Precalculus Examples

$x^{2} - 12 > - 2 x^{2} - 16 x$

Step 1

Rewrite so $x$ is on the left side of the inequality.

$- 2 x^{2} - 16 x < x^{2} - 12$

Step 2

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

Tap for more steps...

Step 2.1

Subtract $x^{2}$ from both sides of the inequality.

$- 2 x^{2} - 16 x - x^{2} < - 12$

Step 2.2

Subtract $x^{2}$ from $- 2 x^{2}$ .

$- 3 x^{2} - 16 x < - 12$

Step 3

Convert the inequality to an equation.

$- 3 x^{2} - 16 x = - 12$

Step 4

Add $12$ to both sides of the equation.

$- 3 x^{2} - 16 x + 12 = 0$

Step 5

Factor the left side of the equation.

Tap for more steps...

Step 5.1

Factor $- 1$ out of $- 3 x^{2} - 16 x + 12$ .

Tap for more steps...

Step 5.1.1

Factor $- 1$ out of $- 3 x^{2}$ .

$- (3 x^{2}) - 16 x + 12 = 0$

Step 5.1.2

Factor $- 1$ out of $- 16 x$ .

$- (3 x^{2}) - (16 x) + 12 = 0$

Step 5.1.3

Rewrite $12$ as $- 1 (- 12)$ .

$- (3 x^{2}) - (16 x) - 1 \cdot - 12 = 0$

Step 5.1.4

Factor $- 1$ out of $- (3 x^{2}) - (16 x)$ .

$- (3 x^{2} + 16 x) - 1 \cdot - 12 = 0$

Step 5.1.5

Factor $- 1$ out of $- (3 x^{2} + 16 x) - 1 (- 12)$ .

$- (3 x^{2} + 16 x - 12) = 0$

Step 5.2

Factor.

Tap for more steps...

Step 5.2.1

Factor by grouping.

Tap for more steps...

Step 5.2.1.1

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 $a \cdot c = 3 \cdot - 12 = - 36$ and whose sum is $b = 16$ .

Tap for more steps...

Step 5.2.1.1.1

Factor $16$ out of $16 x$ .

$- (3 x^{2} + 16 (x) - 12) = 0$

Step 5.2.1.1.2

Rewrite $16$ as $- 2$ plus $18$

$- (3 x^{2} + (- 2 + 18) x - 12) = 0$

Step 5.2.1.1.3

Apply the distributive property.

$- (3 x^{2} - 2 x + 18 x - 12) = 0$

Step 5.2.1.2

Factor out the greatest common factor from each group.

Tap for more steps...

Step 5.2.1.2.1

Group the first two terms and the last two terms.

$- ((3 x^{2} - 2 x) + 18 x - 12) = 0$

Step 5.2.1.2.2

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

$- (x (3 x - 2) + 6 (3 x - 2)) = 0$

Step 5.2.1.3

Factor the polynomial by factoring out the greatest common factor, $3 x - 2$ .

$- ((3 x - 2) (x + 6)) = 0$

Step 5.2.2

Remove unnecessary parentheses.

$- (3 x - 2) (x + 6) = 0$

Step 6

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

$3 x - 2 = 0$

$x + 6 = 0$

Step 7

Set $3 x - 2$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 7.1

Set $3 x - 2$ equal to $0$ .

$3 x - 2 = 0$

Step 7.2

Solve $3 x - 2 = 0$ for $x$ .

Tap for more steps...

Step 7.2.1

Add $2$ to both sides of the equation.

$3 x = 2$

Step 7.2.2

Divide each term in $3 x = 2$ by $3$ and simplify.

Tap for more steps...

Step 7.2.2.1

Divide each term in $3 x = 2$ by $3$ .

$\frac{3 x}{3} = \frac{2}{3}$

Step 7.2.2.2

Simplify the left side.

Tap for more steps...

Step 7.2.2.2.1

Cancel the common factor of $3$ .

Tap for more steps...

Step 7.2.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{2}{3}$

Step 7.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{2}{3}$

Step 8

Set $x + 6$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 8.1

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Step 8.2

Subtract $6$ from both sides of the equation.

$x = - 6$

Step 9

The final solution is all the values that make $- (3 x - 2) (x + 6) = 0$ true.

$x = \frac{2}{3}, - 6$

Step 10

Use each root to create test intervals.

$x < - 6$

$- 6 < x < \frac{2}{3}$

$x > \frac{2}{3}$

Step 11

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Tap for more steps...

Step 11.1

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

Tap for more steps...

Step 11.1.1

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

$x = - 8$

Step 11.1.2

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

${(- 8)}^{2} - 12 > - 2 {(- 8)}^{2} - 16 \cdot - 8$

Step 11.1.3

The left side $52$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 11.2

Test a value on the interval $- 6 < x < \frac{2}{3}$ to see if it makes the inequality true.

Tap for more steps...

Step 11.2.1

Choose a value on the interval $- 6 < x < \frac{2}{3}$ and see if this value makes the original inequality true.

$x = 0$

Step 11.2.2

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

${(0)}^{2} - 12 > - 2 {(0)}^{2} - 16 \cdot 0$

Step 11.2.3

The left side $- 12$ is not greater than the right side $0$ , which means that the given statement is false.

False

Step 11.3

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

Tap for more steps...

Step 11.3.1

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

$x = 3$

Step 11.3.2

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

${(3)}^{2} - 12 > - 2 {(3)}^{2} - 16 \cdot 3$

Step 11.3.3

The left side $- 3$ is greater than the right side $- 66$ , which means that the given statement is always true.

True

Step 11.4

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

$x < - 6$ True

$- 6 < x < \frac{2}{3}$ False

$x > \frac{2}{3}$ True

$x < - 6$ True

$- 6 < x < \frac{2}{3}$ False

$x > \frac{2}{3}$ True

Step 12

The solution consists of all of the true intervals.

$x < - 6$ or $x > \frac{2}{3}$

Step 13

Convert the inequality to interval notation.

$(- \infty, - 6) \cup (\frac{2}{3}, \infty)$

Step 14