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 - Page 623: 2



Work Step by Step

$Given : $$\sqrt (r+5)=$$2\sqrt (r-1)$ Squaring both sides, we get : $r+5=4(r-1)$r+5 $Thus,$$r+5=4r-4$ $r+5+4=4r-4+4$ $r+9=4r$ $r+9-r=4r-r$ $9=3r$ $Thus,$$r=9\div3=3$ Then,check by substituting r = 3 in the original equation : $L.H.S=$$\sqrt (3+5)=\sqrt 8=2\sqrt 2$ $R.H.S=$$2\sqrt (3-1)=2\sqrt 2$ Since L.H.S = R.H.S,r = 3 is a solution
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