Solve the Inequality for x x^2+7x>=0

$x^{2} + 7 x \geq 0$

Convert the inequality to an equation.

$x^{2} + 7 x = 0$

Factor $x$ out of $x^{2} + 7 x$ .

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

$x \cdot x + 7 x = 0$

Factor $x$ out of $7 x$ .

$x \cdot x + x \cdot 7 = 0$

Factor $x$ out of $x \cdot x + x \cdot 7$ .

$x (x + 7) = 0$

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

$x = 0$

$x + 7 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x + 7$ equal to $0$ and solve for $x$ .

Set $x + 7$ equal to $0$ .

$x + 7 = 0$

Subtract $7$ from both sides of the equation.

$x = - 7$

The final solution is all the values that make $x (x + 7) = 0$ true.

$x = 0, - 7$

Use each root to create test intervals.

$x < - 7$

$- 7 < x < 0$

$x > 0$

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 $x < - 7$ to see if it makes the inequality true.

Choose a value on the interval $x < - 7$ and see if this value makes the original inequality true.

$x = - 10$

Replace $x$ with $- 10$ in the original inequality.

${(- 10)}^{2} + 7 (- 10) \geq 0$

The left side $30$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- 7 < x < 0$ to see if it makes the inequality true.

Choose a value on the interval $- 7 < x < 0$ and see if this value makes the original inequality true.

$x = - 4$

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

${(- 4)}^{2} + 7 (- 4) \geq 0$

The left side $- 12$ is less 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.

${(2)}^{2} + 7 (2) \geq 0$

The left side $18$ 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.

$x < - 7$ True

$- 7 < x < 0$ False

$x > 0$ True

$x < - 7$ True

$- 7 < x < 0$ False

$x > 0$ True

The solution consists of all of the true intervals.

$x \leq - 7$ or $x \geq 0$

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 7 or x \geq 0$

Interval Notation:

$(- \infty, - 7] \cup [0, \infty)$

Solve the Inequality for x x^2+7x>=0

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