If `x = 7 + 4 sqrt3` then value of `sqrtx + (1)/(x)` is

Options

A. `4`
B. `sqrtx`
C. `sqrt3`
D. `2`

Answer

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

### Step-by-step Solution:

1. **Substitute the value of x**:
\[ \sqrt{x} + \frac{1}{x} = \sqrt{7 + 4\sqrt{3}} + \frac{1}{7 + 4\sqrt{3}} \]

2. **Rationalize the second term**:
To simplify \( \frac{1}{7 + 4\sqrt{3}} \), we multiply the numerator and denominator by the conjugate \( 7 - 4\sqrt{3} \):
\[ \frac{1}{7 + 4\sqrt{3}} \cdot \frac{7 - 4\sqrt{3}}{7 - 4\sqrt{3}} = \frac{7 - 4\sqrt{3}}{(7 + 4\sqrt{3})(7 - 4\sqrt{3})} \]

3. **Calculate the denominator**:
Using the difference of squares:
\[ (7 + 4\sqrt{3})(7 - 4\sqrt{3}) = 7^2 - (4\sqrt{3})^2 = 49 - 48 = 1 \]
Therefore, the expression simplifies to:
\[ \frac{1}{7 + 4\sqrt{3}} = 7 - 4\sqrt{3} \]

4. **Combine the terms**:
Now we can write:
\[ \sqrt{7 + 4\sqrt{3}} + (7 - 4\sqrt{3}) \]

5. **Calculate \( \sqrt{7 + 4\sqrt{3}} \)**:
We can express \( \sqrt{7 + 4\sqrt{3}} \) in a simpler form. Let's assume:
\[ \sqrt{7 + 4\sqrt{3}} = a + b\sqrt{3} \]
Squaring both sides:
\[ 7 + 4\sqrt{3} = a^2 + 3b^2 + 2ab\sqrt{3} \]
This gives us two equations:
\[ a^2 + 3b^2 = 7 \quad \text{(1)} \]
\[ 2ab = 4 \quad \text{(2)} \]
From equation (2), we find:
\[ ab = 2 \quad \Rightarrow \quad b = \frac{2}{a} \]
Substituting \( b \) in equation (1):
\[ a^2 + 3\left(\frac{2}{a}\right)^2 = 7 \]
\[ a^2 + \frac{12}{a^2} = 7 \]
Multiplying through by \( a^2 \):
\[ a^4 - 7a^2 + 12 = 0 \]
Letting \( y = a^2 \):
\[ y^2 - 7y + 12 = 0 \]
Factoring:
\[ (y - 3)(y - 4) = 0 \quad \Rightarrow \quad y = 3 \quad \text{or} \quad y = 4 \]
Thus, \( a^2 = 3 \) or \( a^2 = 4 \), giving \( a = \sqrt{3} \) or \( a = 2 \).

If \( a = 2 \), then \( b = 1 \) (since \( ab = 2 \)). Thus:
\[ \sqrt{7 + 4\sqrt{3}} = 2 + \sqrt{3} \]

6. **Final Calculation**:
Now substituting back:
\[ \sqrt{7 + 4\sqrt{3}} + (7 - 4\sqrt{3}) = (2 + \sqrt{3}) + (7 - 4\sqrt{3}) = 9 - 3\sqrt{3} \]

### Final Answer:
\[ \sqrt{x} + \frac{1}{x} = 9 - 3\sqrt{3} \]