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 7.1 - Radical Expressions and Functions - Exercise Set - Page 514: 105

Answer

See the explanation below.

Work Step by Step

As per the definition of square root property, $\sqrt x= \pm x$ Thus, the roots of $36$ are $\sqrt{36}= \pm 6$ Or, $6$ and $-6$ Since, $(6)^2=36$ and $(-6)^2=36$
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