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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