If `a=7-4sqrt(3)` , the value of `a^((1)/(2))+a^(-(1)/(2))` is

Options

A. `3sqrt(3)`
B. `4`
C. `7`
D. `2sqrt(3)`

Answer

To solve the problem, we need to find the value of \( a^{\frac{1}{2}} + a^{-\frac{1}{2}} \) where \( a = 7 - 4\sqrt{3} \).

### Step-by-Step Solution:

1. **Identify the expression**: We are given \( a = 7 - 4\sqrt{3} \). We need to find \( \sqrt{a} + \frac{1}{\sqrt{a}} \).

2. **Calculate \( \sqrt{a} \)**:
\[ \sqrt{a} = \sqrt{7 - 4\sqrt{3}} \]
To simplify this expression, we can assume \( \sqrt{a} = \sqrt{x} - \sqrt{y} \) where \( x \) and \( y \) are numbers we need to determine.

3. **Square both sides**:
\[ a = (\sqrt{x} - \sqrt{y})^2 = x + y - 2\sqrt{xy} \]
We want this to equal \( 7 - 4\sqrt{3} \). This gives us the equations:
\[ x + y = 7 \quad \text{(1)} \]
\[ -2\sqrt{xy} = -4\sqrt{3} \quad \Rightarrow \quad \sqrt{xy} = 2\sqrt{3} \quad \Rightarrow \quad xy = 12 \quad \text{(2)} \]

4. **Solve the system of equations**:
From equations (1) and (2):
- \( x + y = 7 \)
- \( xy = 12 \)

Let \( x \) and \( y \) be the roots of the quadratic equation:
\[ t^2 - (x+y)t + xy = 0 \quad \Rightarrow \quad t^2 - 7t + 12 = 0 \]

We can find the roots using the quadratic formula:
\[ t = \frac{7 \pm \sqrt{7^2 - 4 \cdot 12}}{2} = \frac{7 \pm \sqrt{49 - 48}}{2} = \frac{7 \pm 1}{2} \]
This gives us:
\[ t = 4 \quad \text{or} \quad t = 3 \]
Thus, \( x = 4 \) and \( y = 3 \).

5. **Find \( \sqrt{a} \)**:
\[ \sqrt{a} = \sqrt{4} - \sqrt{3} = 2 - \sqrt{3} \]

6. **Calculate \( \frac{1}{\sqrt{a}} \)**:
\[ \frac{1}{\sqrt{a}} = \frac{1}{2 - \sqrt{3}} \]
To rationalize the denominator, multiply by the conjugate:
\[ \frac{1}{2 - \sqrt{3}} \cdot \frac{2 + \sqrt{3}}{2 + \sqrt{3}} = \frac{2 + \sqrt{3}}{(2 - \sqrt{3})(2 + \sqrt{3})} = \frac{2 + \sqrt{3}}{4 - 3} = 2 + \sqrt{3} \]

7. **Combine the results**:
Now we have:
\[ \sqrt{a} + \frac{1}{\sqrt{a}} = (2 - \sqrt{3}) + (2 + \sqrt{3}) = 4 \]

### Final Answer:
The value of \( a^{\frac{1}{2}} + a^{-\frac{1}{2}} \) is \( 4 \).