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: 47


The solution is $y=\frac{3}{2}$.

Work Step by Step

$\sqrt y+12=3\sqrt y$ Square both sides: $(\sqrt y+12)^2=(3\sqrt y)^2$ $y+12=9y$ $8y=12$ $y=\frac{3}{2}$ With $y=\frac{3}{2}$ then $\sqrt \frac{3}{2}+12=3\sqrt \frac{3}{2}$ $3\sqrt \frac{3}{2}=3\sqrt \frac{3}{2}$ Hence, the solution is $y=\frac{3}{2}$.
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