Answer
not a real number
Work Step by Step
$\sqrt (-16)$, is not a real number, because there is no number $a$, such that $a^{2}=-16$.
For example, $4^{2}=16$ and $(-4)^{2}=16$
Intermediate Algebra (6th Edition)
by Martin-Gay, Elayn
Published by
Pearson
ISBN 10:
0321785045
ISBN 13:
978-0-32178-504-6
Chapter 7 - Section 7.1 - Radicals and Radical Functions - Exercise Set - Page 417: 32
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