Solve x^2-3xleq0}

$$x ^ { 2 } - 3 x \leq 0 \}$$

Solve for x

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

Solution Steps

Factor out x.

$$x\left(x-3\right)\leq 0$$

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

$$x\geq 0 x-3\leq 0$$

The solution satisfying both inequalities is x\in \left[0,3\right].

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

Consider the case when x\leq 0 and x-3\geq 0.

$$x-3\geq 0 x\leq 0$$

This is false for any x.

$$x\in \emptyset$$

The final solution is the union of the obtained solutions.

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