JEE Main 2026 — 28 January, Morning Shift — Question 10

$\frac{\beta-\alpha}{\alpha \beta}=\frac{1}{3}, \quad \alpha+\beta=\frac{\lambda+3}{\lambda}, \alpha \beta=\frac{3}{\lambda}$

$\beta-\alpha=\frac{\alpha \beta}{3}=\frac{1}{\lambda}$

on squaring

$\alpha^{2}+\beta^{2}-2 \alpha \beta=\frac{1}{\lambda^{2}}$

$\alpha^{2}+\beta^{2}+2 \alpha \beta=\frac{(\lambda+3)^{2}}{\lambda^{2}} \ldots . (2)$

$4 \alpha \beta=\frac{(\lambda+3)^{2}-1}{\lambda^{2}}$

$\frac{12}{\lambda}=\frac{\lambda^{2}+6 \lambda+8}{\lambda^{2}}$

$\Rightarrow \lambda^{3}-6 \lambda^{2}+8 \lambda=0$

\tag\ldots \Rightarrow \lambda=0,2,4

Sum of possible values of $\lambda$ is $=6$