If \(x^2-32x+1=0\) and \(x>1\), then the value of \(x^2+\displaystyle\frac{1}{x^2}\) is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 1022
Dividing the equation by \(x\) gives \(x+\frac{1}{x}\) directly; squaring it then gives \(x^2+\frac{1}{x^2}\).
\(x-32+\displaystyle\frac{1}{x}=0\), so \(x+\displaystyle\frac{1}{x}=32\)
\(x^2+\displaystyle\frac{1}{x^2}=\left(x+\frac{1}{x}\right)^2-2=1024-2=1022\)