Quadratic Equations – Properties of Roots – JEE Main 28 Jan 2026 Shift 1

Solution:

Given the quadratic equation:
$$\lambda x^2 – (\lambda + 3)x + 3 = 0$$
The sum of roots and product of roots are:
$$\alpha + \beta = \frac{\lambda + 3}{\lambda} = 1 + \frac{3}{\lambda}$$
$$\alpha \beta = \frac{3}{\lambda}$$

We are given the condition:
$$\frac{\beta – \alpha}{\alpha\beta} = \frac{1}{3}$$
Substituting the value of $\alpha\beta$:
$$\frac{\beta – \alpha}{3/\lambda} = \frac{1}{3} \Rightarrow \beta – \alpha = \frac{1}{\lambda}$$

We know the identity $(\beta – \alpha)^2 = (\alpha + \beta)^2 – 4\alpha\beta$.
Substitute the expressions in terms of $\lambda$:
$$\left(\frac{1}{\lambda}\right)^2 = \left(\frac{\lambda + 3}{\lambda}\right)^2 – 4\left(\frac{3}{\lambda}\right)$$
$$\frac{1}{\lambda^2} = \frac{\lambda^2 + 6\lambda + 9}{\lambda^2} – \frac{12}{\lambda}$$

Multiply the entire equation by $\lambda^2$ (since $\lambda \ne 0$ for a quadratic):
$$1 = \lambda^2 + 6\lambda + 9 – 12\lambda$$
$$1 = \lambda^2 – 6\lambda + 9$$
$$\lambda^2 – 6\lambda + 8 = 0$$

Factor the quadratic equation:
$$(\lambda – 2)(\lambda – 4) = 0$$
Possible values for $\lambda$ are $2$ and $4$.

The sum of all possible values of $\lambda$ is:
$$Sum = 2 + 4 = 6$$

Ans. (1)