Chapter 10 - Radical Expressions and Equations - 10-4 Solving Radical Equations - Practice and Problem-Solving Exercises - Page 624: 30

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

$Given,$ $-t=\sqrt (-6t-5)$ 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