If \({\rm x} + \frac{1}{{\rm x}} = - 14 \), and x < -1, what will be the value of \({{\rm x}^2} - \frac{1}{{{{\rm x}^2}}}\) ?

Question

Question

This question was previously asked in

SSC CGL 2022 Tier-I Official Paper (Held On : 05 Dec 2022 Shift 1)

Answer 1

Given:

\({\rm x} + \frac{1}{{\rm x}} = - 14\)

Concept used:

(a² - b²) = (a + b)(a - b)

Calculation:

\({\rm x} + \frac{1}{{\rm x}} = - 14 \)

⇒ \({\rm x^2} + \frac{1}{{\rm x^2}} +2= 196 \)

⇒ \({\rm x^2} + \frac{1}{{\rm x^2}} +2-4= 196-4 \)

⇒ \({\rm x^2} + \frac{1}{{\rm x^2}} -2= 192 \)

⇒ \(({\rm x} - \frac{1}{{\rm x}})^2= 192 \)

⇒ \({\rm x} - \frac{1}{{\rm x}} =\pm 8√3 \)

As x < -1 So, \({\rm x} - \frac{1}{{\rm x}} =- 8\sqrt3 \)

Now,

\({{\rm x}^2} - \frac{1}{{{{\rm x}^2}}}\) = (- 14) × (\(-8\sqrt 3 \))

⇒ \(112\sqrt 3 \)

∴ The required answer is \(112\sqrt 3 \).

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