Precalculus Examples

x^{2} + 8 x < 33

$x^{2} + 8 x < 33$

Step 1

Convert the inequality to an equation.

x^{2} + 8 x = 33

$x^{2} + 8 x = 33$

Step 2

Subtract

33

$33$ from both sides of the equation.

x^{2} + 8 x - 33 = 0

$x^{2} + 8 x - 33 = 0$

Step 3

Factor

x^{2} + 8 x - 33

$x^{2} + 8 x - 33$ using the AC method.

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Step 3.1

Consider the form

x^{2} + b x + c

$x^{2} + b x + c$ . Find a pair of integers whose product is

c

$c$ and whose sum is

b

$b$ . In this case, whose product is

- 33

$- 33$ and whose sum is

8

$8$ .

- 3, 11

$- 3, 11$

Step 3.2

Write the factored form using these integers.

(x - 3) (x + 11) = 0

$(x - 3) (x + 11) = 0$

(x - 3) (x + 11) = 0

$(x - 3) (x + 11) = 0$

Step 4

If any individual factor on the left side of the equation is equal to

0

$0$ , the entire expression will be equal to

0

$0$ .

x - 3 = 0

$x - 3 = 0$

x + 11 = 0

$x + 11 = 0$

Step 5

Set

x - 3

$x - 3$ equal to

0

$0$ and solve for

x

$x$ .

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Step 5.1

Set

x - 3

$x - 3$ equal to

0

$0$ .

x - 3 = 0

$x - 3 = 0$

Step 5.2

Add

3

$3$ to both sides of the equation.

x = 3

$x = 3$

x = 3

$x = 3$

Step 6

Set

x + 11

$x + 11$ equal to

0

$0$ and solve for

x

$x$ .

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Step 6.1

Set

x + 11

$x + 11$ equal to

0

$0$ .

x + 11 = 0

$x + 11 = 0$

Step 6.2

Subtract

11

$11$ from both sides of the equation.

x = - 11

$x = - 11$

x = - 11

$x = - 11$

Step 7

The final solution is all the values that make

(x - 3) (x + 11) = 0

$(x - 3) (x + 11) = 0$ true.

x = 3, - 11

$x = 3, - 11$

Step 8

Use each root to create test intervals.

x < - 11

$x < - 11$

- 11 < x < 3

$- 11 < x < 3$

x > 3

$x > 3$

Step 9

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 9.1

Test a value on the interval

x < - 11

$x < - 11$ to see if it makes the inequality true.

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Step 9.1.1

Choose a value on the interval

x < - 11

$x < - 11$ and see if this value makes the original inequality true.

x = - 14

$x = - 14$

Step 9.1.2

Replace

x

$x$ with

- 14

$- 14$ in the original inequality.

{(- 14)}^{2} + 8 (- 14) < 33

${(- 14)}^{2} + 8 (- 14) < 33$

Step 9.1.3

The left side

84

$84$ is not less than the right side

33

$33$ , which means that the given statement is false.

False

Step 9.2

Test a value on the interval

- 11 < x < 3

$- 11 < x < 3$ to see if it makes the inequality true.

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Step 9.2.1

Choose a value on the interval

- 11 < x < 3

$- 11 < x < 3$ and see if this value makes the original inequality true.

x = 0

$x = 0$

Step 9.2.2

Replace

x

$x$ with

0

$0$ in the original inequality.

{(0)}^{2} + 8 (0) < 33

${(0)}^{2} + 8 (0) < 33$

Step 9.2.3

The left side

0

$0$ is less than the right side

33

$33$ , which means that the given statement is always true.

True

Step 9.3

Test a value on the interval

x > 3

$x > 3$ to see if it makes the inequality true.

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Step 9.3.1

Choose a value on the interval

x > 3

$x > 3$ and see if this value makes the original inequality true.

x = 6

$x = 6$

Step 9.3.2

Replace

x

$x$ with

6

$6$ in the original inequality.

{(6)}^{2} + 8 (6) < 33

${(6)}^{2} + 8 (6) < 33$

Step 9.3.3

The left side

84

$84$ is not less than the right side

33

$33$ , which means that the given statement is false.

False

Step 9.4

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

x < - 11

$x < - 11$ False

- 11 < x < 3

$- 11 < x < 3$ True

x > 3

$x > 3$ False

x < - 11

$x < - 11$ False

- 11 < x < 3

$- 11 < x < 3$ True

x > 3

$x > 3$ False

Step 10

The solution consists of all of the true intervals.

- 11 < x < 3

$- 11 < x < 3$

Step 11

The result can be shown in multiple forms.

Inequality Form:

- 11 < x < 3

$- 11 < x < 3$

Interval Notation:

(- 11, 3)

$(- 11, 3)$

Step 12

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