Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 1 - Equations and Inequalities - 1.3 Complex Numbers - 1.3 Exercises - Page 111: 29

Answer

$$-13$$

Work Step by Step

Recall that the definition $\sqrt{-a}=i\sqrt{a}$ must be applied BEFORE the other rules regarding radicals, therefore, $\sqrt{-13}* \sqrt{-13} = i\sqrt{13} * i\sqrt{13}$ now you may multiply: $i^2(13)$ $(-1)(13)$ $$-13$$ which checks out because $(\sqrt{a})^2 = a$
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