if `|sqrt(x)-2|+sqrt(x)(sqrt(x)-4)+2=0` then find the sum of roots of equation (A) 12 (B) 8 (C) 4 (D) 10

Options

A. 9
B. 12
C. 4
D. 10

Answer

To solve the equation \( |\sqrt{x} - 2| + \sqrt{x}(\sqrt{x} - 4) + 2 = 0 \), we will follow these steps:

### Step 1: Analyze the equation
We start with the equation:
\[ |\sqrt{x} - 2| + \sqrt{x}(\sqrt{x} - 4) + 2 = 0 \]
This equation consists of an absolute value term and a quadratic term in \(\sqrt{x}\).

### Step 2: Simplify the equation
Let's rewrite the quadratic term:
\[ \sqrt{x}(\sqrt{x} - 4) = x - 4\sqrt{x} \]
Thus, the equation becomes:
\[ |\sqrt{x} - 2| + x - 4\sqrt{x} + 2 = 0 \]

### Step 3: Consider cases for the absolute value
We need to consider two cases for the absolute value \( |\sqrt{x} - 2| \).

**Case 1:** \( \sqrt{x} - 2 \geq 0 \) (i.e., \( \sqrt{x} \geq 2 \) or \( x \geq 4 \)):
In this case, \( |\sqrt{x} - 2| = \sqrt{x} - 2 \). The equation becomes:
\[ \sqrt{x} - 2 + x - 4\sqrt{x} + 2 = 0 \]
This simplifies to:
\[ x - 3\sqrt{x} = 0 \]
Factoring out \(\sqrt{x}\):
\[ \sqrt{x}(x - 3) = 0 \]
This gives us two solutions:
1. \( \sqrt{x} = 0 \) (which gives \( x = 0 \), but does not satisfy \( x \geq 4 \))
2. \( x - 3 = 0 \) (which gives \( x = 9 \), and satisfies \( x \geq 4 \))

**Case 2:** \( \sqrt{x} - 2 < 0 \) (i.e., \( \sqrt{x} < 2 \) or \( x < 4 \)):
In this case, \( |\sqrt{x} - 2| = 2 - \sqrt{x} \). The equation becomes:
\[ 2 - \sqrt{x} + x - 4\sqrt{x} + 2 = 0 \]
This simplifies to:
\[ x - 5\sqrt{x} + 4 = 0 \]
Letting \( y = \sqrt{x} \), we rewrite this as:
\[ y^2 - 5y + 4 = 0 \]
Factoring gives:
\[ (y - 4)(y - 1) = 0 \]
This gives us:
1. \( y = 4 \) (which gives \( x = 16 \), but does not satisfy \( x < 4 \))
2. \( y = 1 \) (which gives \( x = 1 \), and satisfies \( x < 4 \))

### Step 4: Collect valid solutions
The valid solutions from both cases are:
- From Case 1: \( x = 9 \)
- From Case 2: \( x = 1 \)

### Step 5: Find the sum of the roots
The sum of the roots is:
\[ 1 + 9 = 10 \]

### Final Answer
The sum of the roots of the equation is \( \boxed{10} \).