Solve the Inequality for x x^2+15x<=6x-18

$x^{2} + 15 x \leq 6 x - 18$

Move all terms containing $x$ to the left side of the inequality.

Subtract $6 x$ from both sides of the inequality.

$x^{2} + 15 x - 6 x \leq - 18$

Subtract $6 x$ from $15 x$ .

$x^{2} + 9 x \leq - 18$

Convert the inequality to an equation.

$x^{2} + 9 x = - 18$

Add $18$ to both sides of the equation.

$x^{2} + 9 x + 18 = 0$

Factor $x^{2} + 9 x + 18$ 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 $18$ and whose sum is $9$ .

$3, 6$

Write the factored form using these integers.

$(x + 3) (x + 6) = 0$

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

$x + 3 = 0$

$x + 6 = 0$

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Set $x + 6$ equal to $0$ and solve for $x$ .

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

The final solution is all the values that make $(x + 3) (x + 6) = 0$ true.

$x = - 3, - 6$

Use each root to create test intervals.

$x < - 6$

$- 6 < x < - 3$

$x > - 3$

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 $x < - 6$ to see if it makes the inequality true.

Choose a value on the interval $x < - 6$ and see if this value makes the original inequality true.

$x = - 8$

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

${(- 8)}^{2} + 15 (- 8) \leq 6 (- 8) - 18$

The left side $- 56$ is greater than the right side $- 66$ , which means that the given statement is false.

False

Test a value on the interval $- 6 < x < - 3$ to see if it makes the inequality true.

Choose a value on the interval $- 6 < x < - 3$ and see if this value makes the original inequality true.

$x = - 4$

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

${(- 4)}^{2} + 15 (- 4) \leq 6 (- 4) - 18$

The left side $- 44$ is less than the right side $- 42$ , which means that the given statement is always true.

True

Test a value on the interval $x > - 3$ to see if it makes the inequality true.

Choose a value on the interval $x > - 3$ and see if this value makes the original inequality true.

$x = 0$

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

${(0)}^{2} + 15 (0) \leq 6 (0) - 18$

The left side $0$ is greater than the right side $- 18$ , which means that the given statement is false.

False

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

$x < - 6$ False

$- 6 < x < - 3$ True

$x > - 3$ False

$x < - 6$ False

$- 6 < x < - 3$ True

$x > - 3$ False

The solution consists of all of the true intervals.

$- 6 \leq x \leq - 3$

The result can be shown in multiple forms.

Inequality Form:

$- 6 \leq x \leq - 3$

Interval Notation:

$[- 6, - 3]$

Solve the Inequality for x x^2+15x<=6x-18

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