Solve 2x^2+18xleq0

$$2 x ^ { 2 } + 18 x \leq 0$$

Solve for x

$$x\in \begin{bmatrix}-9,0\end{bmatrix}$$

Solution Steps

Factor out x.

$$2x\left(x+9\right)\leq 0$$

For the product to be ≤0, one of the values x+9 and x has to be ≥0 and the other has to be ≤0. Consider the case when x+9\geq 0 and x\leq 0.

$$x+9\geq 0 x\leq 0$$

The solution satisfying both inequalities is x\in \left[-9,0\right].

$$x\in \begin{bmatrix}-9,0\end{bmatrix}$$

Consider the case when x+9\leq 0 and x\geq 0.

$$x\geq 0 x+9\leq 0$$

This is false for any x.

$$x\in \emptyset$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}-9,0\end{bmatrix}$$