If `x= sqrt3/2` , then the value of `sqrt(1+x)+sqrt(1-x)` is

Options

A. `1/sqrt3`
B. `2sqrt3`
C. `sqrt3`
D. 2

Answer

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

### Step-by-step Solution:

1. **Substitute the value of \( x \)**:
\[ x = \frac{\sqrt{3}}{2} \]
Therefore, we need to evaluate:
\[ \sqrt{1 + \frac{\sqrt{3}}{2}} + \sqrt{1 - \frac{\sqrt{3}}{2}} \]

2. **Calculate \( 1 + \frac{\sqrt{3}}{2} \)**:
\[ 1 + \frac{\sqrt{3}}{2} = \frac{2}{2} + \frac{\sqrt{3}}{2} = \frac{2 + \sqrt{3}}{2} \]

3. **Calculate \( 1 - \frac{\sqrt{3}}{2} \)**:
\[ 1 - \frac{\sqrt{3}}{2} = \frac{2}{2} - \frac{\sqrt{3}}{2} = \frac{2 - \sqrt{3}}{2} \]

4. **Substitute back into the expression**:
Now we have:
\[ \sqrt{\frac{2 + \sqrt{3}}{2}} + \sqrt{\frac{2 - \sqrt{3}}{2}} \]

5. **Factor out the common term**:
\[ = \sqrt{\frac{1}{2}(2 + \sqrt{3})} + \sqrt{\frac{1}{2}(2 - \sqrt{3})} \]
\[ = \frac{1}{\sqrt{2}} \left( \sqrt{2 + \sqrt{3}} + \sqrt{2 - \sqrt{3}} \right) \]

6. **Simplify \( \sqrt{2 + \sqrt{3}} + \sqrt{2 - \sqrt{3}} \)**:
To simplify \( \sqrt{2 + \sqrt{3}} + \sqrt{2 - \sqrt{3}} \), we can use the identity:
\[ \sqrt{a} + \sqrt{b} = \sqrt{(a+b) + 2\sqrt{ab}} \]
where \( a = 2 + \sqrt{3} \) and \( b = 2 - \sqrt{3} \).

First, calculate \( a + b \):
\[ (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4 \]
Next, calculate \( ab \):
\[ (2 + \sqrt{3})(2 - \sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1 \]

Now we can substitute back:
\[ \sqrt{2 + \sqrt{3}} + \sqrt{2 - \sqrt{3}} = \sqrt{4 + 2\sqrt{1}} = \sqrt{4 + 2} = \sqrt{6} \]

7. **Final expression**:
Substitute back into our expression:
\[ \frac{1}{\sqrt{2}} \cdot \sqrt{6} = \frac{\sqrt{6}}{\sqrt{2}} = \sqrt{3} \]

### Final Answer:
Thus, the value of \( \sqrt{1+x} + \sqrt{1-x} \) is \( \sqrt{3} \).