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: $62$
Answer (1)
25th Jan, 2024
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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