JEE Mathematics 2025 Question with Solution

Given:

$$(x^2 - 9x + 11)^2 - (x-4)(x-5) = 3$$

Find: The product of all rational roots.

First simplify

$$(x-4)(x-5) = x^2 - 9x + 20$$

So the equation becomes

$$(x^2 - 9x + 11)^2 - (x^2 - 9x + 20) = 3$$ $$(x^2 - 9x + 11)^2 - (x^2 - 9x + 20) - 3 = 0$$

Let

$$y = x^2 - 9x + 11$$

Then

$$x^2 - 9x = y - 11$$

Substituting into the equation,

$$y^2 - (y - 11 + 20) - 3 = 0$$ $$y^2 - (y + 9) - 3 = 0$$ $$y^2 - y - 12 = 0$$

Factor the quadratic:

$$(y-4)(y+3) = 0$$

Hence,

$$y = 4 \text{ or } y = -3$$

For $y = 4$,

$$x^2 - 9x + 11 = 4$$ $$x^2 - 9x + 7 = 0$$

Its roots are

$$x = \frac{9 \pm \sqrt{53}}{2}$$

which are irrational.

For $y = -3$,

$$x^2 - 9x + 11 = -3$$ $$x^2 - 9x + 14 = 0$$ $$(x-7)(x-2) = 0$$

So the rational roots are $x = 7$ and $x = 2$.

Their product is

$$7 \times 2 = 14$$

Therefore, the product of all rational roots is $14$. The correct option is A.

Given:

$$(x^2 - 9x + 11)^2 - (x - 4)(x - 5) = 3$$

Find: Product of all rational roots.

A useful substitution is

$$y = x^2 - 9x + 11$$

Also,

$$(x-4)(x-5) = x^2 - 9x + 20$$

Since

$$x^2 - 9x = y - 11$$

we get

$$(x-4)(x-5) = (y-11) + 20 = y + 9$$

Now substitute into the original equation:

$$y^2 - (y+9) = 3$$ $$y^2 - y - 12 = 0$$ $$(y-4)(y+3) = 0$$

So,

$$y = 4 \text{ or } y = -3$$

If $y = 4$, then

$$x^2 - 9x + 11 = 4$$ $$x^2 - 9x + 7 = 0$$

The discriminant is

$$81 - 28 = 53$$

which is not a perfect square, so these roots are irrational.

If $y = -3$, then

$$x^2 - 9x + 11 = -3$$ $$x^2 - 9x + 14 = 0$$ $$(x-7)(x-2) = 0$$

Thus the rational roots are $2$ and $7$.

Therefore,

$$2 \cdot 7 = 14$$

So the required product is $14$.