Solve the Inequality for x -2x^2+39<=-16x-1

$- 2 x^{2} + 39 \leq - 16 x - 1$

Step 1

Add $16 x$ to both sides of the inequality.

$- 2 x^{2} + 39 + 16 x \leq - 1$

Step 2

Convert the inequality to an equation.

$- 2 x^{2} + 39 + 16 x = - 1$

Step 3

Add $1$ to both sides of the equation.

$- 2 x^{2} + 39 + 16 x + 1 = 0$

Step 4

Add $39$ and $1$ .

$- 2 x^{2} + 16 x + 40 = 0$

Step 5

Factor the left side of the equation.

Factor $- 2$ out of $- 2 x^{2} + 16 x + 40$ .

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

$- 2 x^{2} + 16 x + 40 = 0$

Factor $- 2$ out of $16 x$ .

$- 2 x^{2} - 2 (- 8 x) + 40 = 0$

Factor $- 2$ out of $40$ .

$- 2 x^{2} - 2 (- 8 x) - 2 \cdot - 20 = 0$

Factor $- 2$ out of $- 2 (x^{2}) - 2 (- 8 x)$ .

$- 2 (x^{2} - 8 x) - 2 \cdot - 20 = 0$

Factor $- 2$ out of $- 2 (x^{2} - 8 x) - 2 (- 20)$ .

$- 2 (x^{2} - 8 x - 20) = 0$

Factor.

Factor $x^{2} - 8 x - 20$ 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 $- 20$ and whose sum is $- 8$ .

$- 10, 2$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

Step 6

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

$x - 10 = 0$

$x + 2 = 0$

Step 7

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.

$x = 10$

Step 8

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.

$x = - 2$

Step 9

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

$x = 10, - 2$

Step 10

Use each root to create test intervals.

$x < - 2$

$- 2 < x < 10$

$x > 10$

Step 11

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.

$- 2 {(- 4)}^{2} + 39 \leq - 16 \cdot - 4 - 1$

The left side $7$ is less than the right side $63$ , which means that the given statement is always true.

True

Test a value on the interval $- 2 < x < 10$ to see if it makes the inequality true.

Choose a value on the interval $- 2 < x < 10$ and see if this value makes the original inequality true.

$x = 0$

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

$- 2 {(0)}^{2} + 39 \leq - 16 \cdot 0 - 1$

The left side $39$ is greater than the right side $- 1$ , 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.

$- 2 {(12)}^{2} + 39 \leq - 16 \cdot 12 - 1$

The left side $- 249$ is less than the right side $- 193$ , which means that the given statement is always true.

True

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

$x < - 2$ True

$- 2 < x < 10$ False

$x > 10$ True

$x < - 2$ True

$- 2 < x < 10$ False

$x > 10$ True

Step 12

The solution consists of all of the true intervals.

$x \leq - 2$ or $x \geq 10$

Step 13

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 2 or x \geq 10$

Interval Notation:

$(- \infty, - 2] \cup [10, \infty)$

Step 14

Solve the Inequality for x -2x^2+39<=-16x-1

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