Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 10 - Radical Expressions and Equations - 10-4 Solving Radical Equations - Lesson Check: 5

Answer

$C$ is the extraneous solution.

Work Step by Step

$Given,$ $s=\sqrt (s+2)$ Squaring both sides, $s^{2}=s+2$ $Thus,$$s^{2}-s-2=0$ $s^{2}-2s+s-2=0$ $s(s-2)+1(s-2)=0$ $(s+1)(s-2)=0$ $Hence,$$s=-1$ $OR$ $s=2$ Checking by putting values in the original equation, for s = -1: $L.H.S=-1$ $and$ $R.H.S=1$ Thus,discarded. for s = 2: $L.H.S=2$ $and$ $R.H.S=\sqrt (2+2)=\sqrt 4=2$
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