Solve the Inequality for x 3x^2-5x-12>2x^2-6x-6

$3 x^{2} - 5 x - 12 > 2 x^{2} - 6 x - 6$

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

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

$3 x^{2} - 5 x - 12 - 2 x^{2} > - 6 x - 6$

Add $6 x$ to both sides of the inequality.

$3 x^{2} - 5 x - 12 - 2 x^{2} + 6 x > - 6$

Subtract $2 x^{2}$ from $3 x^{2}$ .

$x^{2} - 5 x - 12 + 6 x > - 6$

Add $- 5 x$ and $6 x$ .

$x^{2} + x - 12 > - 6$

Convert the inequality to an equation.

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

Add $6$ to both sides of the equation.

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

Add $- 12$ and $6$ .

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

Factor $x^{2} + x - 6$ using the AC method.

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 $- 6$ and whose sum is $1$ .

$- 2, 3$

Write the factored form using these integers.

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

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

$x - 2 = 0$

$x + 3 = 0$

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

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

$x = 2, - 3$

Use each root to create test intervals.

$x < - 3$

$- 3 < x < 2$

$x > 2$

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

Test a value on the interval $x < - 3$ to see if it makes the inequality true.

Choose a value on the interval $x < - 3$ and see if this value makes the original inequality true.

$x = - 6$

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

$3 {(- 6)}^{2} - 5 \cdot - 6 - 12 > 2 {(- 6)}^{2} - 6 \cdot - 6 - 6$

The left side $126$ is greater than the right side $102$ , which means that the given statement is always true.

True

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

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

$x = 0$

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

$3 {(0)}^{2} - 5 \cdot 0 - 12 > 2 {(0)}^{2} - 6 \cdot 0 - 6$

The left side $- 12$ is not greater than the right side $- 6$ , which means that the given statement is false.

False

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

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

$x = 4$

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

$3 {(4)}^{2} - 5 \cdot 4 - 12 > 2 {(4)}^{2} - 6 \cdot 4 - 6$

The left side $16$ is greater than the right side $2$ , which means that the given statement is always true.

True

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

$x < - 3$ True

$- 3 < x < 2$ False

$x > 2$ True

$x < - 3$ True

$- 3 < x < 2$ False

$x > 2$ True

The solution consists of all of the true intervals.

$x < - 3$ or $x > 2$

The result can be shown in multiple forms.

Inequality Form:

$x < - 3 or x > 2$

Interval Notation:

$(- \infty, - 3) \cup (2, \infty)$

Solve the Inequality for x 3x^2-5x-12>2x^2-6x-6

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