If \(x^2-10x+17=0\), then find the value of \((x-2)^2+\displaystyle\frac{1}{(x-2)^2}+6\).
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 40
The expression is built on \(x-2\), so the equation is rewritten in terms of \(t=x-2\). Dividing by \(t\) then gives \(t+\frac{1}{t}\), and squaring it gives \(t^2+\frac{1}{t^2}\).
Put \(x=t+2\): \((t+2)^2-10(t+2)+17=t^2+4t+4-10t-20+17=t^2-6t+1\)
So \(t^2-6t+1=0\); dividing by \(t\): \(t+\displaystyle\frac{1}{t}=6\)
\(t^2+\displaystyle\frac{1}{t^2}=\left(t+\frac{1}{t}\right)^2-2=36-2=34\)
Required value \(=34+6=40\)