JEE Main 2025 — 24 January, Morning Shift. Previous Year Question.
Problem
The product of all the rational roots of the equation $(x^2-9x+11)^2-(x-4)(x-5)=3$ is equal to:
Key insight. Expanding $(x-4)(x-5)$ gives $x^2-9x+20$ — which is exactly $x^2-9x+11$, just shifted by a constant $9$. That means the whole equation can be rewritten purely in terms of the single expression $y=x^2-9x+11$, collapsing a quartic-looking equation into an ordinary quadratic in $y$.
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Approach
Substitute $y=x^2-9x+11$, noting that $(x-4)(x-5)=x^2-9x+20=y+9$. This turns the original equation into a simple quadratic in $y$. Solve for $y$, then substitute back to get two quadratics in $x$ — one with rational roots and one without — and take the product of only the rational ones.
Solution
Step 1 — Rewrite (x−4)(x−5) in terms of y
Let $y=x^2-9x+11$. Expanding $(x-4)(x-5)=x^2-9x+20$, and since $x^2-9x=y-11$:
$$(x-4)(x-5) = (y-11)+20 = y+9$$
Step 2 — Rewrite the equation purely in y
$$y^2 – (y+9) = 3 \implies y^2-y-9=3 \implies y^2-y-12=0$$
Step 3 — Solve for y
$$(y-4)(y+3)=0 \implies y=4 \text{ or } y=-3$$
Step 4 — Substitute back for y = 4
$$x^2-9x+11=4 \implies x^2-9x+7=0$$
Discriminant $=81-28=53$, which is not a perfect square — these roots are irrational, so they don’t contribute to the product.
Step 5 — Substitute back for y = −3
$$x^2-9x+11=-3 \implies x^2-9x+14=0$$
$$(x-2)(x-7)=0 \implies x=2 \text{ or } x=7$$
These are rational roots.
Step 6 — Compute the product of rational roots
$$2\times7=14$$
Answer
$$14$$
Common mistakes
- Forgetting to check the discriminant of both resulting quadratics. It’s tempting to assume all four roots (from both $y=4$ and $y=-3$) are valid — but the question specifically asks for rational roots, so each quadratic’s discriminant must be checked for being a perfect square.
- Sign error when rewriting $(x-4)(x-5)$ in terms of $y$. Since $x^2-9x=y-11$ (not $y+11$), the constant shift must be added carefully: $(x-4)(x-5)=x^2-9x+20=(y-11)+20=y+9$.
Practise next
- Find the product of all rational roots of $(x^2-7x+9)^2-(x-3)(x-4)=3$, using the same substitution method.
Show answer
$6$. Put $u=x^2-7x+9$. Then $(x-3)(x-4)=x^2-7x+12=u+3$, so the equation is $u^2-(u+3)=3$, that is $u^2-u-6=0$, giving $u=3$ or $u=-2$.
$u=3$ gives $x^2-7x+6=0$, so $x=1$ or $x=6$ — both rational. $u=-2$ gives $x^2-7x+11=0$, whose discriminant is $5$, so those two roots are irrational and are excluded. The product of the rational roots is $1\times6=6$.
The whole trick is spotting that $(x-3)(x-4)$ is $u$ plus a constant. Without that the equation is a quartic.
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