Chapter 10 - Radical Expressions and Equations - 10-4 Solving Radical Equations - Practice and Problem-Solving Exercises - Page 623: 15

$v=-2$

$Given,$ $1=\sqrt (-2v-3) $ Squaring,we get: $1=-2v-3$ $1+3=-2v-3+3$ $4=-2v$ $v=4\div-2=-2$ We need to check by putting the obtained value in the original equation : $R.H.S=\sqrt (-2X-2-3)=\sqrt (4-3)=\sqrt 1=1=L.H.S$ Hence,$v=-2$ is a valid solution