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