If $ X^4 +\frac{16}{x^4}= 15617, x > 0$, then find the value of $ X +\frac{2}{x}$

$\sqrt{129}$

$ X^4 +\frac{16}{x^4}= 15617, x > 0$

If x⁴ + \(\frac{1}{x^4}\) = a

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

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

x² + \(\frac{4}{x^2}\) = \(\sqrt {15617 + 2 × 4}\) =125

x + \(\frac{2}{x}\) = \(\sqrt {125 + 2 × 2}\) =  \(\sqrt {129}\)