Answer
$d=4$
Work Step by Step
$Given,$ $\sqrt (d+12)=d$
Squaring both sides,we get:
$d+12=d^{2}$
$d^{2}-d-12=0$
$d^{2}-4d+3d-12$
$(d-4)(d+3)=0$
$d=4,d=-3$
We need to check by the putting the obtained solutions in
the original equation.
$d=4$
$L.H.S=\sqrt (4+12)=\sqrt 16=4$
$R.H.S=4$
$L.H.S=R.H.S$,hence it is a valid solution
$d=-3$
$L.H.S=\sqrt(-3+12)=\sqrt 9=3$
$R.H.S=-3$
$L.H.S$ does not equal $R.H.S$,hence it is an extraneous solution.
