Graph x^2-4x-12<=0

$x^{2} - 4 x - 12 \leq 0$

Convert the inequality to an equation.

$x^{2} - 4 x - 12 = 0$

Factor $x^{2} - 4 x - 12$ 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 $- 12$ and whose sum is $- 4$ .

$- 6, 2$

Write the factored form using these integers.

$(x - 6) (x + 2) = 0$

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

$x - 6 = 0$

$x + 2 = 0$

Set $x - 6$ equal to $0$ and solve for $x$ .

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

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 $(x - 6) (x + 2) = 0$ true.

$x = 6, - 2$

Use each root to create test intervals.

$x < - 2$

$- 2 < x < 6$

$x > 6$

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.

${(- 4)}^{2} - 4 \cdot - 4 - 12 \leq 0$

The left side $20$ is greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 2 < x < 6$ to see if it makes the inequality true.

Choose a value on the interval $- 2 < x < 6$ and see if this value makes the original inequality true.

$x = 0$

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

${(0)}^{2} - 4 \cdot 0 - 12 \leq 0$

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

True

Test a value on the interval $x > 6$ to see if it makes the inequality true.

Choose a value on the interval $x > 6$ and see if this value makes the original inequality true.

$x = 8$

Replace $x$ with $8$ in the original inequality.

${(8)}^{2} - 4 \cdot 8 - 12 \leq 0$

The left side $20$ 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.

$x < - 2$ False

$- 2 < x < 6$ True

$x > 6$ False

$x < - 2$ False

$- 2 < x < 6$ True

$x > 6$ False

The solution consists of all of the true intervals.

$- 2 \leq x \leq 6$

Graph x^2-4x-12<=0

Related Questions

Support LCMGCF.com by disabling your adblocker.