Chapter 10 - Radical Expressions and Equations - 10-4 Solving Radical Equations - Practice and Problem-Solving Exercises - Page 624: 27

$None$ $of$ $the$ $solutions$ $are$ $extraneous$

$Given,$ $y=\sqrt (2y)$ Checking for $y=0$ : $L.H.S=0$ $R.H.S=\sqrt(2X0)=\sqrt 0=0$ $L.H.S=R.H.S$,hence $y=0$ is not extraneous Checking for $y=2$: $L.H.S=2$ $R.H.S=\sqrt (2X2)=\sqrt 4=2$ $L.H.S=R.H.S$,hence $y=2$ is not extraneous