Precalculus Examples

x^{2} - 4 x - 12 < 0

$x^{2} - 4 x - 12 < 0$

Step 1

Convert the inequality to an equation.

x^{2} - 4 x - 12 = 0

$x^{2} - 4 x - 12 = 0$

Step 2

Factor

x^{2} - 4 x - 12

$x^{2} - 4 x - 12$ using the AC method.

Tap for more steps...

Step 2.1

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

- 12

$- 12$ and whose sum is

- 4

$- 4$ .

- 6, 2

$- 6, 2$

Step 2.2

Write the factored form using these integers.

(x - 6) (x + 2) = 0

$(x - 6) (x + 2) = 0$

(x - 6) (x + 2) = 0

$(x - 6) (x + 2) = 0$

Step 3

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 - 6 = 0

$x - 6 = 0$

x + 2 = 0

$x + 2 = 0$

Step 4

Set

x - 6

$x - 6$ equal to

0

$0$ and solve for

x

$x$ .

Tap for more steps...

Step 4.1

Set

x - 6

$x - 6$ equal to

0

$0$ .

x - 6 = 0

$x - 6 = 0$

Step 4.2

Add

6

$6$ to both sides of the equation.

x = 6

$x = 6$

x = 6

$x = 6$

Step 5

Set

x + 2

$x + 2$ equal to

0

$0$ and solve for

x

$x$ .

Tap for more steps...

Step 5.1

Set

x + 2

$x + 2$ equal to

0

$0$ .

x + 2 = 0

$x + 2 = 0$

Step 5.2

Subtract

2

$2$ from both sides of the equation.

x = - 2

$x = - 2$

x = - 2

$x = - 2$

Step 6

The final solution is all the values that make

(x - 6) (x + 2) = 0

$(x - 6) (x + 2) = 0$ true.

x = 6, - 2

$x = 6, - 2$

Step 7

Use each root to create test intervals.

x < - 2

$x < - 2$

- 2 < x < 6

$- 2 < x < 6$

x > 6

$x > 6$

Step 8

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 8.1

Test a value on the interval

x < - 2

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

Tap for more steps...

Step 8.1.1

Choose a value on the interval

x < - 2

$x < - 2$ and see if this value makes the original inequality true.

x = - 4

$x = - 4$

Step 8.1.2

Replace

x

$x$ with

- 4

$- 4$ in the original inequality.

{(- 4)}^{2} - 4 \cdot - 4 - 12 < 0

${(- 4)}^{2} - 4 \cdot - 4 - 12 < 0$

Step 8.1.3

The left side

20

$20$ is not less than the right side

0

$0$ , which means that the given statement is false.

False

Step 8.2

Test a value on the interval

- 2 < x < 6

$- 2 < x < 6$ to see if it makes the inequality true.

Tap for more steps...

Step 8.2.1

Choose a value on the interval

- 2 < x < 6

$- 2 < x < 6$ and see if this value makes the original inequality true.

x = 0

$x = 0$

Step 8.2.2

Replace

x

$x$ with

0

$0$ in the original inequality.

{(0)}^{2} - 4 \cdot 0 - 12 < 0

${(0)}^{2} - 4 \cdot 0 - 12 < 0$

Step 8.2.3

The left side

- 12

$- 12$ is less than the right side

0

$0$ , which means that the given statement is always true.

True

Step 8.3

Test a value on the interval

x > 6

$x > 6$ to see if it makes the inequality true.

Tap for more steps...

Step 8.3.1

Choose a value on the interval

x > 6

$x > 6$ and see if this value makes the original inequality true.

x = 8

$x = 8$

Step 8.3.2

Replace

x

$x$ with

8

$8$ in the original inequality.

{(8)}^{2} - 4 \cdot 8 - 12 < 0

${(8)}^{2} - 4 \cdot 8 - 12 < 0$

Step 8.3.3

The left side

20

$20$ is not less than the right side

0

$0$ , which means that the given statement is false.

False

Step 8.4

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

x < - 2

$x < - 2$ False

- 2 < x < 6

$- 2 < x < 6$ True

x > 6

$x > 6$ False

x < - 2

$x < - 2$ False

- 2 < x < 6

$- 2 < x < 6$ True

x > 6

$x > 6$ False

Step 9

The solution consists of all of the true intervals.

- 2 < x < 6

$- 2 < x < 6$

Step 10

The result can be shown in multiple forms.

Inequality Form:

- 2 < x < 6

$- 2 < x < 6$

Interval Notation:

(- 2, 6)

$(- 2, 6)$

Step 11

Enter YOUR Problem