If `x = 2 + sqrt3` then the value of `sqrtx + 1/sqrtx` is

Options

A. `sqrt6`
B. `2sqrt6`
C. 6
D. `sqrt3`

Answer

To solve the problem, we need to find the value of \( \sqrt{x} + \frac{1}{\sqrt{x}} \) given \( x = 2 + \sqrt{3} \).

### Step-by-Step Solution:

1. **Substituting the value of \( x \)**:
We know that \( x = 2 + \sqrt{3} \). We need to find \( \sqrt{x} + \frac{1}{\sqrt{x}} \).

2. **Let \( y = \sqrt{x} + \frac{1}{\sqrt{x}} \)**:
We will square both sides to simplify our calculations:
\[ y^2 = \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)^2 \]

3. **Expanding the square**:
Using the formula \( (a + b)^2 = a^2 + b^2 + 2ab \):
\[ y^2 = x + \frac{1}{x} + 2 \]

4. **Finding \( x + \frac{1}{x} \)**:
We first need to calculate \( \frac{1}{x} \):
\[ \frac{1}{x} = \frac{1}{2 + \sqrt{3}} \]
To rationalize the denominator, multiply the numerator and denominator by the conjugate \( 2 - \sqrt{3} \):
\[ \frac{1}{2 + \sqrt{3}} \cdot \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{2 - \sqrt{3}}{(2 + \sqrt{3})(2 - \sqrt{3})} \]
The denominator simplifies to:
\[ 2^2 - (\sqrt{3})^2 = 4 - 3 = 1 \]
Thus, \( \frac{1}{x} = 2 - \sqrt{3} \).

5. **Calculating \( x + \frac{1}{x} \)**:
Now we can find \( x + \frac{1}{x} \):
\[ x + \frac{1}{x} = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4 \]

6. **Substituting back into the equation for \( y^2 \)**:
Now substitute \( x + \frac{1}{x} \) back into the equation for \( y^2 \):
\[ y^2 = 4 + 2 = 6 \]

7. **Finding \( y \)**:
Taking the square root of both sides:
\[ y = \sqrt{6} \]

### Final Answer:
Thus, the value of \( \sqrt{x} + \frac{1}{\sqrt{x}} \) is \( \sqrt{6} \).