If the quadratic equation $$(\lambda + 2)x^2 - 3\lambda x + 4\lambda = 0$$, $$\lambda \neq -2$$, has two positive roots, then the number of possible integral values of $$\lambda$$ is :

For a quadratic equation $$Ax^2 + Bx + C = 0$$ to have two positive real roots ($$\alpha, \beta > 0$$), it must satisfy three conditions simultaneously:

Real Roots: The discriminant must be non-negative $$\implies D \ge 0$$

Positive Sum of Roots: $$\alpha + \beta = -\frac{B}{A} > 0$$

Positive Product of Roots: $$\alpha\beta = \frac{C}{A} > 0$$

Condition 1:

$$D = B^2 - 4AC = (-3\lambda)^2 - 4(\lambda + 2)(4\lambda) \ge 0$$

$$-7\lambda^2 - 32\lambda \ge 0$$

$$\lambda \in \left[-\frac{32}{7}, 0\right] \quad \text{--- (Inequality 1)}$$

Condition 2:

$$\alpha + \beta = \frac{3\lambda}{\lambda + 2} > 0$$

$$\lambda \in (-\infty, -2) \cup (0, \infty) \quad \text{--- (Inequality 2)}$$

Condition 3:

$$\alpha\beta = \frac{4\lambda}{\lambda + 2} > 0 \implies \frac{\lambda}{\lambda + 2} > 0$$

$$\lambda \in (-\infty, -2) \cup (0, \infty) \quad \text{--- (Inequality 3)}$$

Intersection: $$\lambda \in \left[-\frac{32}{7}, -2\right)$$

The integers that lie within the interval $$[-4.57, -2)$$ are $$\lambda = -4, -3$$

There are exactly $$2$$ possible integral values.