Solve quad(8-2x)(2+3x)geq0quad

$$\quad ( 8 - 2 x ) ( 2 + 3 x ) \geq 0 \quad$$

Solve for x

$$x\in \begin{bmatrix}-\frac{2}{3},4\end{bmatrix}$$

Solution Steps

For the product to be ≥0, 3x+2 and 8-2x have to be both ≤0 or both ≥0. Consider the case when 3x+2 and 8-2x are both ≤0.

$$3x+2\leq 0 8-2x\leq 0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when 3x+2 and 8-2x are both ≥0.

$$8-2x\geq 0 3x+2\geq 0$$

The solution satisfying both inequalities is x\in \left[-\frac{2}{3},4\right].

$$x\in \begin{bmatrix}-\frac{2}{3},4\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}-\frac{2}{3},4\end{bmatrix}$$