If `a = (2 + sqrt3)/(2 - sqrt3) ` and `b = (2 - sqrt3)/(2 + sqrt3)` , then the value of `(a^(2) + b^(2) +ab)` is :

Options

A. 185
B. 195
C. 200
D. 175

Answer

To solve for the value of \( a^2 + b^2 + ab \) given \( a = \frac{2 + \sqrt{3}}{2 - \sqrt{3}} \) and \( b = \frac{2 - \sqrt{3}}{2 + \sqrt{3}} \), we can follow these steps:

### Step 1: Calculate \( a \) and \( b \)

We start by calculating \( a \) and \( b \).

\[ a = \frac{2 + \sqrt{3}}{2 - \sqrt{3}} \]

To simplify \( a \), we can multiply the numerator and denominator by the conjugate of the denominator:

\[ a = \frac{(2 + \sqrt{3})(2 + \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3})} \]

Calculating the denominator:

\[ (2 - \sqrt{3})(2 + \sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1 \]

Calculating the numerator:

\[ (2 + \sqrt{3})^2 = 2^2 + 2 \cdot 2 \cdot \sqrt{3} + (\sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3} \]

Thus, we have:

\[ a = 7 + 4\sqrt{3} \]

Now, calculating \( b \):

\[ b = \frac{2 - \sqrt{3}}{2 + \sqrt{3}} \]

Similarly, we multiply by the conjugate:

\[ b = \frac{(2 - \sqrt{3})(2 - \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})} \]

The denominator is still 1, and the numerator is:

\[ (2 - \sqrt{3})^2 = 2^2 - 2 \cdot 2 \cdot \sqrt{3} + (\sqrt{3})^2 = 4 - 4\sqrt{3} + 3 = 7 - 4\sqrt{3} \]

Thus, we have:

\[ b = 7 - 4\sqrt{3} \]

### Step 2: Calculate \( a^2 + b^2 + ab \)

Now we need to find \( a^2 + b^2 + ab \).

Using the identity \( a^2 + b^2 = (a + b)^2 - 2ab \):

1. Calculate \( a + b \):

\[ a + b = (7 + 4\sqrt{3}) + (7 - 4\sqrt{3}) = 14 \]

2. Calculate \( ab \):

\[ ab = \left( \frac{2 + \sqrt{3}}{2 - \sqrt{3}} \right) \left( \frac{2 - \sqrt{3}}{2 + \sqrt{3}} \right) = \frac{(2 + \sqrt{3})(2 - \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3})} = \frac{4 - 3}{1} = 1 \]

3. Now substitute into the identity:

\[ a^2 + b^2 = (a + b)^2 - 2ab = 14^2 - 2 \cdot 1 = 196 - 2 = 194 \]

4. Finally, calculate \( a^2 + b^2 + ab \):

\[ a^2 + b^2 + ab = 194 + 1 = 195 \]

### Final Answer:

The value of \( a^2 + b^2 + ab \) is \( \boxed{195} \).