If $x^8-433 x^4+16=0, x>0$, then what is the value of $\left(x+\frac{2}{x}\right)$ ?

5

If x⁴ + \(\frac{1}{x^4}\) = a

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

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

If $x^8-433 x^4+16=0, x>0$

$\left(x+\frac{2}{x}\right)$

$x^8-433 x^4+16=0, x>0$

Divide by x⁴

x⁴ + \(\frac{16}{x^4}\) = 433

x² + \(\frac{4}{x^2}\) = \(\sqrt {433 + 2 × 4}\) = 21

x + \(\frac{2}{x}\) = \(\sqrt {21 + 2 × 2}\) = 5