Answer
$f=21$
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 $f=21$
Algebra 1: Common Core (15th Edition)
by Charles, Randall I.
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
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