If x⁴ + \(\frac{1}{x^4}\) = 20162, the positive value of x + \(\frac{1}{x}\) is?

12

If x⁴ + \(\frac{1}{x^4}\) = a

then x² + \(\frac{1}{x^2}\) = \(\sqrt {a + 2}\) = b

and x + \(\frac{1}{x}\) = \(\sqrt {b + 2}\)

ATQ,

x⁴ + \(\frac{1}{x^4}\) = 20162

x² + \(\frac{1}{x^2}\) = \(\sqrt {20162 + 2}\) = 142

So, x + \(\frac{1}{x}\) = \(\sqrt {142 + 2}\) = 12