The product of all the rational roots of the equation $(x^2-9x+11)^2-(x-4)(x-5) = 3$ is
Step-by-Step Solution
Key Concept: Rewrite $(x-4)(x-5) = x^2-9x+20 = (x^2-9x+11)+9$, then substitute $t = x^2-9x+11$ to get a simple quadratic in $t$.
Note $(x-4)(x-5)=x^2-9x+20$. Let $t=x^2-9x+11$, so $(x-4)(x-5)=t+9$.
The equation becomes $t^2-(t+9)=3 \Rightarrow t^2-t-12=0 \Rightarrow (t-4)(t+3)=0$.
**$t=4$:** $x^2-9x+7=0 \Rightarrow x=\dfrac{9\pm\sqrt{53}}{2}$ (irrational — discard).
**$t=-3$:** $x^2-9x+14=0 \Rightarrow (x-7)(x-2)=0 \Rightarrow x=2,7$ (rational).
Product of rational roots $= 2\times7 = 14$.
Correct Answer: 1