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: 34


The solution is $y=-\frac{3}{4}$

Work Step by Step

$2y=\sqrt 5y+6$ Square both sides: $(2y)^2=(\sqrt 5y+6)^2$ $4y^2=5y+6$ $4y^2-5n-6=0$ $(y-2)(4y+3)=0$ $y-2=0$ or $4y+3=0$ $y=2$ or $y=-\frac{3}{4}$ With $y=2$ then $2(2)=\sqrt 5(2)+6$ $4=4$ With $y=-\frac{3}{4}$ then $2(-\frac{3}{4})=\sqrt 5(-\frac{3}{4})+6$ $-\frac{3}{2} \ne \frac{3}{3}$ Hence, the solution is $y=-\frac{3}{4}$
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