JEE Main 2026 — Quadratic Equation Question with Solution

For the quadratic equation $(\lambda +2)x^2-3\lambda x+4\lambda =0$ to have two positive roots, the following three conditions must be satisfied:

1. Discriminant $D\ge 0$
$(-3\lambda {)}^2-4(\lambda +2)(4\lambda )\ge 0$
$9{\lambda }^2-16{\lambda }^2-32\lambda \ge 0$
$-7{\lambda }^2-32\lambda \ge 0$
$7{\lambda }^2+32\lambda \le 0$
$\lambda (7\lambda +32)\le 0$
$\lambda \in \left[-\frac{32}{7},0\right]$

2. Sum of roots $>0$
$\alpha +\beta =\frac{3\lambda }{\lambda +2}>0$
$\lambda \in (-\infty ,-2)\cup (0,\infty )$

3. Product of roots $>0$
$\alpha \beta =\frac{4\lambda }{\lambda +2}>0$
$\lambda \in (-\infty ,-2)\cup (0,\infty )$

Taking the intersection of all three conditions, we get:
$\lambda \in \left[-\frac{32}{7},-2\right)$

Since $-\frac{32}{7}\approx -4.57$, the possible integral values of $\lambda$ in this interval are $-4$ and $-3$.

Thus, there are $2$ possible integral values of $\lambda$.

Answer: $2$