Question:medium

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: Rationalise each reciprocal separately.
Start from $\frac{1}{x}$ where $x = 2+\sqrt{3}$. Multiply top and bottom by the conjugate $2-\sqrt{3}$ to clear the surd from the denominator:
\[ \frac{1}{2+\sqrt{3}} = \frac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})} = \frac{2-\sqrt{3}}{4-3} = 2-\sqrt{3} \]
So $\frac{1}{x} = y$. By the same working,
\[ \frac{1}{2-\sqrt{3}} = \frac{2+\sqrt{3}}{4-3} = 2+\sqrt{3} \]
so $\frac{1}{y} = x$. The two numbers are reciprocals of each other.

Step 2: Rewrite the expression.
Squaring those results, $\frac{1}{x^2} = y^2$ and $\frac{1}{y^2} = x^2$. So
\[ \frac{1}{x^2} + \frac{1}{y^2} = y^2 + x^2 \]
The problem has turned into finding $x^2 + y^2$.

Step 3: Square each surd out in full.
Use $(a+b)^2 = a^2 + 2ab + b^2$ with $a = 2$ and $b = \sqrt{3}$:
\[ (2+\sqrt{3})^2 = 4 + 2(2)(\sqrt{3}) + 3 = 7 + 4\sqrt{3} \]
Use $(a-b)^2 = a^2 - 2ab + b^2$:
\[ (2-\sqrt{3})^2 = 4 - 2(2)(\sqrt{3}) + 3 = 7 - 4\sqrt{3} \]

Step 4: Add and watch the surds die.
\[ (7 + 4\sqrt{3}) + (7 - 4\sqrt{3}) = 14 \]
The $4\sqrt{3}$ terms cancel because they carry opposite signs, leaving a clean whole number. That cancellation is exactly why conjugate pairs are used in such questions.

Step 5: Numerical sanity check.
Taking $\sqrt{3} \approx 1.732$, we get $x \approx 3.732$ and $y \approx 0.268$. Then $\frac{1}{x^2} \approx \frac{1}{13.93} \approx 0.072$ and $\frac{1}{y^2} \approx \frac{1}{0.0718} \approx 13.93$. Their sum is about $14.0$, matching the exact answer.

Step 6: Rule out the rest.
16 is $(x+y)^2$, the value you stop at if you skip subtracting $2xy$.
12 needs $x^2+y^2 = 16 - 4$, which assumes the product $xy$ is 2 rather than 1.
10 has no valid route at all and is well below the estimate of about 14 found above.

Final Answer:
Since $x$ and $y$ are reciprocals, the expression equals $x^2+y^2 = 14$.
\[ \boxed{14} \]
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