Solve x^2+x>0

$$x ^ { 2 } + x > 0$$

Solve for x

$$x\in \left(-\infty,-1\right)\cup \left(0,\infty\right)$$

Solution Steps

Factor out x.

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

For the product to be positive, x+1 and x have to be both negative or both positive. Consider the case when x+1 and x are both negative.

$$x+1<0 x<0$$

The solution satisfying both inequalities is x<-1.

$$x<-1$$

Consider the case when x+1 and x are both positive.

$$x>0 x+1>0$$

The solution satisfying both inequalities is x>0.

$$x>0$$

The final solution is the union of the obtained solutions.

$$x<-1\text{; }x>0$$