If `x=(sqrt(3)+1)/(sqrt3-1)andy=(sqrt3-1)/(sqrt3+1),"then "x^(2)+y^(2)` is equal to

Options

A. 14
B. 13
C. 15
D. 10

Answer

To solve the problem, we need to find the value of \( x^2 + y^2 \) given:

\[ x = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \quad \text{and} \quad y = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \]

### Step 1: Simplify \( x \)

To simplify \( x \), we will rationalize the denominator:

\[ x = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \cdot \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{(\sqrt{3} + 1)^2}{(\sqrt{3} - 1)(\sqrt{3} + 1)} \]

Calculating the denominator:

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

Calculating the numerator:

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

Thus,

\[ x = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3} \]

### Step 2: Simplify \( y \)

Now, we will simplify \( y \) in a similar manner:

\[ y = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \cdot \frac{\sqrt{3} - 1}{\sqrt{3} - 1} = \frac{(\sqrt{3} - 1)^2}{(\sqrt{3} + 1)(\sqrt{3} - 1)} \]

Calculating the denominator (already done):

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

Calculating the numerator:

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

Thus,

\[ y = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3} \]

### Step 3: Calculate \( x^2 + y^2 \)

Now we can find \( x^2 + y^2 \):

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

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

Now, adding \( x^2 \) and \( y^2 \):

\[ x^2 + y^2 = (7 + 4\sqrt{3}) + (7 - 4\sqrt{3}) = 7 + 7 + 4\sqrt{3} - 4\sqrt{3} = 14 \]

### Final Answer

Thus, the value of \( x^2 + y^2 \) is:

\[ \boxed{14} \]