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 - Practice and Problem-Solving Exercises - Page 624: 30

Answer

$none$ $of$ $the$ $solutions$ $are$ $extraneous$

Work Step by Step

$Given,$ $-t=\sqrt (-6t-5)$ Checking for $t=-5$: $L.H.S=-(-5)=5$ $R.H.S=\sqrt ((-6X-5)-5)=\sqrt (30-5)=\sqrt 25=5$ $L.H.S=R.H.S$,hence not extraneous Checking for $t=-1$: $L.H.S=-(-1)=1$ $R.H.S=\sqrt ((-6X-1)-5)=\sqrt (6-5)=\sqrt 1=1$ $L.H.S=R.H.S$,hence not extraneous
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