If $x^4 + \frac{1}{x^4}=6887$, then the positive value of $ x -\frac{1}{x}$ is :
Answer & explanation
Correct answer: option 4
We know that,
If x4 + \(\frac{1}{x^4}\) = a
then x2 + \(\frac{1}{x^2}\) = \(\sqrt {a + 2}\) = b
and x - \(\frac{1}{x}\) = \(\sqrt {b - 2}\)
So,
If $x^4 + \frac{1}{x^4}=6887$,
then x2 + \(\frac{1}{x^2}\) = \(\sqrt {6887 + 2}\) = 83
then the positive value of $ x -\frac{1}{x}$ = \(\sqrt {83 - 2}\) = 9