If $x^4 + x^{− 4} = 47$, x > 0, then what is the value of $x+\frac{1}{x}-2$? |
1 0 5 3 |
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 |