Find the solution set of the inequality ax² + (3a − 4)x − 12 ≥ 0, discussing all cases of the parameter a.

Step-by-step solution

1. 1

   Separate the degenerate case a = 0 first. The coefficient of $x^2$ is the parameter itself, so the inequality is not always quadratic. If $a = 0$ it reduces to a linear inequality:

   $-4x - 12 \ge 0 \quad \Longrightarrow \quad x \le -3$

   Solution set: $(-\infty, -3]$. Every remaining case assumes $a \ne 0$.

2. 2

   Compute the discriminant and notice it is a perfect square. For $a \ne 0$,

   $\Delta = (3a-4)^2 + 48a = 9a^2 - 24a + 16 + 48a = 9a^2 + 24a + 16 = (3a+4)^2$

   Because $\Delta \ge 0$ for every real $a$, the quadratic always has real roots — no "no real root" branch is needed. That is the key structural fact this problem is built on.

3. 3

   Extract the two roots exactly.

   $x = \frac{4 - 3a \pm |3a+4|}{2a} \quad \Longrightarrow \quad x_1 = \frac{8}{2a} = \frac{4}{a}, \qquad x_2 = \frac{-6a}{2a} = -3$

   So one root is pinned at $x = -3$ for every $a$, and the other slides with $a$. Hence

   $ax^2 + (3a-4)x - 12 = a(x+3)\left(x - \frac{4}{a}\right)$

   which expands back to $ax^2 + 3ax - 4x - 12$ — the original expression.

4. 4

   Case a > 0: parabola opens upward. Then $4/a > 0 > -3$, so the roots are ordered $-3 < 4/a$ and the expression is non-negative outside them:

   $x \le -3 \quad \text{or} \quad x \ge \frac{4}{a}, \qquad \text{i.e.} \quad (-\infty,-3] \cup \left[\tfrac{4}{a}, +\infty\right)$

5. 5

   Case a < 0: parabola opens downward, so compare the roots. Now the expression is non-negative between the roots, and the order of $-3$ and $4/a$ depends on $a$. Since $4/a = -3 \iff a = -\tfrac43$:

   $-\tfrac43 < a < 0: \ \ \frac4a < -3 \ \Longrightarrow \ \left[\tfrac4a,\, -3\right]$

   $a = -\tfrac43: \ \ \frac4a = -3 \ \Longrightarrow \ \{-3\} \ \ (\Delta = 0,\ \text{a single point})$

   $a < -\tfrac43: \ \ -3 < \frac4a \ \Longrightarrow \ \left[-3,\, \tfrac4a\right]$

   The middle case is easy to miss: at $a = -\tfrac43$ the two roots merge and the solution set shrinks to one point.

6. 6

   Collect the complete answer.

   $a = 0:\ (-\infty,-3] \qquad a > 0:\ (-\infty,-3]\cup\left[\tfrac4a,\infty\right)$

   $a < 0:\ \left[\min\left(-3,\tfrac4a\right),\ \max\left(-3,\tfrac4a\right)\right]$

   Spot check with $a = 2$: the inequality is $2x^2+2x-12 \ge 0$, i.e. $2(x+3)(x-2) \ge 0$, giving $x \le -3$ or $x \ge 2 = 4/a$ as predicted.