If \(x^2+x=19\), then find the value of \((x+5)^2+\displaystyle\frac{1}{(x+5)^2}\).
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 79
The expression is built on \(x+5\), so the equation is rewritten in terms of \(t=x+5\). Dividing by \(t\) then gives \(t+\frac{1}{t}\), and squaring it gives \(t^2+\frac{1}{t^2}\).
Put \(x=t-5\): \((t-5)^2+(t-5)=t^2-10t+25+t-5=t^2-9t+20\)
So \(t^2-9t+20=19\), that is \(t^2-9t+1=0\)
Dividing by \(t\): \(t+\displaystyle\frac{1}{t}=9\)
\(t^2+\displaystyle\frac{1}{t^2}=\left(t+\frac{1}{t}\right)^2-2=81-2=79\)