Question

\( \frac{\sqrt{7} + \sqrt{5}}{\sqrt{7} - \sqrt{5}} + \frac{\sqrt{7} - \sqrt{5}}{\sqrt{7} + \sqrt{5}} = \)

• A. \( 2\sqrt{35} \)
• B. \( -2\sqrt{35} \)
• C. 12
• D. -12

Hint:

Rationalizing the denominators simplifies expressions involving square roots.

Answer 1

The Correct Option is C

Let the expression be: \[ \frac{\sqrt{7} + \sqrt{5}}{\sqrt{7} - \sqrt{5}} + \frac{\sqrt{7} - \sqrt{5}}{\sqrt{7} + \sqrt{5}}. \] We can simplify this expression by rationalizing both the denominators: \[ \frac{\sqrt{7} + \sqrt{5}}{\sqrt{7} - \sqrt{5}} = \frac{(\sqrt{7} + \sqrt{5})^2}{7 - 5} = \frac{7 + 5 + 2\sqrt{35}}{2} = \frac{12 + 2\sqrt{35}}{2} = 6 + \sqrt{35}. \] Similarly, for the second term: \[ \frac{\sqrt{7} - \sqrt{5}}{\sqrt{7} + \sqrt{5}} = \frac{(\sqrt{7} - \sqrt{5})^2}{7 - 5} = \frac{7 + 5 - 2\sqrt{35}}{2} = \frac{12 - 2\sqrt{35}}{2} = 6 - \sqrt{35}. \] Adding both terms: \[ (6 + \sqrt{35}) + (6 - \sqrt{35}) = 1(2) \]