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SSC CGL 8 Aug 2017 Shift 1 Paper

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Question : 62 of 100

Marks: +1, -0

If

x = 6 + 2\sqrt{6}

, then what is the value of

\sqrt{x-1} + \frac{1}{\sqrt{x-1}}

?

$2\sqrt{3}$
$3\sqrt{2}$
$2\sqrt{2}$
$3\sqrt{3}$

Solution:

x = 6 + 2\sqrt{6}

On subtracting 1 from both sides,

x - 1 = 6 + 2\sqrt{6} - 1

⇒

x-1 = 5 + 2\sqrt{6}

⇒

\sqrt{x-1} = \sqrt{5 + 2\sqrt{6}}

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

and

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

= \sqrt{3} - \sqrt{2}

∴

\sqrt{x-1} + \frac{1}{\sqrt{x-1}}

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

= 2\sqrt{3}

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