Solve the Inequality for x -x^2+2x<=0

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

Convert the inequality to an equation.

$- x^{2} + 2 x = 0$

Factor $- x$ out of $- x^{2} + 2 x$ .

Factor $- x$ out of $- x^{2}$ .

$- x \cdot x + 2 x = 0$

Factor $- x$ out of $2 x$ .

$- x \cdot x - x \cdot - 2 = 0$

Factor $- x$ out of $- x (x) - x (- 2)$ .

$- x (x - 2) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$x - 2 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

The final solution is all the values that make $- x (x - 2) = 0$ true.

$x = 0, 2$

Use each root to create test intervals.

$x < 0$

$0 < x < 2$

$x > 2$

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $x < 0$ to see if it makes the inequality true.

Choose a value on the interval $x < 0$ and see if this value makes the original inequality true.

$x = - 2$

Replace $x$ with $- 2$ in the original inequality.

$- {(- 2)}^{2} + 2 (- 2) \leq 0$

The left side $- 8$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $0 < x < 2$ to see if it makes the inequality true.

Choose a value on the interval $0 < x < 2$ and see if this value makes the original inequality true.

$x = 1$

Replace $x$ with $1$ in the original inequality.

$- {(1)}^{2} + 2 (1) \leq 0$

The left side $1$ is greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > 2$ to see if it makes the inequality true.

Choose a value on the interval $x > 2$ and see if this value makes the original inequality true.

$x = 4$

Replace $x$ with $4$ in the original inequality.

$- {(4)}^{2} + 2 (4) \leq 0$

The left side $- 8$ is less than the right side $0$ , which means that the given statement is always true.

True

Compare the intervals to determine which ones satisfy the original inequality.

$x < 0$ True

$0 < x < 2$ False

$x > 2$ True

$x < 0$ True

$0 < x < 2$ False

$x > 2$ True

The solution consists of all of the true intervals.

$x \leq 0$ or $x \geq 2$

The result can be shown in multiple forms.

Inequality Form:

$x \leq 0 or x \geq 2$

Interval Notation:

$(- \infty, 0] \cup [2, \infty)$

Solve the Inequality for x -x^2+2x<=0

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