If a = `(sqrt(3) + sqrt(2))/(sqrt(3) - sqrt(2)) and b = (sqrt(3) - sqrt(2))/(sqrt(3) + sqrt(2))`, then value of `a^(2) + b^(2) ` is

Options

A. 28
B. 58
C. 10
D. 98

Answer

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

### Step 1: Calculate \( a^2 \)
First, we calculate \( a^2 \):
\[ a^2 = \left( \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} \right)^2 = \frac{(\sqrt{3} + \sqrt{2})^2}{(\sqrt{3} - \sqrt{2})^2} \]
Calculating the numerator:
\[ (\sqrt{3} + \sqrt{2})^2 = 3 + 2 + 2\sqrt{3}\sqrt{2} = 5 + 2\sqrt{6} \]
Calculating the denominator:
\[ (\sqrt{3} - \sqrt{2})^2 = 3 + 2 - 2\sqrt{3}\sqrt{2} = 1 - 2\sqrt{6} \]
Thus,
\[ a^2 = \frac{5 + 2\sqrt{6}}{1 - 2\sqrt{6}} \]

### Step 2: Calculate \( b^2 \)
Now, we calculate \( b^2 \):
\[ b^2 = \left( \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}} \right)^2 = \frac{(\sqrt{3} - \sqrt{2})^2}{(\sqrt{3} + \sqrt{2})^2} \]
Calculating the numerator:
\[ (\sqrt{3} - \sqrt{2})^2 = 3 + 2 - 2\sqrt{3}\sqrt{2} = 1 - 2\sqrt{6} \]
Calculating the denominator:
\[ (\sqrt{3} + \sqrt{2})^2 = 3 + 2 + 2\sqrt{3}\sqrt{2} = 5 + 2\sqrt{6} \]
Thus,
\[ b^2 = \frac{1 - 2\sqrt{6}}{5 + 2\sqrt{6}} \]

### Step 3: Calculate \( a^2 + b^2 \)
Now, we can find \( a^2 + b^2 \):
\[ a^2 + b^2 = \frac{5 + 2\sqrt{6}}{1 - 2\sqrt{6}} + \frac{1 - 2\sqrt{6}}{5 + 2\sqrt{6}} \]
To add these fractions, we need a common denominator:
\[ = \frac{(5 + 2\sqrt{6})^2 + (1 - 2\sqrt{6})^2}{(1 - 2\sqrt{6})(5 + 2\sqrt{6})} \]

Calculating the numerator:
\[ (5 + 2\sqrt{6})^2 = 25 + 20\sqrt{6} + 24 = 49 + 20\sqrt{6} \]
\[ (1 - 2\sqrt{6})^2 = 1 - 4\sqrt{6} + 24 = 25 - 4\sqrt{6} \]
Thus,
\[ 49 + 20\sqrt{6} + 25 - 4\sqrt{6} = 74 + 16\sqrt{6} \]

Calculating the denominator:
\[ (1 - 2\sqrt{6})(5 + 2\sqrt{6}) = 5 + 2\sqrt{6} - 10\sqrt{6} - 24 = -19 - 8\sqrt{6} \]

Thus,
\[ a^2 + b^2 = \frac{74 + 16\sqrt{6}}{-19 - 8\sqrt{6}} \]

### Step 4: Simplify the expression
To simplify this, we can multiply the numerator and denominator by the conjugate of the denominator:
\[ = \frac{(74 + 16\sqrt{6})(-19 + 8\sqrt{6})}{(-19 - 8\sqrt{6})(-19 + 8\sqrt{6})} \]
Calculating the denominator:
\[ (-19)^2 - (8\sqrt{6})^2 = 361 - 384 = -23 \]

Calculating the numerator:
\[ = 74 \cdot (-19) + 74 \cdot 8\sqrt{6} + 16\sqrt{6} \cdot (-19) + 16\sqrt{6} \cdot 8\sqrt{6} \]
\[ = -1406 + 592\sqrt{6} - 304\sqrt{6} + 768 = -638 + 288\sqrt{6} \]

Thus, we have:
\[ a^2 + b^2 = \frac{-638 + 288\sqrt{6}}{-23} \]

### Final Result
The final value of \( a^2 + b^2 \) is:
\[ \frac{638 - 288\sqrt{6}}{23} \]