Mathematics - Quadratic Equation Question with Solution | TestHub

$\begin{array}{rl}& {\left(x^2-9x+11\right)}^2-(x-4)(x-5)=3\\ & {\left(x^2-9x+11\right)}^2-\left(x^2-9x+20\right)=3\end{array}$
Let $x^2-9x+11=t$
$\begin{array}{rl}& t^2-(t+9)=3\\ & \Rightarrow t^2-t-12=0\\ & \Rightarrow t^2-4t+3t-12=0\\ & \Rightarrow t(t-4)+3(t-4)=0\\ & \Rightarrow t=4\text{ or }-3\\ & x^2-9x+11=4\\ & x^2-9x+7=0\end{array}$
Here, we will get irrational roots
$\begin{array}{rl}& x^2-9x+11=-3\\ & x^2-9x+14=0\\ & x^2-7x-2x+14=0\\ & \Rightarrow x=7,2\\ & \Rightarrow \text{ Product of all rational roots }=14\end{array}$