If `n = 7 + 4sqrt(3)` then value of `(sqrt(n) + 1/(sqrtn))` is

Options

A. `2`
B. `2sqrt(3)`
C. `4`
D. `4sqrt(3)`

Answer

To solve the problem, we need to find the value of \( \sqrt{n} + \frac{1}{\sqrt{n}} \) given that \( n = 7 + 4\sqrt{3} \).

### Step-by-step Solution:

1. **Identify \( n \)**:
\[ n = 7 + 4\sqrt{3} \]

2. **Rewrite \( n \)**:
We can express \( n \) in a different form. Notice that:
\[ n = (2 + \sqrt{3})^2 \]
This is because:
\[ (2 + \sqrt{3})^2 = 2^2 + 2 \cdot 2 \cdot \sqrt{3} + (\sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3} \]

3. **Calculate \( \sqrt{n} \)**:
Since \( n = (2 + \sqrt{3})^2 \), we have:
\[ \sqrt{n} = 2 + \sqrt{3} \]

4. **Calculate \( \frac{1}{\sqrt{n}} \)**:
We need to find \( \frac{1}{\sqrt{n}} \):
\[ \frac{1}{\sqrt{n}} = \frac{1}{2 + \sqrt{3}} \]
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 as follows:
\[ (2 + \sqrt{3})(2 - \sqrt{3}) = 4 - 3 = 1 \]
Thus:
\[ \frac{1}{\sqrt{n}} = 2 - \sqrt{3} \]

5. **Combine \( \sqrt{n} \) and \( \frac{1}{\sqrt{n}} \)**:
Now we can find \( \sqrt{n} + \frac{1}{\sqrt{n}} \):
\[ \sqrt{n} + \frac{1}{\sqrt{n}} = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 2 + \sqrt{3} + 2 - \sqrt{3} \]
The \( \sqrt{3} \) terms cancel out:
\[ = 2 + 2 = 4 \]

### Final Answer:
The value of \( \sqrt{n} + \frac{1}{\sqrt{n}} \) is \( 4 \).