Question:

The value of \( \frac{1}{x^2} + \frac{1}{y^2} \) where \( x = 2 + \sqrt{3} \) and \( y = 2 - \sqrt{3} \) is

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Use the identity \((a + b)(a - b) = a^2 - b^2\) to simplify surd expressions quickly.
Updated On: Aug 11, 2025
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The Correct Option is C

Solution and Explanation

First, find \(x^2\) and \(y^2\): \[ x^2 = (2 + \sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3} \] \[ y^2 = (2 - \sqrt{3})^2 = 4 - 4\sqrt{3} + 3 = 7 - 4\sqrt{3} \] Now, \[ \frac{1}{x^2} + \frac{1}{y^2} = \frac{y^2 + x^2}{x^2 y^2} \] Since \(x^2 + y^2 = (7 + 4\sqrt{3}) + (7 - 4\sqrt{3}) = 14\) and \(x^2 y^2 = (7 + 4\sqrt{3})(7 - 4\sqrt{3}) = 49 - 48 = 1\), we get: \[ \frac{1}{x^2} + \frac{1}{y^2} = \frac{14}{1} = 14 \]
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