If \(x^2-13x+1=0\) 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) → 167
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 \(x^2-13x+1=0\) by \(x\): \(x-13+\displaystyle\frac{1}{x}=0\), so \(x+\displaystyle\frac{1}{x}=13\)
\(x^2+\displaystyle\frac{1}{x^2}=\left(x+\frac{1}{x}\right)^2-2=13^2-2\)
\(=169-2=167\)