Algebra 1: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281140
ISBN 13: 978-0-13328-114-9

Chapter 10 - Radical Expressions and Equations - Chapter Review - Page 655: 40



Work Step by Step

You are given $\sqrt {f+4}=5$. Solve for f: $\sqrt {f+4}=5$ $\sqrt {(f+4)^2} = {5}^{2}$ $f+4=25$ $f=21$ Check solution: $\sqrt {21+4}=5$ $\sqrt {25}=5$ $5=5$ The solution is $f=21$
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