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Question

The value of $\sqrt{11 - 2\sqrt{30}} - \frac{1}{\sqrt{11 - 2\sqrt{30}}}$ is:

This question was previously asked in
SSC CGL 2025 Tier 2 Paper 1 Question Paper (19-Jan-2026)
The correct answer is
$-2\sqrt{5}$

Simplifying the Radical Expression

We need to evaluate the expression $\sqrt{11 - 2\sqrt{30}} - \frac{1}{\sqrt{11 - 2\sqrt{30}}}$.

Step 1: Simplify the radical term $\sqrt{11 - 2\sqrt{30}}$

To simplify $\sqrt{11 - 2\sqrt{30}}$, we look for two numbers whose sum is 11 and product is 30. These numbers are 6 and 5.

Therefore, we can rewrite the expression under the square root as:

$11 - 2\sqrt{30} = 6 + 5 - 2\sqrt{6 \times 5} = (\sqrt{6})^2 + (\sqrt{5})^2 - 2\sqrt{6}\sqrt{5} = (\sqrt{6} - \sqrt{5})^2$

So, $\sqrt{11 - 2\sqrt{30}} = \sqrt{(\sqrt{6} - \sqrt{5})^2}$. Since $\sqrt{6} > \sqrt{5}$, the absolute value is $\sqrt{6} - \sqrt{5}$.

Let $x = \sqrt{11 - 2\sqrt{30}} = \sqrt{6} - \sqrt{5}$.

Step 2: Calculate the reciprocal $\frac{1}{x}$

Now, we find the reciprocal of $x$:

$ \frac{1}{x} = \frac{1}{\sqrt{6} - \sqrt{5}} $

Rationalize the denominator by multiplying the numerator and denominator by the conjugate $(\sqrt{6} + \sqrt{5})$:

$ \frac{1}{x} = \frac{1}{\sqrt{6} - \sqrt{5}} \times \frac{\sqrt{6} + \sqrt{5}}{\sqrt{6} + \sqrt{5}} = \frac{\sqrt{6} + \sqrt{5}}{(\sqrt{6})^2 - (\sqrt{5})^2} = \frac{\sqrt{6} + \sqrt{5}}{6 - 5} = \frac{\sqrt{6} + \sqrt{5}}{1} = \sqrt{6} + \sqrt{5} $

Step 3: Evaluate the expression $x - \frac{1}{x}$

Substitute the values of $x$ and $\frac{1}{x}$ back into the original expression form:

$ x - \frac{1}{x} = (\sqrt{6} - \sqrt{5}) - (\sqrt{6} + \sqrt{5}) $

$ = \sqrt{6} - \sqrt{5} - \sqrt{6} - \sqrt{5} $

$ = (\sqrt{6} - \sqrt{6}) + (-\sqrt{5} - \sqrt{5}) $

$ = 0 - 2\sqrt{5} $

$ = -2\sqrt{5} $

Conclusion

The value of the expression $\sqrt{11 - 2\sqrt{30}} - \frac{1}{\sqrt{11 - 2\sqrt{30}}}$ is $-2\sqrt{5}$.

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