Solve the Inequality for x x^2+6x>=-5

$x^{2} + 6 x \geq - 5$

Convert the inequality to an equation.

$x^{2} + 6 x = - 5$

Add $5$ to both sides of the equation.

$x^{2} + 6 x + 5 = 0$

Factor $x^{2} + 6 x + 5$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $5$ and whose sum is $6$ .

$1, 5$

Write the factored form using these integers.

$(x + 1) (x + 5) = 0$

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

$x + 1 = 0$

$x + 5 = 0$

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.

$x = - 1$

Set $x + 5$ equal to $0$ and solve for $x$ .

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

The final solution is all the values that make $(x + 1) (x + 5) = 0$ true.

$x = - 1, - 5$

Use each root to create test intervals.

$x < - 5$

$- 5 < x < - 1$

$x > - 1$

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 $x < - 5$ to see if it makes the inequality true.

Choose a value on the interval $x < - 5$ and see if this value makes the original inequality true.

$x = - 8$

Replace $x$ with $- 8$ in the original inequality.

${(- 8)}^{2} + 6 (- 8) \geq - 5$

The left side $16$ is greater than the right side $- 5$ , which means that the given statement is always true.

True

Test a value on the interval $- 5 < x < - 1$ to see if it makes the inequality true.

Choose a value on the interval $- 5 < x < - 1$ and see if this value makes the original inequality true.

$x = - 3$

Replace $x$ with $- 3$ in the original inequality.

${(- 3)}^{2} + 6 (- 3) \geq - 5$

The left side $- 9$ is less than the right side $- 5$ , which means that the given statement is false.

False

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.

$x = 0$

Replace $x$ with $0$ in the original inequality.

${(0)}^{2} + 6 (0) \geq - 5$

The left side $0$ is greater than the right side $- 5$ , which means that the given statement is always true.

True

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

$x < - 5$ True

$- 5 < x < - 1$ False

$x > - 1$ True

$x < - 5$ True

$- 5 < x < - 1$ False

$x > - 1$ True

The solution consists of all of the true intervals.

$x \leq - 5$ or $x \geq - 1$

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 5 or x \geq - 1$

Interval Notation:

$(- \infty, - 5] \cup [- 1, \infty)$

Solve the Inequality for x x^2+6x>=-5

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