## Algebra 1

$C$ is the extraneous solution.
$Given,$ $s=\sqrt (s+2)$ 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: $L.H.S=2$ $and$ $R.H.S=\sqrt (2+2)=\sqrt 4=2$