Solve the Inequality for x (x^2+x-6)/(x^2-3x-4)<=0

$\frac{x^{2} + x - 6}{x^{2} - 3 x - 4} \leq 0$

Step 1

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

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

$x^{2} - 3 x - 4 = 0$

Step 2

Factor $x^{2} + x - 6$ 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 $1$ .

$- 2, 3$

Write the factored form using these integers.

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

Step 3

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

$x - 2 = 0$

$x + 3 = 0$

Step 4

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 5

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.

$x = - 3$

Step 6

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

$x = 2, - 3$

Step 7

Factor $x^{2} - 3 x - 4$ 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 $- 4$ and whose sum is $- 3$ .

$- 4, 1$

Write the factored form using these integers.

$(x - 4) (x + 1) = 0$

Step 8

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

$x - 4 = 0$

$x + 1 = 0$

Step 9

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Step 10

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.

$x = - 1$

Step 11

The final solution is all the values that make $(x - 4) (x + 1) = 0$ true.

$x = 4, - 1$

Step 12

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

$x = 2, - 3$

$x = 4, - 1$

Step 13

Consolidate the solutions.

$x = 2, - 3, 4, - 1$

Step 14

Find the domain of $\frac{x^{2} + x - 6}{x^{2} - 3 x - 4}$ .

Set the denominator in $\frac{x^{2} + x - 6}{x^{2} - 3 x - 4}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 3 x - 4 = 0$

Solve for $x$ .

Factor $x^{2} - 3 x - 4$ 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 $- 4$ and whose sum is $- 3$ .

$- 4, 1$

Write the factored form using these integers.

$(x - 4) (x + 1) = 0$

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

$x - 4 = 0$

$x + 1 = 0$

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

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 $(x - 4) (x + 1) = 0$ true.

$x = 4, - 1$

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

$(- \infty, - 1) \cup (- 1, 4) \cup (4, \infty)$

Step 15

Use each root to create test intervals.

$x < - 3$

$- 3 < x < - 1$

$- 1 < x < 2$

$2 < x < 4$

$x > 4$

Step 16

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 $x < - 3$ 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.

$\frac{{(- 6)}^{2} - 6 - 6}{{(- 6)}^{2} - 3 \cdot - 6 - 4} \leq 0$

The left side $0.48$ is greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 3 < x < - 1$ to see if it makes the inequality true.

Choose a value on the interval $- 3 < x < - 1$ and see if this value makes the original inequality true.

$x = - 2$

Replace $x$ with $- 2$ in the original inequality.

$\frac{{(- 2)}^{2} - 2 - 6}{{(- 2)}^{2} - 3 \cdot - 2 - 4} \leq 0$

The left side $- 0.\overline{6}$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- 1 < x < 2$ to see if it makes the inequality true.

Choose a value on the interval $- 1 < x < 2$ and see if this value makes the original inequality true.

$x = 0$

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

$\frac{{(0)}^{2} + 0 - 6}{{(0)}^{2} - 3 \cdot 0 - 4} \leq 0$

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 $2 < x < 4$ to see if it makes the inequality true.

Choose a value on the interval $2 < x < 4$ and see if this value makes the original inequality true.

$x = 3$

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

$\frac{{(3)}^{2} + 3 - 6}{{(3)}^{2} - 3 \cdot 3 - 4} \leq 0$

The left side $- 1.5$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $x > 4$ to see if it makes the inequality true.

Choose a value on the interval $x > 4$ and see if this value makes the original inequality true.

$x = 6$

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

$\frac{{(6)}^{2} + 6 - 6}{{(6)}^{2} - 3 \cdot 6 - 4} \leq 0$

The left side $2.\overline{571428}$ 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.

$x < - 3$ False

$- 3 < x < - 1$ True

$- 1 < x < 2$ False

$2 < x < 4$ True

$x > 4$ False

$x < - 3$ False

$- 3 < x < - 1$ True

$- 1 < x < 2$ False

$2 < x < 4$ True

$x > 4$ False

Step 17

The solution consists of all of the true intervals.

$- 3 \leq x < - 1$ or $2 \leq x < 4$

Step 18

The result can be shown in multiple forms.

Inequality Form:

$- 3 \leq x < - 1 or 2 \leq x < 4$

Interval Notation:

$[- 3, - 1) \cup [2, 4)$

Step 19

Solve the Inequality for x (x^2+x-6)/(x^2-3x-4)<=0

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