If $x^2 +\frac{1}{x^2}= 38,$ then what is the value of $\begin{vmatrix}x-\frac{1}{x}\end{vmatrix}$ ?
Answer & explanation
Correct answer: option 3
If x2 + \(\frac{1}{x^2}\) = b
Then, x + \(\frac{1}{x}\) = \(\sqrt {b - 2}\)
If $x^2 +\frac{1}{x^2}= 38,$
Then, x - \(\frac{1}{x}\) = \(\sqrt {38 - 2}\) = 6