Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 7 - Section 7.1 - Radicals and Radical Functions - Exercise Set: 8



Work Step by Step

$\sqrt 9=3$, because $(3)^{2}=(3\times3)=9$ By adding a negative sign in front of $\sqrt 9$, we are taking the negative of $\sqrt 9$, which is $-(3)=-3$
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