Precalculus Examples

$x^{2} < 10 - 3 x$

Step 1

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

$x^{2} + 3 x < 10$

Step 2

Convert the inequality to an equation.

$x^{2} + 3 x = 10$

Step 3

Subtract $10$ from both sides of the equation.

$x^{2} + 3 x - 10 = 0$

Step 4

Factor $x^{2} + 3 x - 10$ 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 $- 10$ and whose sum is $3$ .

$- 2, 5$

Step 4.2

Write the factored form using these integers.

$(x - 2) (x + 5) = 0$

Step 5

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

$x - 2 = 0$

$x + 5 = 0$

Step 6

Set $x - 2$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Step 6.2

Add $2$ to both sides of the equation.

$x = 2$

Step 7

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

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Step 7.1

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

$x + 5 = 0$

Step 7.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 8

The final solution is all the values that make $(x - 2) (x + 5) = 0$ true.

$x = 2, - 5$

Step 9

Use each root to create test intervals.

$x < - 5$

$- 5 < x < 2$

$x > 2$

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 $x < - 5$ to see if it makes the inequality true.

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Step 10.1.1

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

$x = - 8$

Step 10.1.2

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

${(- 8)}^{2} < 10 - 3 \cdot - 8$

Step 10.1.3

The left side $64$ is not less than the right side $34$ , which means that the given statement is false.

False

Step 10.2

Test a value on the interval $- 5 < x < 2$ to see if it makes the inequality true.

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Step 10.2.1

Choose a value on the interval $- 5 < x < 2$ and see if this value makes the original inequality true.

$x = 0$

Step 10.2.2

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

${(0)}^{2} < 10 - 3 \cdot 0$

Step 10.2.3

The left side $0$ is less than the right side $10$ , which means that the given statement is always true.

True

Step 10.3

Test a value on the interval $x > 2$ to see if it makes the inequality true.

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Step 10.3.1

Choose a value on the interval $x > 2$ and see if this value makes the original inequality true.

$x = 4$

Step 10.3.2

Replace $x$ with $4$ in the original inequality.

${(4)}^{2} < 10 - 3 \cdot 4$

Step 10.3.3

The left side $16$ is not less than the right side $- 2$ , which means that the given statement is false.

False

Step 10.4

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

$x < - 5$ False

$- 5 < x < 2$ True

$x > 2$ False

$x < - 5$ False

$- 5 < x < 2$ True

$x > 2$ False

Step 11

The solution consists of all of the true intervals.

$- 5 < x < 2$

Step 12

Convert the inequality to interval notation.

$(- 5, 2)$

Step 13