Algebra Examples

$\frac{x^{2} - 1}{x^{2} + 5 x + 4} \leq 0$

Step 1

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

$x^{2} - 1 = 0$

$x^{2} + 5 x + 4 = 0$

Step 2

Add $1$ to both sides of the equation.

$x^{2} = 1$

Step 3

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{1}$

Step 4

Any root of $1$ is $1$ .

$x = \pm 1$

Step 5

The complete solution is the result of both the positive and negative portions of the solution.

Tap for more steps...

Step 5.1

First, use the positive value of the $\pm$ to find the first solution.

$x = 1$

Step 5.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 1$

Step 5.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = 1, - 1$

Step 6

Factor $x^{2} + 5 x + 4$ using the AC method.

Tap for more steps...

Step 6.1

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 $5$ .

$1, 4$

Step 6.2

Write the factored form using these integers.

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

Step 7

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

$x + 1 = 0$

$x + 4 = 0$

Step 8

Set $x + 1$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 8.1

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 8.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 9

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

Tap for more steps...

Step 9.1

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

$x + 4 = 0$

Step 9.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 10

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

$x = - 1, - 4$

Step 11

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

$x = 1, - 1$

$x = - 1, - 4$

Step 12

Consolidate the solutions.

$x = 1, - 1, - 1, - 4$

Step 13

Find the domain of $\frac{x^{2} - 1}{x^{2} + 5 x + 4}$ .

Tap for more steps...

Step 13.1

Set the denominator in $\frac{x^{2} - 1}{x^{2} + 5 x + 4}$ equal to $0$ to find where the expression is undefined.

$x^{2} + 5 x + 4 = 0$

Step 13.2

Solve for $x$ .

Tap for more steps...

Step 13.2.1

Factor $x^{2} + 5 x + 4$ using the AC method.

Tap for more steps...

Step 13.2.1.1

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 $5$ .

$1, 4$

Step 13.2.1.2

Write the factored form using these integers.

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

Step 13.2.2

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

$x + 1 = 0$

$x + 4 = 0$

Step 13.2.3

Set $x + 1$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 13.2.3.1

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 13.2.3.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 13.2.4

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

Tap for more steps...

Step 13.2.4.1

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

$x + 4 = 0$

Step 13.2.4.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 13.2.5

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

$x = - 1, - 4$

Step 13.3

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

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

Step 14

Use each root to create test intervals.

$x < - 4$

$- 4 < x < - 1$

$- 1 < x < 1$

$x > 1$

Step 15

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Tap for more steps...

Step 15.1

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

Tap for more steps...

Step 15.1.1

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

$x = - 6$

Step 15.1.2

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

$\frac{{(- 6)}^{2} - 1}{{(- 6)}^{2} + 5 (- 6) + 4} \leq 0$

Step 15.1.3

The left side $3.5$ is greater than the right side $0$ , which means that the given statement is false.

False

Step 15.2

Test a value on the interval $- 4 < x < - 1$ to see if it makes the inequality true.

Tap for more steps...

Step 15.2.1

Choose a value on the interval $- 4 < x < - 1$ and see if this value makes the original inequality true.

$x = - 2$

Step 15.2.2

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

$\frac{{(- 2)}^{2} - 1}{{(- 2)}^{2} + 5 (- 2) + 4} \leq 0$

Step 15.2.3

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

True

Step 15.3

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

Tap for more steps...

Step 15.3.1

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

$x = 0$

Step 15.3.2

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

$\frac{{(0)}^{2} - 1}{{(0)}^{2} + 5 (0) + 4} \leq 0$

Step 15.3.3

The left side $- 0.25$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 15.4

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

Tap for more steps...

Step 15.4.1

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

$x = 4$

Step 15.4.2

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

$\frac{{(4)}^{2} - 1}{{(4)}^{2} + 5 (4) + 4} \leq 0$

Step 15.4.3

The left side $0.375$ is greater than the right side $0$ , which means that the given statement is false.

False

Step 15.5

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

$x < - 4$ False

$- 4 < x < - 1$ True

$- 1 < x < 1$ True

$x > 1$ False

$x < - 4$ False

$- 4 < x < - 1$ True

$- 1 < x < 1$ True

$x > 1$ False

Step 16

The solution consists of all of the true intervals.

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

Step 17

The result can be shown in multiple forms.

Inequality Form:

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

Interval Notation:

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

Step 18