Solve the Inequality for x x^2+6x-16<=0

$x^{2} + 6 x - 16 \leq 0$

Step 1

Convert the inequality to an equation.

$x^{2} + 6 x - 16 = 0$

Step 2

Factor $x^{2} + 6 x - 16$ 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 $- 16$ and whose sum is $6$ .

$- 2, 8$

Write the factored form using these integers.

$(x - 2) (x + 8) = 0$

Step 3

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

$x - 2 = 0$

$x + 8 = 0$

Step 4

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.

$x = 2$

Step 5

Set $x + 8$ equal to $0$ and solve for $x$ .

Set $x + 8$ equal to $0$ .

$x + 8 = 0$

Subtract $8$ from both sides of the equation.

$x = - 8$

Step 6

The final solution is all the values that make $(x - 2) (x + 8) = 0$ true.

$x = 2, - 8$

Step 7

Use each root to create test intervals.

$x < - 8$

$- 8 < x < 2$

$x > 2$

Step 8

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 $x < - 8$ to see if it makes the inequality true.

Choose a value on the interval $x < - 8$ and see if this value makes the original inequality true.

$x = - 10$

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

${(- 10)}^{2} + 6 (- 10) - 16 \leq 0$

The left side $24$ is greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 8 < x < 2$ to see if it makes the inequality true.

Choose a value on the interval $- 8 < x < 2$ and see if this value makes the original inequality true.

$x = 0$

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

${(0)}^{2} + 6 (0) - 16 \leq 0$

The left side $- 16$ is less than the right side $0$ , which means that the given statement is always true.

True

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.

${(4)}^{2} + 6 (4) - 16 \leq 0$

The left side $24$ is greater than the right side $0$ , which means that the given statement is false.

False

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

$x < - 8$ False

$- 8 < x < 2$ True

$x > 2$ False

$x < - 8$ False

$- 8 < x < 2$ True

$x > 2$ False

Step 9

The solution consists of all of the true intervals.

$- 8 \leq x \leq 2$

Step 10

The result can be shown in multiple forms.

Inequality Form:

$- 8 \leq x \leq 2$

Interval Notation:

$[- 8, 2]$

Step 11

Solve the Inequality for x x^2+6x-16<=0

Related Questions

Support LCMGCF.com by disabling your adblocker.