Algebra Examples

$2 x^{2} + 6 > 13 x$

Step 1

Subtract $13 x$ from both sides of the inequality.

$2 x^{2} + 6 - 13 x > 0$

Step 2

Convert the inequality to an equation.

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

Step 3

Factor by grouping.

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Step 3.1

Reorder terms.

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

Step 3.2

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 $a \cdot c = 2 \cdot 6 = 12$ and whose sum is $b = - 13$ .

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Step 3.2.1

Factor $- 13$ out of $- 13 x$ .

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

Step 3.2.2

Rewrite $- 13$ as $- 1$ plus $- 12$

$2 x^{2} + (- 1 - 12) x + 6 = 0$

Step 3.2.3

Apply the distributive property.

$2 x^{2} - 1 x - 12 x + 6 = 0$

Step 3.3

Factor out the greatest common factor from each group.

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Step 3.3.1

Group the first two terms and the last two terms.

$(2 x^{2} - 1 x) - 12 x + 6 = 0$

Step 3.3.2

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

$x (2 x - 1) - 6 (2 x - 1) = 0$

Step 3.4

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

$(2 x - 1) (x - 6) = 0$

Step 4

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

$2 x - 1 = 0$

$x - 6 = 0$

Step 5

Set $2 x - 1$ equal to $0$ and solve for $x$ .

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Step 5.1

Set $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Step 5.2

Solve $2 x - 1 = 0$ for $x$ .

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Step 5.2.1

Add $1$ to both sides of the equation.

$2 x = 1$

Step 5.2.2

Divide each term in $2 x = 1$ by $2$ and simplify.

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Step 5.2.2.1

Divide each term in $2 x = 1$ by $2$ .

$\frac{2 x}{2} = \frac{1}{2}$

Step 5.2.2.2

Simplify the left side.

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Step 5.2.2.2.1

Cancel the common factor of $2$ .

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Step 5.2.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{1}{2}$

Step 5.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{1}{2}$

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

The final solution is all the values that make $(2 x - 1) (x - 6) = 0$ true.

$x = \frac{1}{2}, 6$

Step 8

Use each root to create test intervals.

$x < \frac{1}{2}$

$\frac{1}{2} < x < 6$

$x > 6$

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 $x < \frac{1}{2}$ to see if it makes the inequality true.

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Step 9.1.1

Choose a value on the interval $x < \frac{1}{2}$ and see if this value makes the original inequality true.

$x = 0$

Step 9.1.2

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

$2 {(0)}^{2} + 6 > 13 (0)$

Step 9.1.3

The left side $6$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 9.2

Test a value on the interval $\frac{1}{2} < x < 6$ to see if it makes the inequality true.

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Step 9.2.1

Choose a value on the interval $\frac{1}{2} < x < 6$ and see if this value makes the original inequality true.

$x = 3$

Step 9.2.2

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

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

Step 9.2.3

The left side $24$ is not greater than the right side $39$ , which means that the given statement is false.

False

Step 9.3

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

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Step 9.3.1

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

$x = 8$

Step 9.3.2

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

$2 {(8)}^{2} + 6 > 13 (8)$

Step 9.3.3

The left side $134$ is greater than the right side $104$ , which means that the given statement is always true.

True

Step 9.4

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

$x < \frac{1}{2}$ True

$\frac{1}{2} < x < 6$ False

$x > 6$ True

$x < \frac{1}{2}$ True

$\frac{1}{2} < x < 6$ False

$x > 6$ True

Step 10

The solution consists of all of the true intervals.

$x < \frac{1}{2}$ or $x > 6$

Step 11

The result can be shown in multiple forms.

Inequality Form:

$x < \frac{1}{2} or x > 6$

Interval Notation:

$(- \infty, \frac{1}{2}) \cup (6, \infty)$

Step 12