Solve the Inequality for x x^2+x>0

$x^{2} + x > 0$

Convert the inequality to an equation.

$x^{2} + x = 0$

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

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

$x \cdot x + x = 0$

Raise $x$ to the power of $1$ .

$x \cdot x + x = 0$

Factor $x$ out of $x^{1}$ .

$x \cdot x + x \cdot 1 = 0$

Factor $x$ out of $x \cdot x + x \cdot 1$ .

$x (x + 1) = 0$

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

$x = 0$

$x + 1 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

The final solution is all the values that make $x (x + 1) = 0$ true.

$x = 0, - 1$

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 0$

$x > 0$

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 $x < - 1$ to see if it makes the inequality true.

Choose a value on the interval $x < - 1$ and see if this value makes the original inequality true.

$x = - 4$

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

${(- 4)}^{2} - 4 > 0$

The left side $12$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- 1 < x < 0$ to see if it makes the inequality true.

Choose a value on the interval $- 1 < x < 0$ and see if this value makes the original inequality true.

$x = - 0.5$

Replace $x$ with $- 0.5$ in the original inequality.

${(- 0.5)}^{2} - 0.5 > 0$

The left side $- 0.25$ is not greater than the right side $0$ , which means that the given statement is false.

False

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.

${(2)}^{2} + 2 > 0$

The left side $6$ is greater 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.

$x < - 1$ True

$- 1 < x < 0$ False

$x > 0$ True

$x < - 1$ True

$- 1 < x < 0$ False

$x > 0$ True

The solution consists of all of the true intervals.

$x < - 1$ or $x > 0$

The result can be shown in multiple forms.

Inequality Form:

$x < - 1 or x > 0$

Interval Notation:

$(- \infty, - 1) \cup (0, \infty)$

Solve the Inequality for x x^2+x>0

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