The product of all the rational roots of the equation $\bigl(x^{2}-9x+11\bigr)^{2}-(x-4)(x-5)=3$, is equal to:
Step-by-Step Solution
Key Concept: Notice $(x-4)(x-5)=x^{2}-9x+20=(x^{2}-9x+11)+9$. Substitute $t=x^{2}-9x+11$ to reduce a quartic to a quadratic in $t$.
Let $t=x^{2}-9x+11$. Since $(x-4)(x-5)=x^{2}-9x+20=t+9$, the equation becomes
$$t^{2}-(t+9)=3\ \Longrightarrow\ t^{2}-t-12=0\ \Longrightarrow\ (t-4)(t+3)=0.$$
\textbf{Branch } $t=4$: $x^{2}-9x+7=0$, discriminant $81-28=53$ — irrational roots, discard.
\textbf{Branch } $t=-3$: $x^{2}-9x+14=0\Rightarrow (x-2)(x-7)=0\Rightarrow x=2,7$ (rational).
Product of all rational roots $=2\cdot 7 = 14$.
Correct Answer: 1