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: 41

Answer

$-2\sqrt{6} - 2\sqrt{10}i$

Work Step by Step

RECALL: (1) $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab} \text{ where } a, b, \gt 0$ (2) $a(b+c) =ab+ac$ (3) $i^2=-1$ (4) For any real number $a \gt 0$, $\sqrt{-a} = i\sqrt{a}$ Use rule (4) above to obtain: $=i\sqrt{8}(i\sqrt{3}-\sqrt5) \\=i\sqrt{4(2)}(i\sqrt{3}-\sqrt{5}) \\=i\sqrt{2^2(2)}(i\sqrt{3}-\sqrt{5}) \\=i\cdot 2\sqrt{2}(i\sqrt{3}-\sqrt{5}) \\=2i\sqrt{2}(i\sqrt{3}-\sqrt{5})$ Use rules (1) and (2) to obtain: $=2i^2\sqrt{6} - 2\sqrt{10}i$ Use rule (3) above to obtain: $=2(-1)\sqrt{6} - 2\sqrt{10}i \\=-2\sqrt{6} - 2\sqrt{10}i$
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