Finite Math Examples

$x^{2} + 3 x + 8 > 3 - 3 x$

Step 1

Move all terms containing $x$ to the left side of the inequality.

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Step 1.1

Add $3 x$ to both sides of the inequality.

$x^{2} + 3 x + 8 + 3 x > 3$

Step 1.2

Add $3 x$ and $3 x$ .

$x^{2} + 6 x + 8 > 3$

Step 2

Convert the inequality to an equation.

$x^{2} + 6 x + 8 = 3$

Step 3

Subtract $3$ from both sides of the equation.

$x^{2} + 6 x + 8 - 3 = 0$

Step 4

Subtract $3$ from $8$ .

$x^{2} + 6 x + 5 = 0$

Step 5

Factor $x^{2} + 6 x + 5$ using the AC method.

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Step 5.1

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 $5$ and whose sum is $6$ .

$1, 5$

Step 5.2

Write the factored form using these integers.

$(x + 1) (x + 5) = 0$

Step 6

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

$x + 1 = 0$

$x + 5 = 0$

Step 7

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

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Step 7.1

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

$x + 1 = 0$

Step 7.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 8

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

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Step 8.1

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

$x + 5 = 0$

Step 8.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 9

The final solution is all the values that make $(x + 1) (x + 5) = 0$ true.

$x = - 1, - 5$

Step 10

Use each root to create test intervals.

$x < - 5$

$- 5 < x < - 1$

$x > - 1$

Step 11

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 11.1

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

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Step 11.1.1

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

$x = - 8$

Step 11.1.2

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

${(- 8)}^{2} + 3 (- 8) + 8 > 3 - 3 \cdot - 8$

Step 11.1.3

The left side $48$ is greater than the right side $27$ , which means that the given statement is always true.

True

Step 11.2

Test a value on the interval $- 5 < x < - 1$ to see if it makes the inequality true.

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Step 11.2.1

Choose a value on the interval $- 5 < x < - 1$ and see if this value makes the original inequality true.

$x = - 3$

Step 11.2.2

Replace $x$ with $- 3$ in the original inequality.

${(- 3)}^{2} + 3 (- 3) + 8 > 3 - 3 \cdot - 3$

Step 11.2.3

The left side $8$ is not greater than the right side $12$ , which means that the given statement is false.

False

Step 11.3

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

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Step 11.3.1

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

$x = 0$

Step 11.3.2

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

${(0)}^{2} + 3 (0) + 8 > 3 - 3 \cdot 0$

Step 11.3.3

The left side $8$ is greater than the right side $3$ , which means that the given statement is always true.

True

Step 11.4

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

$x < - 5$ True

$- 5 < x < - 1$ False

$x > - 1$ True

$x < - 5$ True

$- 5 < x < - 1$ False

$x > - 1$ True

Step 12

The solution consists of all of the true intervals.

$x < - 5$ or $x > - 1$

Step 13

Convert the inequality to interval notation.

$(- \infty, - 5) \cup (- 1, \infty)$

Step 14