Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 7 - Section 7.7 - Complex Numbers - Exercise Set - Page 570: 10



Work Step by Step

Factor the radicand so that one factor $-1$ to obtain: $=-\sqrt{300(-1)} \\=-\sqrt{100(3)(-1)} \\=-\sqrt{10^2(3)(-1)}$ RECALL: (1) $\sqrt{abc} = \sqrt{a} \cdot \sqrt{b} \cdot \sqrt{c}$ (2) $\sqrt{-1} = i$ Use rule (1) above to obtain: $=-\sqrt{10^2} \cdot \sqrt{3} \cdot \sqrt{-1} \\=-10 \cdot \sqrt{3} \cdot \sqrt{-1} \\=-10\sqrt{3} \cdot \sqrt{-1}$ Use rule (2) above to obtain: $=-10\sqrt{3} \cdot i \\=-10i\sqrt{3}$
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