If the quadratic equation (lambda + 2)x^2 - 3 | vedclass

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Question

(B) For a quadratic equation $ax^2 + bx + c = 0$ to have two positive roots, the following conditions must be satisfied:$1$. Discriminant $D \ge 0$: $

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) For a quadratic equation $ax^2 + bx + c = 0$ to have two positive roots, the following conditions must be satisfied:$1$. Discriminant $D \ge 0$: $D = (-3\lambda)^2 - 4(\lambda + 2)(4\lambda) = 9\lambda^2 - 16\lambda^2 - 32\lambda = -7\lambda^2 - 32\lambda \ge 0$. This implies $\lambda(7\lambda + 32) \le 0$, so $\lambda \in [-\frac{32}{7}, 0]$.$2$. Product of roots $P = \frac{c}{a} > 0$: $\frac{4\lambda}{\lambda + 2} > 0$. This implies $\lambda \in (-\infty, -2) \cup (0, \infty)$.$3$. Sum of roots $S = -\frac{b}{a} > 0$: $\frac{3\lambda}{\lambda + 2} > 0$. This implies $\lambda \in (-\infty, -2) \cup (0, \infty)$.Combining these conditions: $\lambda \in [-\frac{32}{7}, -2)$.The integral values of $\lambda$ in this interval are $-4$ and $-3$.Thus, there are $2$ possible integral values.

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:

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