Solve for x (x+9)/(x^2-5x+6)<=0

$\frac{x + 9}{x^{2} - 5 x + 6} \leq 0$

Step 1

Find all the values where the expression switches from negative to positive by setting each factor equal to $0$ and solving.

$x + 9 = 0$

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

Step 2

Subtract $9$ from both sides of the equation.

$x = - 9$

Step 3

Factor $x^{2} - 5 x + 6$ 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 $6$ and whose sum is $- 5$ .

$- 3, - 2$

Write the factored form using these integers.

$(x - 3) (x - 2) = 0$

Step 4

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

$x - 3 = 0$

$x - 2 = 0$

Step 5

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Step 6

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.

$x = 2$

Step 7

The final solution is all the values that make $(x - 3) (x - 2) = 0$ true.

$x = 3, 2$

Step 8

Solve for each factor to find the values where the absolute value expression goes from negative to positive.

$x = - 9$

$x = 3, 2$

Step 9

Consolidate the solutions.

$x = - 9, 3, 2$

Step 10

Find the domain of $\frac{x + 9}{x^{2} - 5 x + 6}$ .

Set the denominator in $\frac{x + 9}{x^{2} - 5 x + 6}$ equal to $0$ to find where the expression is undefined.

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

Solve for $x$ .

Factor $x^{2} - 5 x + 6$ 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 $6$ and whose sum is $- 5$ .

$- 3, - 2$

Write the factored form using these integers.

$(x - 3) (x - 2) = 0$

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

$x - 3 = 0$

$x - 2 = 0$

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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 $(x - 3) (x - 2) = 0$ true.

$x = 3, 2$

The domain is all values of $x$ that make the expression defined.

$(- \infty, 2) \cup (2, 3) \cup (3, \infty)$

Step 11

Use each root to create test intervals.

$x < - 9$

$- 9 < x < 2$

$2 < x < 3$

$x > 3$

Step 12

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 $x < - 9$ to see if it makes the inequality true.

Choose a value on the interval $x < - 9$ and see if this value makes the original inequality true.

$x = - 12$

Replace $x$ with $- 12$ in the original inequality.

$\frac{(- 12) + 9}{{(- 12)}^{2} - 5 \cdot - 12 + 6} \leq 0$

The left side $- 0.0\overline{142857}$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- 9 < x < 2$ to see if it makes the inequality true.

Choose a value on the interval $- 9 < x < 2$ and see if this value makes the original inequality true.

$x = 0$

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

$\frac{(0) + 9}{{(0)}^{2} - 5 \cdot 0 + 6} \leq 0$

The left side $1.5$ is greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $2 < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $2 < x < 3$ and see if this value makes the original inequality true.

$x = 2.5$

Replace $x$ with $2.5$ in the original inequality.

$\frac{(2.5) + 9}{{(2.5)}^{2} - 5 \cdot 2.5 + 6} \leq 0$

The left side $- 46$ is less than the right side $0$ , 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.

$x = 6$

Replace $x$ with $6$ in the original inequality.

$\frac{(6) + 9}{{(6)}^{2} - 5 \cdot 6 + 6} \leq 0$

The left side $1.25$ is greater than the right side $0$ , which means that the given statement is false.

False

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

$x < - 9$ True

$- 9 < x < 2$ False

$2 < x < 3$ True

$x > 3$ False

$x < - 9$ True

$- 9 < x < 2$ False

$2 < x < 3$ True

$x > 3$ False

Step 13

The solution consists of all of the true intervals.

$x \leq - 9$ or $2 < x < 3$

Step 14

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 9 or 2 < x < 3$

Interval Notation:

$(- \infty, - 9] \cup (2, 3)$

Step 15

Solve for x (x+9)/(x^2-5x+6)<=0

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