Solve x^2+xgeq0

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

Solve for x

$$x\in (-\infty,-1]\cup [0,\infty)$$

Solution Steps

Factor out x.

$$x\left(x+1\right)\geq 0$$

For the product to be ≥0, x+1 and x have to be both ≤0 or both ≥0. Consider the case when x+1 and x are both ≤0.

$$x+1\leq 0 x\leq 0$$

The solution satisfying both inequalities is x\leq -1.

$$x\leq -1$$

Consider the case when x+1 and x are both ≥0.

$$x\geq 0 x+1\geq 0$$

The solution satisfying both inequalities is x\geq 0.

$$x\geq 0$$

The final solution is the union of the obtained solutions.

$$x\leq -1\text{; }x\geq 0$$