$\begin{aligned}
& \left(x^2-9 x+11\right)^2-(x-4)(x-5)=3 \
& \left(x^2-9 x+11\right)^2-\left(x^2-9 x+20\right)=3
\end{aligned}Letx^2-9 x+11=t\begin{aligned}
& t^2-(t+9)=3 \
& \Rightarrow t^2-t-12=0 \
& \Rightarrow t^2-4 t+3 t-12=0 \
& \Rightarrow t(t-4)+3(t-4)=0 \
& \Rightarrow t=4 \text { or }-3 \
& x^2-9 x+11=4 \
& x^2-9 x+7=0
\end{aligned}Here,wewillgetirrationalroots\begin{aligned}
& x^2-9 x+11=-3 \
& x^2-9 x+14=0 \
& x^2-7 x-2 x+14=0 \
& \Rightarrow x=7,2 \
& \Rightarrow \text { Product of all rational roots }=14
\end{aligned}$
The product of all the rational roots of the equation (x^2-9 x+11 )^2-(x-4)(x-5)=3, is equal to
JEE Main 2025 — Mathematics Algebra
The product of all the rational roots of the equation (x2−9x+11)2−(x−4)(x−5)=3, is equal to
Held on 24 Jan 2025 · Verified 6 Jul 2026.
Options
- A
14
- B
21
- C
28
- D
7
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