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UPSC CDS I 2025 Mathematics Paper

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Question : 41 of 100

Marks: +1, -0

What is

\sqrt{17-4\sqrt{15}}+\sqrt{8-2\sqrt{15}}

equal to?

$\sqrt{3}$
$2\sqrt{3}$
$2(\sqrt{5}-\sqrt{3})$
$2(\sqrt{5}+\sqrt{3})$

Solution:

Concept:

Simplify expressions of the form

\sqrt{a \pm 2\sqrt{b}}

by writing them as squares of binomials.

Explanation:

First term:

\sqrt{17-4\sqrt{15}} = \sqrt{(2\sqrt{3})^2 + (\sqrt{5})^2 - 2(2\sqrt{3})(\sqrt{5})}

.

This equals

\sqrt{(2\sqrt{3} - \sqrt{5})^2} = 2\sqrt{3} - \sqrt{5}

(since

2\sqrt{3} > \sqrt{5}

).

Second term:

\sqrt{8-2\sqrt{15}} = \sqrt{(\sqrt{3})^2 + (\sqrt{5})^2 - 2(\sqrt{3})(\sqrt{5})}

.

This equals

\sqrt{(\sqrt{5} - \sqrt{3})^2} = \sqrt{5} - \sqrt{3}

(since

\sqrt{5} > \sqrt{3}

).

Add both results:

(2\sqrt{3} - \sqrt{5}) + (\sqrt{5} - \sqrt{3}) = \sqrt{3}

.

Answer:

\sqrt{3}

(Option A)

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