Chapter 10 - Radical Expressions and Equations - 10-4 Solving Radical Equations - Practice and Problem-Solving Exercises - Page 623: 22

$s=-2$

$Given,$ $\sqrt (s+10)=\sqrt (6-s)$ Squaring,we get: $s+10=6-s$ $s+10+s=6-s+s$ $2s+10-10=6-10$ $2s=-4$ $s=-4\div2=-2$ We need to check by putting the obtained value in the original equation : $L.H.S=\sqrt(-2+10) =\sqrt 8=\sqrt (6-(-2))=R.H.S$ Hence,$s=-2$ is a valid solution