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:
$=i\sqrt{64}-i\sqrt{25}
\\=i\sqrt{8^2} - i\sqrt{5^2}
\\=i(8) - i(5)
\\=8i-5i
\\=3i$
Precalculus (6th Edition) Blitzer
by Blitzer, Robert F.
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 - Page 314: 29
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