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: 38



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: $=\dfrac{-12+i\sqrt{28}}{32} \\=\dfrac{-12+i\sqrt{4(7)}}{32} \\=\dfrac{-12+i\sqrt{2^2(7)}}{32} \\=\dfrac{-12+i\cdot 2\sqrt{7}}{32} \\=\dfrac{-12+2i\sqrt{7}}{32}$ Factor out 2 in the numerator then cancel the common factors to obtain: $\require{cancel} \\=\dfrac{2(-6+i\sqrt{7})}{32} \\=\dfrac{\cancel{2}(-6+i\sqrt{7})}{\cancel{32}16} \\=\dfrac{-6+i\sqrt{7}}{16}$
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