If $x = \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}$ and $y = \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}}$, then $(x + y)$ equals

### Concept & Formula The core insight is recognizing that $x$ and $y$ are reciprocal conjugates. Instead of rationalizing $x$ and $y$ separately, we can add them directly using a common denominator and a standard algebraic identity. $$ \frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}} + \frac{\sqrt{a} - \sqrt{b}}{\sqrt{a} + \sqrt{b}} = \frac{2(a + b)}{a - b} $$ ### Step-by-Step Solution Given: $x = \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}$ and $y = \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}}$ We need to find $x + y$: $x + y = \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} + \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}}$ Find the common denominator by multiplying the two denominators: Denominator $= (\sqrt{5} - \sqrt{3})(\sqrt{5} + \sqrt{3})$ Using the identity $(a-b)(a+b) = a^2 - b^2$: Denominator $= (\sqrt{5})^2 - (\sqrt{3})^2 = 5 - 3 = 2$ Now, cross-multiply to find the numerator: Numerator $= (\sqrt{5} + \sqrt{3})^2 + (\sqrt{5} - \sqrt{3})^2$ Using the algebraic expansion $(a+b)^2 + (a-b)^2 = 2(a^2 + b^2)$: Numerator $= 2((\sqrt{5})^2 + (\sqrt{3})^2)$ Numerator $= 2(5 + 3) = 2(8) = 16$ Substitute the numerator and denominator back into the fraction: $x + y = \frac{16}{2} = 8$ ### Exam Strategy & Shortcut Memorize the standard result for the sum of inverse conjugate surds: If $x = \frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}}$ and $y = \frac{\sqrt{a} - \sqrt{b}}{\sqrt{a} + \sqrt{b}}$, then $x + y = \frac{2(a + b)}{a - b}$. Here, $a = 5$ and $b = 3$. $x + y = \frac{2(5 + 3)}{5 - 3} = \frac{16}{2} = 8$. This completely bypasses intermediate calculations and solves the problem instantly without writing anything down. ### Common Pitfall A frequent mistake is attempting to rationalize $x$ and $y$ individually first. While it yields the correct answer ($x = 4 + \sqrt{15}$ and $y = 4 - \sqrt{15}$, so $x+y=8$), it requires double the calculation effort and time compared to using the direct identity. ### Final Answer **Therefore, the correct answer is 8.**