Solve 4a^2-12ageq0

$$4 a ^ { 2 } - 12 a \geq 0$$

Solve for a

$$a\in (-\infty,0]\cup [3,\infty)$$

Solution Steps

Factor out a.

$$4a\left(a-3\right)\geq 0$$

For the product to be ≥0, a and a-3 have to be both ≤0 or both ≥0. Consider the case when a and a-3 are both ≤0.

$$a\leq 0 a-3\leq 0$$

The solution satisfying both inequalities is a\leq 0.

$$a\leq 0$$

Consider the case when a and a-3 are both ≥0.

$$a-3\geq 0 a\geq 0$$

The solution satisfying both inequalities is a\geq 3.

$$a\geq 3$$

The final solution is the union of the obtained solutions.

$$a\leq 0\text{; }a\geq 3$$