Question : If $x^2-8x+1=0$, what is the value of $(x^2+\frac{1}{x^2})$?
Option 1: $18$
Option 2: $34$
Option 3: $40$
Option 4: $62$
Correct Answer: $62$
Solution : Given: $x^2-8x+1=0$ Dividing both sides by $x$, we get, $⇒x-8+\frac{1}{x}=0$ $⇒x+\frac{1}{x}=8$ Squaring both sides, we get, $⇒(x+\frac{1}{x})^2=8^2$ $⇒x^2+\frac{1}{x^2}+2=64$ $\therefore x^2+\frac{1}{x^2}=64-2=62$ Hence, the correct answer is $62$.
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