If \(x - \frac{1}{x}\) = 13, what will be the value of \( {x^4}+ \frac{1}{x^4}\)?

Question

Question

This question was previously asked in

SSC CGL 2022 Tier-I Official Paper (Held On : 03 Dec 2022 Shift 1)

Answer 1

Given:

x - \(\frac{1}{x}\) = 13

Concept used:

(x - \(\frac{1}{x}\))2 = x2 + \(\frac{1}{x^2}\) - 2

(x + \(\frac{1}{x}\))2 = x2 + \(\frac{1}{x^2}\) + 2

Calculation:

x - \(\frac{1}{x}\) = 13

⇒ (x - \(\frac{1}{x}\))² = 13²

⇒ x²+ \(\frac{1}{x^2}\) - 2 = 169

⇒ x2 + \(\frac{1}{x^2}\) = 171

Now,

(x2 + \(\frac{1}{x^2}\))² = 171²

⇒ x4 + \(\frac{1}{x^4}\) + 2 = 29241

⇒ x4 + \(\frac{1}{x^4}\) = 29241 - 2

⇒ x4 + \(\frac{1}{x^4}\) = 29239

∴ The required answer is 29239.

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