtain: $=\dfrac{-8+i\sqrt{32}}{24} \\=\dfrac{-8+i\sqrt{16(2)}}{24} \\=\dfrac{-8+i\sqrt{4^2(2)}}{24} \\=\dfrac{-8+i\cdot 4\sqrt{2}}{24} \\=\dfrac{-8+4i\sqrt{2}}{24}$ Factor out 4 in the numerator then cancel the common factors to obtain: $\require{cancel} \\=\dfrac{4(-2+i\sqrt{2})}{24} \\=\dfrac{\cancel{4}(-2+i\sqrt{2})}{\cancel{24}6} \\=\dfrac{-2+i\sqrt{2}}{6}$
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