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: 39

Answer

$\dfrac{-3+i\sqrt{3}}{24}$

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: $=\dfrac{-6+i\sqrt{12}}{48} \\=\dfrac{-6+i\sqrt{4(3)}}{48} \\=\dfrac{-6+i\sqrt{2^2(3)}}{48} \\=\dfrac{-6+i\cdot 2\sqrt{3}}{48} \\=\dfrac{-6+2i\sqrt{3}}{48}$ Factor out 2 in the numerator then cancel the common factors to obtain: $\require{cancel} \\=\dfrac{2(-3+i\sqrt{3})}{48} \\=\dfrac{\cancel{2}(-3+i\sqrt{3})}{\cancel{48}24} \\=\dfrac{-3+i\sqrt{3}}{24}$
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