If $x^4 + x^{− 4} = 47$, x > 0, then what is the value of $x+\frac{1}{x}-2$?

1

If x⁴ + \(\frac{1}{x^4}\) = a

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

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

According to the question,

$x^4 + x^{− 4} = 47$

x² + \(\frac{1}{x^2}\) = \(\sqrt {47 + 2}\) = 7

and x + \(\frac{1}{x}\) = \(\sqrt {7 + 2}\) = 3

$x+\frac{1}{x}-2$ = 3 - 2 = 1