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If $x^2 +\frac{1}{x^2}= 38,$ then what is the value of $\begin{vmatrix}x-\frac{1}{x}\end{vmatrix}$ ? |
5 9 6 4 |
6 |
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 |
