If \(x^2-23x+1=0\) and \(x>1\), then the value of \(x^2+\displaystyle\frac{1}{x^2}\) is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 527
Dividing the equation by \(x\) gives \(x+\frac{1}{x}\) directly, and squaring it uses \(\left(x+\frac{1}{x}\right)^2=x^2+\frac{1}{x^2}+2\).
Dividing \(x^2-23x+1=0\) by \(x\): \(x-23+\displaystyle\frac{1}{x}=0\), so \(x+\displaystyle\frac{1}{x}=23\)
\(x^2+\displaystyle\frac{1}{x^2}=\left(x+\frac{1}{x}\right)^2-2=23^2-2\)
\(=529-2=527\)