Algebra – Quadratic Equations – JEE Main 22 Jan 2026 Shift 2

Solution:

For the quadratic equation $ax^2 + bx + c = 0$, the difference of roots is given by $|\beta – \alpha| = \frac{\sqrt{D}}{|a|}$.
Here $a=12, b=-20, c=3\lambda$.
$$ D = b^2 – 4ac = (-20)^2 – 4(12)(3\lambda) = 400 – 144\lambda $$
$$ |\beta – \alpha| = \frac{\sqrt{400 – 144\lambda}}{12} $$

Given the inequality:
$$ \frac{1}{2} \le \frac{\sqrt{400 – 144\lambda}}{12} \le \frac{3}{2} $$
Multiply by 12:
$$ 6 \le \sqrt{400 – 144\lambda} \le 18 $$
Squaring all sides:
$$ 36 \le 400 – 144\lambda \le 324 $$

Subtract 400 from all parts:
$$ 36 – 400 \le -144\lambda \le 324 – 400 $$
$$ -364 \le -144\lambda \le -76 $$
Divide by -144 and reverse the inequalities:
$$ \frac{76}{144} \le \lambda \le \frac{364}{144} $$

Calculating the approximate values:
$$ \frac{76}{144} \approx 0.52 $$
$$ \frac{364}{144} \approx 2.52 $$
So, $0.52 \le \lambda \le 2.52$.

Since $\lambda \in \mathbb{Z}$ (integer), the possible values for $\lambda$ are 1 and 2.
The sum of all possible values is:
$$ \text{Sum} = 1 + 2 = 3 $$

Ans. (3)