Solve the Inequality for x 2x^2-2>=2x+10

$2 x^{2} - 2 \geq 2 x + 10$

Subtract $2 x$ from both sides of the inequality.

$2 x^{2} - 2 - 2 x \geq 10$

Convert the inequality to an equation.

$2 x^{2} - 2 - 2 x = 10$

Subtract $10$ from both sides of the equation.

$2 x^{2} - 2 - 2 x - 10 = 0$

Subtract $10$ from $- 2$ .

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

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.

$x = 3, - 2$

Use each root to create test intervals.

$x < - 2$

$- 2 < x < 3$

$x > 3$

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.

$2 {(- 4)}^{2} - 2 \geq 2 (- 4) + 10$

The left side $30$ is greater than the right side $2$ , which means that the given statement is always true.

True

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

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

$x = 0$

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

$2 {(0)}^{2} - 2 \geq 2 (0) + 10$

The left side $- 2$ is less than the right side $10$ , 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.

$2 {(6)}^{2} - 2 \geq 2 (6) + 10$

The left side $70$ is greater than the right side $22$ , which means that the given statement is always true.

True

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

$x < - 2$ True

$- 2 < x < 3$ False

$x > 3$ True

$x < - 2$ True

$- 2 < x < 3$ False

$x > 3$ True

The solution consists of all of the true intervals.

$x \leq - 2$ or $x \geq 3$

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 2 or x \geq 3$

Interval Notation:

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

Solve the Inequality for x 2x^2-2>=2x+10

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