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: 29

Answer

$3i$

Work Step by Step

RECALL: (1) $\sqrt{-1}=i$ (2) For any real number $a\gt0$, $\sqrt{-a} = i\sqrt{a}$. Use rule (2) above to obtain: $=i\sqrt{64}-i\sqrt{25} \\=i\sqrt{8^2} - i\sqrt{5^2} \\=i(8) - i(5) \\=8i-5i \\=3i$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.