Quadratic Equations – JEE Main 24 January 2025 Shift 1

Solution:

Simplify the term $(x-4)(x-5) = x^2 – 9x + 20$.
Let $t = x^2 – 9x$.
The equation becomes:
$$ (t+11)^2 – (t+20) = 3 $$
$$ t^2 + 22t + 121 – t – 20 – 3 = 0 $$
$$ t^2 + 21t + 98 = 0 $$
$$ (t+7)(t+14) = 0 $$
So, $t = -7$ or $t = -14$.

Case 1: $x^2 – 9x = -7 \Rightarrow x^2 – 9x + 7 = 0$.
Discriminant $D = (-9)^2 – 4(7) = 81 – 28 = 53$ (Not a perfect square).
Roots are irrational.

Case 2: $x^2 – 9x = -14 \Rightarrow x^2 – 9x + 14 = 0$.
Discriminant $D = (-9)^2 – 4(14) = 81 – 56 = 25$ (Perfect square).
Roots are rational:
$$ (x-2)(x-7) = 0 \Rightarrow x = 2, 7 $$

The rational roots are 2 and 7.
Product of rational roots = $2 \times 7 = 14$.

Ans. (1)