Question:medium

If \(x = \frac{2\sqrt{2}-\sqrt{7}}{2\sqrt{2}+\sqrt{7}}\), then \(x + x^{-1}\) is equal to

Show Hint

Use symmetry identities for expressions like \(x + 1/x\).
Updated On: Apr 18, 2026
  • 28
  • 32
  • 30
  • 24
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept:
We are asked to find the sum of a value \( x \) and its reciprocal \( x^{-1} \). Given \( x \) is a fraction of surds, \( x^{-1} \) will simply be the inverted fraction.
: Key Formula or Approach:
1. \( x + \frac{1}{x} = \frac{N}{D} + \frac{D}{N} = \frac{N^2 + D^2}{ND} \).
2. \( (A - B)^2 + (A + B)^2 = 2(A^2 + B^2) \).
3. \( (A + B)(A - B) = A^2 - B^2 \).
Step 2: Detailed Explanation:
Let \( N = 2\sqrt{2} - \sqrt{7} \) and \( D = 2\sqrt{2} + \sqrt{7} \).
Then \( x = \frac{N}{D} \) and \( x^{-1} = \frac{D}{N} \). \[ x + x^{-1} = \frac{N}{D} + \frac{D}{N} = \frac{N^2 + D^2}{ND} \] Numerator: \[ N^2 + D^2 = (2\sqrt{2} - \sqrt{7})^2 + (2\sqrt{2} + \sqrt{7})^2 \] Using the identity \( (a - b)^2 + (a + b)^2 = 2(a^2 + b^2) \): \[ N^2 + D^2 = 2((2\sqrt{2})^2 + (\sqrt{7})^2) = 2(8 + 7) = 2(15) = 30 \] Denominator: \[ ND = (2\sqrt{2} + \sqrt{7})(2\sqrt{2} - \sqrt{7}) \] Using \( (a + b)(a - b) = a^2 - b^2 \): \[ ND = (2\sqrt{2})^2 - (\sqrt{7})^2 = 8 - 7 = 1 \] Calculation: \[ x + x^{-1} = \frac{30}{1} = 30 \].
Step 3: Final Answer:
The value of \( x + x^{-1} \) is 30.
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