If $x^4 + \frac{1}{x^4}=6887$, then the positive value of $ x -\frac{1}{x}$ is :

9

We know that,

If x⁴ + \(\frac{1}{x^4}\) = a

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

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

So,

If $x^4 + \frac{1}{x^4}=6887$,

then x² + \(\frac{1}{x^2}\) = \(\sqrt {6887 + 2}\) = 83

then the positive value of $ x -\frac{1}{x}$ = \(\sqrt {83 - 2}\) = 9