If \({\rm X} = 3 + 2\sqrt 2 \) , x > 0, then the value of \(\sqrt {\rm X} - \frac{1}{{\sqrt {\rm X} }}\) is:

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Question

This question was previously asked in

SSC CGL 2022 Tier-I Official Paper (Held On : 03 Dec 2022 Shift 2)

Answer 1

Given:

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

Concept used:

\((a-\frac{1}{a})^2 = a^2 + \frac{1}{a^2} - 2\)

Calculation:

\({\rm X} = 3 + 2√ 2 \)

⇒ \({\rm \frac{1}{X}} = \frac{1}{3 + 2√ 2 }\)

⇒ \({\rm \frac{1}{X}} = \frac{3 -2√ 2 }{(3 + 2√ 2)(3 - 2√ 2 ) }\)

⇒ \({\rm \frac{1}{X}} = \frac{3 - 2√ 2 }{3^2 - (2√ 2 )^2 }\)

⇒ \({\rm \frac{1}{X}} = \frac{3 - 2√ 2 }{9 - 8}\)

⇒ \({\rm \frac{1}{X}} = {3 - 2√ 2 }\)

Now,

Let \(√ {\rm X} - \frac{1}{{√ {\rm X} }}\) be a

So, \(\left(√ {\rm X} - \frac{1}{{√ {\rm X} }}\right)^2\) = a²

⇒ \(√ {\rm X}^2 + \frac{1}{{√ {\rm X^2} }} -2\) = a2

⇒ \(\rm X + \frac{1}{{X} } -2\) = a2

⇒ 3 + 2√2 + 3 - 2√2 - 2 = a2

⇒ 6 - 2 = a²

⇒ 4 = a2

⇒ a = 2 = \(\sqrt {\rm X} - \frac{1}{{\sqrt {\rm X} }}\)

∴ The required answer is 2.

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