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 636: 19



Work Step by Step

We are given: $\sqrt 3x+1=\sqrt 5x-8$ Square each side: $(\sqrt 3x+1)^2=(\sqrt 5x-8)^2$ $3x+1=5x-8$ $2x=9$ $x=\frac{9}{2}$ Check: $\sqrt 3(\frac{9}{2})+1=\sqrt 5(\frac{9}{2})-8$ $\frac{\sqrt 58}{2}=\frac{\sqrt 58}{2}$ Therefore, the solution is $x=\frac{9}{2}$
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