Precalculus Examples

$5 x^{2} + 17 x < 12$

Step 1

Convert the inequality to an equation.

$5 x^{2} + 17 x = 12$

Step 2

Subtract $12$ from both sides of the equation.

$5 x^{2} + 17 x - 12 = 0$

Step 3

Factor by grouping.

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Step 3.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 5 \cdot - 12 = - 60$ and whose sum is $b = 17$ .

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Step 3.1.1

Factor $17$ out of $17 x$ .

$5 x^{2} + 17 (x) - 12 = 0$

Step 3.1.2

Rewrite $17$ as $- 3$ plus $20$

$5 x^{2} + (- 3 + 20) x - 12 = 0$

Step 3.1.3

Apply the distributive property.

$5 x^{2} - 3 x + 20 x - 12 = 0$

Step 3.2

Factor out the greatest common factor from each group.

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Step 3.2.1

Group the first two terms and the last two terms.

$(5 x^{2} - 3 x) + 20 x - 12 = 0$

Step 3.2.2

Factor out the greatest common factor (GCF) from each group.

$x (5 x - 3) + 4 (5 x - 3) = 0$

Step 3.3

Factor the polynomial by factoring out the greatest common factor, $5 x - 3$ .

$(5 x - 3) (x + 4) = 0$

Step 4

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

$5 x - 3 = 0$

$x + 4 = 0$

Step 5

Set $5 x - 3$ equal to $0$ and solve for $x$ .

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Step 5.1

Set $5 x - 3$ equal to $0$ .

$5 x - 3 = 0$

Step 5.2

Solve $5 x - 3 = 0$ for $x$ .

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Step 5.2.1

Add $3$ to both sides of the equation.

$5 x = 3$

Step 5.2.2

Divide each term in $5 x = 3$ by $5$ and simplify.

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Step 5.2.2.1

Divide each term in $5 x = 3$ by $5$ .

$\frac{5 x}{5} = \frac{3}{5}$

Step 5.2.2.2

Simplify the left side.

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Step 5.2.2.2.1

Cancel the common factor of $5$ .

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Step 5.2.2.2.1.1

Cancel the common factor.

$\frac{5 x}{5} = \frac{3}{5}$

Step 5.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{3}{5}$

Step 6

Set $x + 4$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Step 6.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 7

The final solution is all the values that make $(5 x - 3) (x + 4) = 0$ true.

$x = \frac{3}{5}, - 4$

Step 8

Use each root to create test intervals.

$x < - 4$

$- 4 < x < \frac{3}{5}$

$x > \frac{3}{5}$

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 $x < - 4$ to see if it makes the inequality true.

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Step 9.1.1

Choose a value on the interval $x < - 4$ and see if this value makes the original inequality true.

$x = - 6$

Step 9.1.2

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

$5 {(- 6)}^{2} + 17 (- 6) < 12$

Step 9.1.3

The left side $78$ is not less than the right side $12$ , which means that the given statement is false.

False

Step 9.2

Test a value on the interval $- 4 < x < \frac{3}{5}$ to see if it makes the inequality true.

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Step 9.2.1

Choose a value on the interval $- 4 < x < \frac{3}{5}$ and see if this value makes the original inequality true.

$x = 0$

Step 9.2.2

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

$5 {(0)}^{2} + 17 (0) < 12$

Step 9.2.3

The left side $0$ is less than the right side $12$ , which means that the given statement is always true.

True

Step 9.3

Test a value on the interval $x > \frac{3}{5}$ to see if it makes the inequality true.

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Step 9.3.1

Choose a value on the interval $x > \frac{3}{5}$ and see if this value makes the original inequality true.

$x = 3$

Step 9.3.2

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

$5 {(3)}^{2} + 17 (3) < 12$

Step 9.3.3

The left side $96$ is not less than the right side $12$ , which means that the given statement is false.

False

Step 9.4

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

$x < - 4$ False

$- 4 < x < \frac{3}{5}$ True

$x > \frac{3}{5}$ False

$x < - 4$ False

$- 4 < x < \frac{3}{5}$ True

$x > \frac{3}{5}$ False

Step 10

The solution consists of all of the true intervals.

$- 4 < x < \frac{3}{5}$

Step 11

Convert the inequality to interval notation.

$(- 4, \frac{3}{5})$

Step 12