Question

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

• A. $4$
• B. $10$
• C. $9$
• D. $12$

Answer 1

The correct option is B

$10$

Explanation of the correct option.

Step 1: Find the solutions.

Given : $\left|\sqrt{x}-2\right|+\sqrt{x}\left(\sqrt{x}-4\right)+2=0,(x>0)$

Let $\sqrt{x}=t$,

$\Rightarrow \left|t-2\right|+t\left(t-4\right)+2=0$

If $t\ge 2$,

$\Rightarrow$ $t-2+t^2-4t+2=0$

$\Rightarrow$ $t^2-3t=0$

$\Rightarrow$ $t(t-3)=0$

Thus, $t=3$ or $t=0$.

Since $t\ge 2$, $t=0$ is not possible.

So if $t=3$ then $x=9$

If $t<2$,

$\Rightarrow$$2-t+t^2-4t+2=0$

$\Rightarrow$ $t^2-5t+4=0$

$\Rightarrow$ $t^2-4t-t+4=0$

$\Rightarrow$ $\left(t-4\right)\left(t-1\right)=0$

Thus, $t=4$ or $t=1$.

Since $t<2$, $t=4$ is not possible.

So if $t=1$ then $x=1$.

Step 2: Compute the required sum.

Since solutions are $9$ and $1$.

Sum of solutions $=9+1$

Sum of solutions $=10$

Hence, option B is the correct option.