Question : If $x^2-8 x-1=0$, what is the value of $x^2+\frac{1}{x^2}?$
Option 1: 68
Option 2: 62
Option 3: 64
Option 4: 66
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Correct Answer: 66
Solution : $x^2-8 x-1=0$ ⇒ $x^2 -1 = 8x$ Dividing both sides by $x$ $x-\frac{1}{x} = 8$ Squaring both sides, $(x-\frac{1}{x})^2 = 8^2$ ⇒ $x^2 + \frac{1}{x^2} - 2×x×\frac{1}{x} = 64$ $\therefore$ $x^2 + \frac{1}{x^2} = 64+2=66$ Hence, the correct answer is 66.
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