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 - Page 314: 40



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: $=\dfrac{-15-i\sqrt{18}}{33} \\=\dfrac{-15-i\sqrt{9(2)}}{33} \\=\dfrac{-15-i\sqrt{3^2(2)}}{33} \\=\dfrac{-15-i\cdot 3\sqrt{2}}{33} \\=\dfrac{-15-3i\sqrt{2}}{33}$ Factor out 3 in the numerator then cancel the common factors to obtain: $\require{cancel} \\=\dfrac{3(-5-i\sqrt{2})}{33} \\=\dfrac{\cancel{3}(-5-i\sqrt{2})}{\cancel{33}11} \\=\dfrac{-5-i\sqrt{2}}{11}$
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