Solve -2x(3x+1)geq0

$$-2x(3x+1) \geq 0$$

Solve for x

$$x\in \begin{bmatrix}-\frac{1}{3},0\end{bmatrix}$$

Solution Steps

Use the distributive property to multiply -2x by 3x+1.

$$-6x^{2}-2x\geq 0$$

Multiply the inequality by -1 to make the coefficient of the highest power in -6x^{2}-2x positive. Since -1 is negative, the inequality direction is changed.

$$6x^{2}+2x\leq 0$$

Factor out x.

$$2x\left(3x+1\right)\leq 0$$

For the product to be ≤0, one of the values x+\frac{1}{3} and x has to be ≥0 and the other has to be ≤0. Consider the case when x+\frac{1}{3}\geq 0 and x\leq 0.

$$x+\frac{1}{3}\geq 0 x\leq 0$$

The solution satisfying both inequalities is x\in \left[-\frac{1}{3},0\right].

$$x\in \begin{bmatrix}-\frac{1}{3},0\end{bmatrix}$$

Consider the case when x+\frac{1}{3}\leq 0 and x\geq 0.

$$x\geq 0 x+\frac{1}{3}\leq 0$$

This is false for any x.

$$x\in \emptyset$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}-\frac{1}{3},0\end{bmatrix}$$