Solve for x (x-1/2)(2x^2-2x-12)>=0

$(x - \frac{1}{2}) (2 x^{2} - 2 x - 12) \geq 0$

Step 1

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

$x - \frac{1}{2} = 0$

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

Step 2

Set $x - \frac{1}{2}$ equal to $0$ and solve for $x$ .

Set $x - \frac{1}{2}$ equal to $0$ .

$x - \frac{1}{2} = 0$

Add $\frac{1}{2}$ to both sides of the equation.

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

Step 3

Set $2 x^{2} - 2 x - 12$ equal to $0$ and solve for $x$ .

Set $2 x^{2} - 2 x - 12$ equal to $0$ .

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

Solve $2 x^{2} - 2 x - 12 = 0$ for $x$ .

Factor the left side of the equation.

Factor $2$ out of $2 x^{2} - 2 x - 12$ .

Factor $2$ out of $2 x^{2}$ .

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

Factor $2$ out of $- 2 x$ .

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

Factor $2$ out of $- 12$ .

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

Factor $2$ out of $2 (x^{2}) + 2 (- x)$ .

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

Factor $2$ out of $2 (x^{2} - x) + 2 (- 6)$ .

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

Factor.

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 $- 1$ .

$- 3, 2$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

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

$x - 3 = 0$

$x + 2 = 0$

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

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

$x = 3, - 2$

Step 4

The final solution is all the values that make $(x - \frac{1}{2}) (2 x^{2} - 2 x - 12) \geq 0$ true.

$x = \frac{1}{2}, 3, - 2$

Step 5

Use each root to create test intervals.

$x < - 2$

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

$\frac{1}{2} < x < 3$

$x > 3$

Step 6

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 $x < - 2$ 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.

$((- 4) - \frac{1}{2}) (2 {(- 4)}^{2} - 2 \cdot - 4 - 12) \geq 0$

The left side $- 126$ is less than the right side $0$ , which means that the given statement is false.

False

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

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

$x = 0$

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

$((0) - \frac{1}{2}) (2 {(0)}^{2} - 2 \cdot 0 - 12) \geq 0$

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

True

Test a value on the interval $\frac{1}{2} < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $\frac{1}{2} < x < 3$ and see if this value makes the original inequality true.

$x = 2$

Replace $x$ with $2$ in the original inequality.

$((2) - \frac{1}{2}) (2 {(2)}^{2} - 2 \cdot 2 - 12) \geq 0$

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

False

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.

$((6) - \frac{1}{2}) (2 {(6)}^{2} - 2 \cdot 6 - 12) \geq 0$

The left side $264$ is greater than the right side $0$ , which means that the given statement is always true.

True

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

$x < - 2$ False

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

$\frac{1}{2} < x < 3$ False

$x > 3$ True

$x < - 2$ False

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

$\frac{1}{2} < x < 3$ False

$x > 3$ True

Step 7

The solution consists of all of the true intervals.

$- 2 \leq x \leq \frac{1}{2}$ or $x \geq 3$

Step 8

The result can be shown in multiple forms.

Inequality Form:

$- 2 \leq x \leq \frac{1}{2} or x \geq 3$

Interval Notation:

$[- 2, \frac{1}{2}] \cup [3, \infty)$

Step 9

Solve for x (x-1/2)(2x^2-2x-12)>=0

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