Solve x^2+5xleq0

$$x ^ { 2 } + 5 x \leq 0$$

Solve for x

$$x\in \begin{bmatrix}-5,0\end{bmatrix}$$

Solution Steps

Factor out x.

$$x\left(x+5\right)\leq 0$$

For the product to be ≤0, one of the values x+5 and x has to be ≥0 and the other has to be ≤0. Consider the case when x+5\geq 0 and x\leq 0.

$$x+5\geq 0 x\leq 0$$

The solution satisfying both inequalities is x\in \left[-5,0\right].

$$x\in \begin{bmatrix}-5,0\end{bmatrix}$$

Consider the case when x+5\leq 0 and x\geq 0.

$$x\geq 0 x+5\leq 0$$

This is false for any x.

$$x\in \emptyset$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}-5,0\end{bmatrix}$$