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: 35

Answer

$2+6i\sqrt{7}$

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: $=(-3-i\sqrt{7})^2$ Use rule (1) above with $a=-3$ and $b=i\sqrt{7}$ to obtain: $=(-3)^2-2(-3)(i\sqrt{7}) + (i\sqrt{7})^2 \\=9+6i\sqrt{7}+i^2(7) \\=9+6i\sqrt{7} + 7i^2$ Use rule (3) above to obtain: $=9+6i\sqrt{7} + 7(-1) \\=9+6i\sqrt{7} -7 \\=2+6i\sqrt{7}$
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