The product of all the rational roots of the equation $\left(x^2-9 x+11\right)^2-(x-4)(x-5)=3$, is equal to
- $14$
- $21$
- $28$
- $7$
Solution
& \left(x^2-9 x+11\right)^2-(x-4)(x-5)=3 \\ & \left(x^2-9 x+11\right)^2-\left(x^2-9 x+20\right)=3
\end{aligned}$
Let $x^2-9 x+11=t$
$\begin{aligned}
& t^2-(t+9)=3 \\ & \Rightarrow t^2-t-12=0 \\ & \Rightarrow t^2-4 t+3 t-12=0 \\ & \Rightarrow t(t-4)+3(t-4)=0 \\ & \Rightarrow t=4 \text { or }-3 \\ & x^2-9 x+11=4 \\ & x^2-9 x+7=0
\end{aligned}$
Here, we will get irrational roots
$\begin{aligned}
& x^2-9 x+11=-3 \\ & x^2-9 x+14=0 \\ & x^2-7 x-2 x+14=0 \\ & \Rightarrow x=7,2 \\ & \Rightarrow \text { Product of all rational roots }=14
\end{aligned}$ ~
Asked in: JEE Main 2025 (24 Jan Shift 1)