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: 42

Answer

$-4\sqrt{3} - 2\sqrt{6}i$

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: $=i\sqrt{12}(i\sqrt{4}-\sqrt{2}) \\=i\sqrt{4(3)}(i\sqrt{2^2}-\sqrt{2}) \\=i\sqrt{2^2(3)}(2i-\sqrt{2}) \\=i\cdot 2\sqrt{3}(2i-\sqrt{2}) \\=2i\sqrt{3}(2i-\sqrt{2})$ Use rules (1) and (2) to obtain: $=4i^2\sqrt{3} - 2\sqrt{6}i$ Use rule (3) above to obtain: $=4(-1)\sqrt{3} - 2\sqrt{6}i \\=-4\sqrt{3} - 2\sqrt{6}i$
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