({3}+{2})+{1}{({3}-{2})}=?

Correct Answer is : 2(3+2)2({3}+{2})2(3+2)

Correct Answer is : 2(3+2)2({3}+{2})2(3+2)

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SSC MTS 6 Feb 2026 Shift 3 Paper

Question : 14 of 150

Marks: +1, -0

(\sqrt{3}+\sqrt{2})+\frac{1}{(\sqrt{3}-\sqrt{2})}=?

Solution:

Concept:

Rationalize the denominator: multiply numerator and denominator by the conjugate

(\sqrt{3} + \sqrt{2})

to simplify the expression.

Explanation:

First, rewrite the term

\frac{1}{\sqrt{3} - \sqrt{2}}

. Its conjugate is

\sqrt{3} + \sqrt{2}

Multiply both numerator and denominator:

\frac{1 \times (\sqrt{3} + \sqrt{2})}{(\sqrt{3} - \sqrt{2})(\sqrt{3} + \sqrt{2})} = \frac{\sqrt{3} + \sqrt{2}}{(\sqrt{3})^2 - (\sqrt{2})^2}

Since

(\sqrt{3})^2 = 3

and

(\sqrt{2})^2 = 2

, the denominator becomes

3 - 2 = 1

Thus

\frac{1}{\sqrt{3} - \sqrt{2}} = \sqrt{3} + \sqrt{2}

Now the whole expression is

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

Answer:

2(\sqrt{3} + \sqrt{2})

(Option D)

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