Algebra Examples

$5 x^{2} + 13 x + 3 > 9$

Step 1

Convert the inequality to an equation.

$5 x^{2} + 13 x + 3 = 9$

Step 2

Subtract $9$ from both sides of the equation.

$5 x^{2} + 13 x + 3 - 9 = 0$

Step 3

Subtract $9$ from $3$ .

$5 x^{2} + 13 x - 6 = 0$

Step 4

Factor by grouping.

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Step 4.1

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 $a \cdot c = 5 \cdot - 6 = - 30$ and whose sum is $b = 13$ .

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Step 4.1.1

Factor $13$ out of $13 x$ .

$5 x^{2} + 13 (x) - 6 = 0$

Step 4.1.2

Rewrite $13$ as $- 2$ plus $15$

$5 x^{2} + (- 2 + 15) x - 6 = 0$

Step 4.1.3

Apply the distributive property.

$5 x^{2} - 2 x + 15 x - 6 = 0$

Step 4.2

Factor out the greatest common factor from each group.

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Step 4.2.1

Group the first two terms and the last two terms.

$(5 x^{2} - 2 x) + 15 x - 6 = 0$

Step 4.2.2

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

$x (5 x - 2) + 3 (5 x - 2) = 0$

Step 4.3

Factor the polynomial by factoring out the greatest common factor, $5 x - 2$ .

$(5 x - 2) (x + 3) = 0$

Step 5

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

$5 x - 2 = 0$

$x + 3 = 0$

Step 6

Set $5 x - 2$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $5 x - 2$ equal to $0$ .

$5 x - 2 = 0$

Step 6.2

Solve $5 x - 2 = 0$ for $x$ .

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Step 6.2.1

Add $2$ to both sides of the equation.

$5 x = 2$

Step 6.2.2

Divide each term in $5 x = 2$ by $5$ and simplify.

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Step 6.2.2.1

Divide each term in $5 x = 2$ by $5$ .

$\frac{5 x}{5} = \frac{2}{5}$

Step 6.2.2.2

Simplify the left side.

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Step 6.2.2.2.1

Cancel the common factor of $5$ .

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Step 6.2.2.2.1.1

Cancel the common factor.

$\frac{5 x}{5} = \frac{2}{5}$

Step 6.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{2}{5}$

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 $(5 x - 2) (x + 3) = 0$ true.

$x = \frac{2}{5}, - 3$

Step 9

Use each root to create test intervals.

$x < - 3$

$- 3 < x < \frac{2}{5}$

$x > \frac{2}{5}$

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.

$5 {(- 6)}^{2} + 13 (- 6) + 3 > 9$

Step 10.1.3

The left side $105$ is greater than the right side $9$ , which means that the given statement is always true.

True

Step 10.2

Test a value on the interval $- 3 < x < \frac{2}{5}$ to see if it makes the inequality true.

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Step 10.2.1

Choose a value on the interval $- 3 < x < \frac{2}{5}$ and see if this value makes the original inequality true.

$x = 0$

Step 10.2.2

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

$5 {(0)}^{2} + 13 (0) + 3 > 9$

Step 10.2.3

The left side $3$ is not greater than the right side $9$ , which means that the given statement is false.

False

Step 10.3

Test a value on the interval $x > \frac{2}{5}$ to see if it makes the inequality true.

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Step 10.3.1

Choose a value on the interval $x > \frac{2}{5}$ and see if this value makes the original inequality true.

$x = 3$

Step 10.3.2

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

$5 {(3)}^{2} + 13 (3) + 3 > 9$

Step 10.3.3

The left side $87$ is greater than the right side $9$ , 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

$- 3 < x < \frac{2}{5}$ False

$x > \frac{2}{5}$ True

$x < - 3$ True

$- 3 < x < \frac{2}{5}$ False

$x > \frac{2}{5}$ True

Step 11

The solution consists of all of the true intervals.

$x < - 3$ or $x > \frac{2}{5}$

Step 12

The result can be shown in multiple forms.

Inequality Form:

$x < - 3 or x > \frac{2}{5}$

Interval Notation:

$(- \infty, - 3) \cup (\frac{2}{5}, \infty)$

Step 13