Solve the Inequality for x 2x^2-6x-133>7

$2 x^{2} - 6 x - 133 > 7$

Convert the inequality to an equation.

$2 x^{2} - 6 x - 133 = 7$

Subtract $7$ from both sides of the equation.

$2 x^{2} - 6 x - 133 - 7 = 0$

Subtract $7$ from $- 133$ .

$2 x^{2} - 6 x - 140 = 0$

Factor the left side of the equation.

Factor $2$ out of $2 x^{2} - 6 x - 140$ .

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

$2 (x^{2}) - 6 x - 140 = 0$

Factor $2$ out of $- 6 x$ .

$2 (x^{2}) + 2 (- 3 x) - 140 = 0$

Factor $2$ out of $- 140$ .

$2 x^{2} + 2 (- 3 x) + 2 \cdot - 70 = 0$

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

$2 (x^{2} - 3 x) + 2 \cdot - 70 = 0$

Factor $2$ out of $2 (x^{2} - 3 x) + 2 \cdot - 70$ .

$2 (x^{2} - 3 x - 70) = 0$

Factor.

Factor $x^{2} - 3 x - 70$ 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 $- 70$ and whose sum is $- 3$ .

$- 10, 7$

Write the factored form using these integers.

$2 ((x - 10) (x + 7)) = 0$

Remove unnecessary parentheses.

$2 (x - 10) (x + 7) = 0$

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

$x - 10 = 0$

$x + 7 = 0$

Set $x - 10$ equal to $0$ and solve for $x$ .

Set $x - 10$ equal to $0$ .

$x - 10 = 0$

Add $10$ to both sides of the equation.

$x = 10$

Set $x + 7$ equal to $0$ and solve for $x$ .

Set $x + 7$ equal to $0$ .

$x + 7 = 0$

Subtract $7$ from both sides of the equation.

$x = - 7$

The final solution is all the values that make $2 (x - 10) (x + 7) = 0$ true.

$x = 10, - 7$

Use each root to create test intervals.

$x < - 7$

$- 7 < x < 10$

$x > 10$

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 $x < - 7$ to see if it makes the inequality true.

Choose a value on the interval $x < - 7$ and see if this value makes the original inequality true.

$x = - 10$

Replace $x$ with $- 10$ in the original inequality.

$2 {(- 10)}^{2} - 6 \cdot - 10 - 133 > 7$

The left side $127$ is greater than the right side $7$ , which means that the given statement is always true.

True

Test a value on the interval $- 7 < x < 10$ to see if it makes the inequality true.

Choose a value on the interval $- 7 < x < 10$ and see if this value makes the original inequality true.

$x = 0$

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

$2 {(0)}^{2} - 6 \cdot 0 - 133 > 7$

The left side $- 133$ is not greater than the right side $7$ , which means that the given statement is false.

False

Test a value on the interval $x > 10$ to see if it makes the inequality true.

Choose a value on the interval $x > 10$ and see if this value makes the original inequality true.

$x = 12$

Replace $x$ with $12$ in the original inequality.

$2 {(12)}^{2} - 6 \cdot 12 - 133 > 7$

The left side $83$ is greater than the right side $7$ , which means that the given statement is always true.

True

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

$x < - 7$ True

$- 7 < x < 10$ False

$x > 10$ True

$x < - 7$ True

$- 7 < x < 10$ False

$x > 10$ True

The solution consists of all of the true intervals.

$x < - 7$ or $x > 10$

The result can be shown in multiple forms.

Inequality Form:

$x < - 7 or x > 10$

Interval Notation:

$(- \infty, - 7) \cup (10, \infty)$

Solve the Inequality for x 2x^2-6x-133>7

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