If \(x^2-12x+33=0\), then find the value of \((x-4)^2+\displaystyle\frac{1}{(x-4)^2}-5\).
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 9
The expression is built on \(x-4\), so the equation is rewritten in terms of \(t=x-4\). Dividing by \(t\) then gives \(t+\frac{1}{t}\), and squaring it gives \(t^2+\frac{1}{t^2}\).
Put \(x=t+4\): \((t+4)^2-12(t+4)+33=t^2+8t+16-12t-48+33=t^2-4t+1\)
So \(t^2-4t+1=0\); dividing by \(t\): \(t+\displaystyle\frac{1}{t}=4\)
\(t^2+\displaystyle\frac{1}{t^2}=\left(t+\frac{1}{t}\right)^2-2=16-2=14\)
Required value \(=14-5=9\)