Animated Solution for Mathematics - Quadratic Equations: The product of all the rational roots of the equation $(x^2-9x+11{)}^2-(x-4)(x-5)=3,$ is equal to

Look closely at the structure of the equation:

If we expand the product $(x-4)(x-5)$, we obtain $x^2-5x-4x+20$, which simplifies to $x^2-9x+20$.

The term $x^2-9x$ is the heartbeat of this equation. By recognizing this common structure, we avoid the trap of expanding a quartic polynomial and instead prepare to solve a simpler quadratic in disguise.

Whenever you see a repeating algebraic block, substitution is your best friend. Let us define a new variable, $t$, such that:

Since $x^2-9x+20$ can be written as $(x^2-9x+11)+9$, it follows that this term is equal to $t+9$. Our equation now transforms into a manageable quadratic:

Simplifying this expression, we get:

This is a classic factorization problem. We seek two numbers that multiply to $-12$ and add to $-1$, which are $-4$ and $3$. Thus, the equation factors as:

This yields two potential cases: $t=4$ and $t=-3$.

Now, we must return to our original variable $x$ for each case.

For $t=4$, we have:

We calculate the discriminant $D=b^2-4ac$:

Since $53$ is not a perfect square, the roots for this case are irrational. We discard them as the problem specifically concerns rational roots.

For $t=-3$, we have:

This quadratic factors perfectly:

The roots are $x=2$ and $x=7$. Both are rational.

The question asks for the product of all rational roots. We have identified them as $2$ and $7$.

Their product is:

By choosing the path of substitution over the path of expansion, we turned a potential nightmare into a moment of clarity. The final answer is 14.