The sum of the solutions of the equation $\left|\sqrt{x}-2\right|+\sqrt{x}\left(\sqrt{x}-4\right)+2=0,\left(x>0\right)$ is equal to

Given,

     $\left|\sqrt{x}-2\right|+\sqrt{x}\left(\sqrt{x}-4\right)+2=0$
$\Rightarrow \left|\sqrt{x}-2\right|+x-4\sqrt{x}+2=0$
$\Rightarrow \left|\sqrt{x}-2\right|+{\left(\sqrt{x}\right)}^2-4\sqrt{x}+4=2$       (Adding $2$ on both sides)
$\Rightarrow {\left(\sqrt{x}-2\right)}^2+\left|\sqrt{x}-2\right|=2$
$\Rightarrow t^2+t-2=0$; $\left|\sqrt{x}-2\right|=t ; \left(t>0\right)$
$\Rightarrow t=1$or $t=-2$ ; Rejected because $t>0$
$\Rightarrow \left|\sqrt{x}-2\right|=1$
$\Rightarrow \sqrt{x}=3$ or $\sqrt{x}=1$
$\Rightarrow x=9$ or $x=1$
Hence, the required sum $=10$.