Solve 2x^2leq8x

$$2 x ^ { 2 } \leq 8 x$$

Solve for x

$$x\in \begin{bmatrix}0,4\end{bmatrix}$$

Solution Steps

Subtract 8x from both sides.

$$2x^{2}-8x\leq 0$$

Factor out x.

$$2x\left(x-4\right)\leq 0$$

For the product to be ≤0, one of the values x and x-4 has to be ≥0 and the other has to be ≤0. Consider the case when x\geq 0 and x-4\leq 0.

$$x\geq 0 x-4\leq 0$$

The solution satisfying both inequalities is x\in \left[0,4\right].

$$x\in \begin{bmatrix}0,4\end{bmatrix}$$

Consider the case when x\leq 0 and x-4\geq 0.

$$x-4\geq 0 x\leq 0$$

This is false for any x.

$$x\in \emptyset$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}0,4\end{bmatrix}$$