Question

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

• A. $4$
• B. $9$
• C. $10$
• D. $12$

Answer 1

The correct option is C $10$

Case $1$: When $\sqrt{x}\ge 2$
$\sqrt{x}-2+x-4\sqrt{x}+2=0\Rightarrow x-3\sqrt{x}=0\Rightarrow \sqrt{x}(\sqrt{x}-3)$
$\Rightarrow \sqrt{x}=0or\sqrt{x}-3=0$
$\therefore x=0,x=9$
As, $\sqrt{x}\ge 2\Rightarrow x=9$

Case $2$: When $\sqrt{x}<2$
$2-\sqrt{x}+x-4\sqrt{x}+2=0\Rightarrow x-5\sqrt{x}+4=0$
$(\sqrt{x}{)}^2-4\sqrt{x}-\sqrt{x}+4=0$
$\Rightarrow \sqrt{x}(\sqrt{x}-4)-1(\sqrt{x}-1)=0$
$\Rightarrow (\sqrt{x}-4)(\sqrt{x}-1)=0\Rightarrow x=1,16$
As, $\sqrt{x}<2\Rightarrow x=1$

Hence, sum $=9+1=10$