Solve -x^2+2xgeq0

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

Solve for x

$$x\in \begin{bmatrix}0,2\end{bmatrix}$$

Solution Steps

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

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

Factor out x.

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

For the product to be ≤0, one of the values x and x-2 has to be ≥0 and the other has to be ≤0. Consider the case when x\geq 0 and x-2\leq 0.

$$x\geq 0 x-2\leq 0$$

The solution satisfying both inequalities is x\in \left[0,2\right].

$$x\in \begin{bmatrix}0,2\end{bmatrix}$$

Consider the case when x\leq 0 and x-2\geq 0.

$$x-2\geq 0 x\leq 0$$

This is false for any x.

$$x\in \emptyset$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}0,2\end{bmatrix}$$