Question

Case-I : x $\in$ [0, 9]
2(3 - $\sqrt{x}$) + x - 6$\sqrt{x}$ + 6 = 0
$\Rightarrow$ x - 8$\sqrt{x}$ + 12 = 0 $\Rightarrow$$\sqrt{x}$ = 4, 2 
x = 16, 4 $\Rightarrow$x = 4
     $\uparrow$
rejected
Case-II : x $\in$ [9, $\infty$]
2($\sqrt{x}$ -3) + x - 6$\sqrt{x}$ + 6 = 0
x - 4$\sqrt{x}$ = 0 $\Rightarrow$x = 16, 0
                               $\uparrow$
                          rejected
So x = 4, 16