Let `S={x in R: x ge 0 and 2|(sqrt(x)-3|+sqrt(x)(sqrt(x)-6)+6=0}` then S (1) is an empty set (2) contains exactly one element (3) contains exact;y two elements (4) contains exactly four elements

Options

A. contains exactly four elements
B. is an empty set
C. contains exactly one element
D. contains exactly two elements

Explanation

To solve the problem, we start with the given equation:

\[ 2|\sqrt{x} - 3| + \sqrt{x}(\sqrt{x} - 6) + 6 = 0 \]

### Step 1: Simplify the equation
First, we can rewrite the equation:

\[ 2|\sqrt{x} - 3| + \sqrt{x}(\sqrt{x} - 6) + 6 = 0 \]

This can be rearranged to:

\[ \sqrt{x}(\sqrt{x} - 6) + 2|\sqrt{x} - 3| + 6 = 0 \]

### Step 2: Analyze the absolute value
The expression \( |\sqrt{x} - 3| \) can be split into two cases based on the value of \( \sqrt{x} \):

1. **Case 1:** \( \sqrt{x} \geq 3 \) (so \( |\sqrt{x} - 3| = \sqrt{x} - 3 \))
2. **Case 2:** \( \sqrt{x} < 3 \) (so \( |\sqrt{x} - 3| = 3 - \sqrt{x} \))

### Step 3: Solve Case 1
For \( \sqrt{x} \geq 3 \):

Substituting into the equation:

\[ 2(\sqrt{x} - 3) + \sqrt{x}(\sqrt{x} - 6) + 6 = 0 \]

This simplifies to:

\[ 2\sqrt{x} - 6 + \sqrt{x}^2 - 6\sqrt{x} + 6 = 0 \]

Combining like terms gives:

\[ \sqrt{x}^2 - 4\sqrt{x} = 0 \]

Factoring out \( \sqrt{x} \):

\[ \sqrt{x}(\sqrt{x} - 4) = 0 \]

This gives us:

1. \( \sqrt{x} = 0 \) → \( x = 0 \)
2. \( \sqrt{x} = 4 \) → \( x = 16 \)

Both values \( x = 0 \) and \( x = 16 \) are valid since \( \sqrt{x} \geq 3 \) is satisfied for \( x = 16 \).

### Step 4: Solve Case 2
For \( \sqrt{x} < 3 \):

Substituting into the equation:

\[ 2(3 - \sqrt{x}) + \sqrt{x}(\sqrt{x} - 6) + 6 = 0 \]

This simplifies to:

\[ 6 - 2\sqrt{x} + \sqrt{x}^2 - 6\sqrt{x} + 6 = 0 \]

Combining like terms gives:

\[ \sqrt{x}^2 - 8\sqrt{x} + 12 = 0 \]

This can be factored as:

\[ (\sqrt{x} - 6)(\sqrt{x} - 2) = 0 \]

This gives us:

1. \( \sqrt{x} = 6 \) → \( x = 36 \) (not valid since \( \sqrt{x} < 3 \))
2. \( \sqrt{x} = 2 \) → \( x = 4 \) (valid since \( \sqrt{x} < 3 \))

### Step 5: Collect all valid solutions
The valid solutions for \( x \) are:

1. \( x = 0 \)
2. \( x = 4 \)
3. \( x = 16 \)

### Conclusion
Thus, the set \( S \) contains exactly three elements: \( \{0, 4, 16\} \).

### Answer
The correct option is (2) contains exactly three elements.