If $x^2 +\frac{1}{x^2}= 66$, the value of $x-\frac{1}{x}$ is _____.
Answer & explanation
Correct answer: option 2
then x2 + \(\frac{1}{x^2}\) = b
Then, x - \(\frac{1}{x}\) = \(\sqrt {b - 2}\)
If $x^2 +\frac{1}{x^2}= 66$
Then, x - \(\frac{1}{x}\) = \(\sqrt {66 - 2}\) = 8