Precalculus Examples

$\frac{x^{2} + 9 x + 20}{x^{2} - x - 20} \geq 0$

Step 1

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

$x^{2} + 9 x + 20 = 0$

$x^{2} - x - 20 = 0$

Step 2

Factor $x^{2} + 9 x + 20$ using the AC method.

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Step 2.1

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 $20$ and whose sum is $9$ .

$4, 5$

Step 2.2

Write the factored form using these integers.

$(x + 4) (x + 5) = 0$

Step 3

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

$x + 4 = 0$

$x + 5 = 0$

Step 4

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

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Step 4.1

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

$x + 4 = 0$

Step 4.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 5

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

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Step 5.1

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

$x + 5 = 0$

Step 5.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 6

The final solution is all the values that make $(x + 4) (x + 5) = 0$ true.

$x = - 4, - 5$

Step 7

Factor $x^{2} - x - 20$ using the AC method.

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Step 7.1

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 $- 20$ and whose sum is $- 1$ .

$- 5, 4$

Step 7.2

Write the factored form using these integers.

$(x - 5) (x + 4) = 0$

Step 8

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

$x - 5 = 0$

$x + 4 = 0$

Step 9

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

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Step 9.1

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

$x - 5 = 0$

Step 9.2

Add $5$ to both sides of the equation.

$x = 5$

Step 10

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

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Step 10.1

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

$x + 4 = 0$

Step 10.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 11

The final solution is all the values that make $(x - 5) (x + 4) = 0$ true.

$x = 5, - 4$

Step 12

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

$x = - 4, - 5$

$x = 5, - 4$

Step 13

Consolidate the solutions.

$x = - 4, - 5, 5, - 4$

Step 14

Find the domain of $\frac{x^{2} + 9 x + 20}{x^{2} - x - 20}$ .

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Step 14.1

Set the denominator in $\frac{x^{2} + 9 x + 20}{x^{2} - x - 20}$ equal to $0$ to find where the expression is undefined.

$x^{2} - x - 20 = 0$

Step 14.2

Solve for $x$ .

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Step 14.2.1

Factor $x^{2} - x - 20$ using the AC method.

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Step 14.2.1.1

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 $- 20$ and whose sum is $- 1$ .

$- 5, 4$

Step 14.2.1.2

Write the factored form using these integers.

$(x - 5) (x + 4) = 0$

Step 14.2.2

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

$x - 5 = 0$

$x + 4 = 0$

Step 14.2.3

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

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Step 14.2.3.1

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

$x - 5 = 0$

Step 14.2.3.2

Add $5$ to both sides of the equation.

$x = 5$

Step 14.2.4

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

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Step 14.2.4.1

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

$x + 4 = 0$

Step 14.2.4.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 14.2.5

The final solution is all the values that make $(x - 5) (x + 4) = 0$ true.

$x = 5, - 4$

Step 14.3

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

$(- \infty, - 4) \cup (- 4, 5) \cup (5, \infty)$

Step 15

Use each root to create test intervals.

$x < - 5$

$- 5 < x < - 4$

$- 4 < x < 5$

$x > 5$

Step 16

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

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Step 16.1

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

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Step 16.1.1

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

$x = - 8$

Step 16.1.2

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

$\frac{{(- 8)}^{2} + 9 (- 8) + 20}{{(- 8)}^{2} - (- 8) - 20} \geq 0$

Step 16.1.3

The left side $0.\overline{230769}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 16.2

Test a value on the interval $- 5 < x < - 4$ to see if it makes the inequality true.

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Step 16.2.1

Choose a value on the interval $- 5 < x < - 4$ and see if this value makes the original inequality true.

$x = - 4.5$

Step 16.2.2

Replace $x$ with $- 4.5$ in the original inequality.

$\frac{{(- 4.5)}^{2} + 9 (- 4.5) + 20}{{(- 4.5)}^{2} - (- 4.5) - 20} \geq 0$

Step 16.2.3

The left side $- 0.05263157$ is less than the right side $0$ , which means that the given statement is false.

False

Step 16.3

Test a value on the interval $- 4 < x < 5$ to see if it makes the inequality true.

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Step 16.3.1

Choose a value on the interval $- 4 < x < 5$ and see if this value makes the original inequality true.

$x = 0$

Step 16.3.2

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

$\frac{{(0)}^{2} + 9 (0) + 20}{{(0)}^{2} - (0) - 20} \geq 0$

Step 16.3.3

The left side $- 1$ is less than the right side $0$ , which means that the given statement is false.

False

Step 16.4

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

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Step 16.4.1

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

$x = 8$

Step 16.4.2

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

$\frac{{(8)}^{2} + 9 (8) + 20}{{(8)}^{2} - (8) - 20} \geq 0$

Step 16.4.3

The left side $4.\overline{3}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 16.5

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

$x < - 5$ True

$- 5 < x < - 4$ False

$- 4 < x < 5$ False

$x > 5$ True

$x < - 5$ True

$- 5 < x < - 4$ False

$- 4 < x < 5$ False

$x > 5$ True

Step 17

The solution consists of all of the true intervals.

$x \leq - 5$ or $x > 5$

Step 18

Convert the inequality to interval notation.

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

Step 19