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SSC CGL Tier I 12 Sep 2025 Shift 2 Paper

Question : 73 of 100

Marks: +1, -0

If x =

\sqrt{\frac{2+\sqrt{3}}{2-√3}}

, then determine the value of x² + x - 9?

Solution:

Concept:

Rationalize the denominator to simplify the square root expression.

Explanation:

Start with

x = \sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}

Multiply numerator and denominator by the conjugate

2+\sqrt{3}

\frac{2+\sqrt{3}}{2-\sqrt{3}} \times \frac{2+\sqrt{3}}{2+\sqrt{3}} = \frac{(2+\sqrt{3})^2}{4-3} = (2+\sqrt{3})^2

Therefore

x = \sqrt{(2+\sqrt{3})^2} = 2+\sqrt{3}

(since it is positive).

Now compute

x^2 = (2+\sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3}

Then

x^2 + x - 9 = (7 + 4\sqrt{3}) + (2+\sqrt{3}) - 9 = (7+2-9) + (4\sqrt{3}+\sqrt{3}) = 0 + 5\sqrt{3}

Answer:

5\sqrt{3}

(Option B)

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