Precalculus Examples

$x^{2} \geq 3 x + 18$

Step 1

Subtract $3 x$ from both sides of the inequality.

$x^{2} - 3 x \geq 18$

Step 2

Convert the inequality to an equation.

$x^{2} - 3 x = 18$

Step 3

Subtract $18$ from both sides of the equation.

$x^{2} - 3 x - 18 = 0$

Step 4

Factor $x^{2} - 3 x - 18$ using the AC method.

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Step 4.1

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 $- 18$ and whose sum is $- 3$ .

$- 6, 3$

Step 4.2

Write the factored form using these integers.

$(x - 6) (x + 3) = 0$

Step 5

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

$x - 6 = 0$

$x + 3 = 0$

Step 6

Set $x - 6$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Step 6.2

Add $6$ to both sides of the equation.

$x = 6$

Step 7

Set $x + 3$ equal to $0$ and solve for $x$ .

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Step 7.1

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Step 7.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 8

The final solution is all the values that make $(x - 6) (x + 3) = 0$ true.

$x = 6, - 3$

Step 9

Use each root to create test intervals.

$x < - 3$

$- 3 < x < 6$

$x > 6$

Step 10

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 10.1

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

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Step 10.1.1

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

$x = - 6$

Step 10.1.2

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

${(- 6)}^{2} \geq 3 (- 6) + 18$

Step 10.1.3

The left side $36$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 10.2

Test a value on the interval $- 3 < x < 6$ to see if it makes the inequality true.

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Step 10.2.1

Choose a value on the interval $- 3 < x < 6$ and see if this value makes the original inequality true.

$x = 0$

Step 10.2.2

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

${(0)}^{2} \geq 3 (0) + 18$

Step 10.2.3

The left side $0$ is less than the right side $18$ , which means that the given statement is false.

False

Step 10.3

Test a value on the interval $x > 6$ to see if it makes the inequality true.

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Step 10.3.1

Choose a value on the interval $x > 6$ and see if this value makes the original inequality true.

$x = 8$

Step 10.3.2

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

${(8)}^{2} \geq 3 (8) + 18$

Step 10.3.3

The left side $64$ is greater than the right side $42$ , which means that the given statement is always true.

True

Step 10.4

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

$x < - 3$ True

$- 3 < x < 6$ False

$x > 6$ True

$x < - 3$ True

$- 3 < x < 6$ False

$x > 6$ True

Step 11

The solution consists of all of the true intervals.

$x \leq - 3$ or $x \geq 6$

Step 12

Convert the inequality to interval notation.

$(- \infty, - 3] \cup [6, \infty)$

Step 13