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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Check the product first: (2 plus root 3)(2 minus root 3) equals 1. So the expression reduces to x squared plus y squared, which is (x plus y) squared minus 2xy.
Updated On: Jul 17, 2026
  • 12
  • 16
  • 14
  • 10
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The Correct Option is C

Solution and Explanation

Step 1: Notice that x and y are conjugates.
The two numbers \( 2+\sqrt{3} \) and \( 2-\sqrt{3} \) differ only in the sign before the surd, so they are conjugate surds. Their sum and product are both free of surds, which makes them easy to work with.
\[ x + y = (2+\sqrt{3}) + (2-\sqrt{3}) = 4 \]
\[ xy = (2+\sqrt{3})(2-\sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1 \]
The product being exactly 1 is the key fact of this question.

Step 2: Combine the two fractions.
\[ \frac{1}{x^2} + \frac{1}{y^2} = \frac{y^2 + x^2}{x^2 y^2} = \frac{x^2+y^2}{(xy)^2} \]
Since \( xy = 1 \), the denominator is \( 1^2 = 1 \). So
\[ \frac{1}{x^2} + \frac{1}{y^2} = x^2 + y^2 \]
The reciprocal-of-squares expression has collapsed into a plain sum of squares.

Step 3: Find the sum of squares using an identity.
The identity \( x^2 + y^2 = (x+y)^2 - 2xy \) avoids expanding the surds at all. Putting in the values from Step 1,
\[ x^2 + y^2 = 4^2 - 2(1) = 16 - 2 = 14 \]

Step 4: Check by direct expansion.
\[ x^2 = (2+\sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3} \]
\[ y^2 = (2-\sqrt{3})^2 = 4 - 4\sqrt{3} + 3 = 7 - 4\sqrt{3} \]
Adding them, the surd parts cancel:
\[ x^2 + y^2 = (7 + 4\sqrt{3}) + (7 - 4\sqrt{3}) = 14 \]
Both methods agree.

Step 5: Look at the wrong options.
Option (B) 16 is \( (x+y)^2 \), which means the student forgot to subtract \( 2xy \).
Option (A) 12 would come from subtracting 4 instead of 2, that is, from using \( xy = 2 \) instead of 1.
Option (D) 10 comes from subtracting \( 2xy \) twice or from a slip in expanding the square.
Only 14 is consistent.

Final Answer:
The value of the expression is 14.
\[ \boxed{14} \]
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