Number of Integral Values of Parameter for Positive Roots

To find the number of possible integral values of the parameter $\lambda$ such that the quadratic equation $(\lambda + 2)x^2 - 3\lambda x + 4\lambda = 0, \quad \lambda \neq -2$ has two positive real roots, we apply the location of roots conditions for a quadratic equation $Ax^2 + Bx + C = 0$.

Condition 1: Real Roots (Discriminant $D \ge 0$)

For the equation to have real roots, the discriminant must be non-negative: $D = (-3\lambda)^2 - 4(\lambda + 2)(4\lambda) \ge 0$ $9\lambda^2 - 16\lambda(\lambda + 2) \ge 0$ $9\lambda^2 - 16\lambda^2 - 32\lambda \ge 0$ $-7\lambda^2 - 32\lambda \ge 0$ $7\lambda^2 + 32\lambda \le 0$ $\lambda(7\lambda + 32) \le 0$

Solving this inequality gives: $\lambda \in \left[-\frac{32}{7}, 0\right]$

Condition 2: Sum of Roots is Positive

For both roots to be positive, their sum must be strictly positive: $\text{Sum of roots} = \frac{3\lambda}{\lambda + 2} > 0$

Solving the inequality $\frac{\lambda}{\lambda + 2} > 0$: $\lambda \in (-\infty, -2) \cup (0, \infty)$

Condition 3: Product of Roots is Positive

For both roots to be positive, their product must also be strictly positive: $\text{Product of roots} = \frac{4\lambda}{\lambda + 2} > 0$

Solving the inequality $\frac{\lambda}{\lambda + 2} > 0$: $\lambda \in (-\infty, -2) \cup (0, \infty)$

---

Intersection of Conditions

Taking the intersection of all three conditions: $\lambda \in \left[-\frac{32}{7}, 0\right] \cap \Big( (-\infty, -2) \cup (0, \infty) \Big)$

Since $-\frac{32}{7} \approx -4.57$, the combined range for $\lambda$ is: $\lambda \in \left[-\frac{32}{7}, -2\right)$

Finding Integral Values

The integers contained within the interval $\left[-4.57, -2\right)$ are: $\lambda = -4, -3$

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

Correct Option: B