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 - 10-4 Solving Radical Equations - Practice and Problem-Solving Exercises - Page 637: 32


The solution is $n=5$

Work Step by Step

$n=\sqrt 4n+5$ Square both sides: $n^2=(\sqrt 4n+5)^2$ $n^2=4n+5$ $n^2-4n-5=0$ $(n-5)(n+1)=0$ $n-5=0$ or $n+1=0$ $n=5$ or $n=-1$ With $n=5$ then $5=\sqrt 4(5)+5$ $5=5$ With $n=-1$ then $-1=\sqrt 4(-1)+5$ $-1=1$ Hence, the solution is $n=5$.
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