Solve x^2+4xgeq0

$$x ^ { 2 } + 4 x \geq 0$$

Solve for x

$$x\in (-\infty,-4]\cup [0,\infty)$$

Solution Steps

Factor out x.

$$x\left(x+4\right)\geq 0$$

For the product to be ≥0, x+4 and x have to be both ≤0 or both ≥0. Consider the case when x+4 and x are both ≤0.

$$x+4\leq 0 x\leq 0$$

The solution satisfying both inequalities is x\leq -4.

$$x\leq -4$$

Consider the case when x+4 and x are both ≥0.

$$x\geq 0 x+4\geq 0$$

The solution satisfying both inequalities is x\geq 0.

$$x\geq 0$$

The final solution is the union of the obtained solutions.

$$x\leq -4\text{; }x\geq 0$$