Question

The sum of the solution of the equation $\left| \sqrt{x}-2 \right|+\sqrt{x}\left( \sqrt{x}-4 \right)+2=0$ is equal to:
A. 4
B. 9
C. 10
D. 12

Answer 1

Hint: To solve the question, first we have to convert this equation into the quadratic equation form. We can start by considering the fact that ${{\left| \sqrt{x}-2 \right|}^{2}}=x-4\sqrt{x}+4$ and representing the given equation such that we can apply the above fact in it and reduce it to a quadratic equation Then only we can use the equation to find the roots of the quadratic equation, which is $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. After finding the roots we have to add it because in question it is asked to find the sum of the solutions.

Complete step by step answer:
The given equation is $\left| \sqrt{x}-2 \right|+\sqrt{x}\left( \sqrt{x}-4 \right)+2=0$.
The first step we have to do is to convert it into the quadratic form. For that we have expanded the equation. So, on expanding we get,
$\left| \sqrt{x}-2 \right|+x-4\sqrt{x}+2=0$……..(1)
Now, we know that ${{\left| \sqrt{x}-2 \right|}^{2}}=x-4\sqrt{x}+4$
So we can rewrite the equation (1) as
$\left| \sqrt{x}-2 \right|+x-4\sqrt{x}+4-2=0$
So, now we can reduce the equation to the quadratic form by replacing $x-4\sqrt{x}+4\text{ as }{{\left| \sqrt{x}-2 \right|}^{2}}$. So, the equation (1) will become,
$\left| \sqrt{x}-2 \right|+{{\left| \sqrt{x}-2 \right|}^{2}}-2=0$
We can rewrite the equation as,
${{\left| \sqrt{x}-2 \right|}^{2}}+\left| \sqrt{x}-2 \right|-2=0$
So, now the equation is reduced to the quadratic form. Now we have to find the roots. So, the general equation to find the roots are,
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
So, we can put $\left| \sqrt{x}-2 \right|=y$
So, the equation becomes,
${{y}^{2}}+y-2=0$
Now, we can use the equation to find the root as
$y=\dfrac{-1\pm \sqrt{{{(-1)}^{2}}-4(1)(-2)}}{2(1)}$
$\Rightarrow y=\dfrac{-1\pm \sqrt{1+8}}{2}$
$\begin{align}
  & \Rightarrow y=\dfrac{-1\pm \sqrt{9}}{2} \\
 & \Rightarrow y=\dfrac{-1\pm 3}{2} \\
\end{align}$
So we get that the value of y can be either
$y=\dfrac{-1+3}{2}=1$
Or
$y=\dfrac{-1-3}{2}=-2$.
So, we have y = 1, -2.
We know that $\left| \sqrt{x}-2 \right|=y$, so we has to find the values of x.
Case 1: If we take y = 1, then
$\begin{align}
  & \left| \sqrt{x}-2 \right|=1 \\
 & \Rightarrow \pm \left( \sqrt{x}-2 \right)=1 \\
\end{align}$
If $\left( \sqrt{x}-2 \right)$is positive then, we have
$\begin{align}
  & \left( \sqrt{x}-2 \right)=1 \\
 & \Rightarrow \sqrt{x}=3 \\
 & \Rightarrow x=9 \\
\end{align}$
If $\left( \sqrt{x}-2 \right)$ is negative then, we have
$\begin{align}
  & -\left( \sqrt{x}-2 \right)=1 \\
 & \Rightarrow \left( \sqrt{x}-2 \right)=-1 \\
 & \Rightarrow \sqrt{x}=1 \\
 & \Rightarrow x=1 \\
\end{align}$
So, the sum of solution when $\left| \sqrt{x}-2 \right|=1$ will be 9 + 1 = 10.
Case 2: If we take y = -2, then
$\begin{align}
  & \left| \sqrt{x}-2 \right|=-2 \\
 & \Rightarrow \pm \left( \sqrt{x}-2 \right)=-2 \\
\end{align}$
If $\left( \sqrt{x}-2 \right)$is positive then, we have
$\begin{align}
  & \left( \sqrt{x}-2 \right)=-2 \\
 & \Rightarrow \sqrt{x}=0 \\
 & \Rightarrow x=0 \\
\end{align}$
If $\left( \sqrt{x}-2 \right)$ is negative then, we have
$\begin{align}
  & -\left( \sqrt{x}-2 \right)=-2 \\
 & \Rightarrow \left( \sqrt{x}-2 \right)=2 \\
 & \Rightarrow \sqrt{x}=4 \\
 & \Rightarrow x=16 \\
\end{align}$
So, the sum of solution when $\left| \sqrt{x}-2 \right|=-2$ will be 0 + 16 = 16.

So, the correct answer is “Option C”.

Note: Sometimes we stop solving the problem when we get the value of ‘y’. But, we have to keep in mind that we are asked to find the solution of the equation. So, we have to find the value of ‘x’ in both the cases. Also, we have to keep in mind that modulus of a function can be either positive or negative. So, we have to solve both conditions.