If $x^4 + x^{-4} = 2599$, then one of the values of $x - x^{-1}$, where x > 0, is equal to:

7

If $K^2+\frac{1}{K^2}$ = n

then, $K+\frac{1}{K}=\sqrt {n + 2}$

If $x^4 + x^{-4} = 2599$

Then one of the values of $x - x^{-1}$ = ?

If $x^4 + x^{-4} = 2599$

then $x^2 + x^{-2}$ = \(\sqrt {2599 + 2}\) = 51

$x - x^{-1}$ = \(\sqrt {51 - 2}\) = 7