Solve -2xgeqx^2

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

Solve for x

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

Solution Steps

Subtract x^{2} from both sides.

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

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

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

Factor out x.

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

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

$$x+2\geq 0 x\leq 0$$

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

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

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

$$x\geq 0 x+2\leq 0$$

This is false for any x.

$$x\in \emptyset$$

The final solution is the union of the obtained solutions.

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