Convert to Interval Notation x^2-8x+12>0

$x^{2} - 8 x + 12 > 0$

Step 1

Convert the inequality to an equation.

$x^{2} - 8 x + 12 = 0$

Step 2

Factor $x^{2} - 8 x + 12$ 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 $- 8$ .

$- 6, - 2$

Write the factored form using these integers.

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

Step 3

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

$x - 6 = 0$

$x - 2 = 0$

Step 4

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.

$x = 6$

Step 5

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 6

The final solution is all the values that make $(x - 6) (x - 2) = 0$ true.

$x = 6, 2$

Step 7

Use each root to create test intervals.

$x < 2$

$2 < x < 6$

$x > 6$

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 $x < 2$ 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.

$x = 0$

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

${(0)}^{2} - 8 \cdot 0 + 12 > 0$

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

True

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.

$x = 4$

Replace $x$ with $4$ in the original inequality.

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

The left side $- 4$ is not greater than the right side $0$ , which means that the given statement is false.

False

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.

${(8)}^{2} - 8 \cdot 8 + 12 > 0$

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

True

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

$x < 2$ True

$2 < x < 6$ False

$x > 6$ True

$x < 2$ True

$2 < x < 6$ False

$x > 6$ True

Step 9

The solution consists of all of the true intervals.

$x < 2$ or $x > 6$

Step 10

Convert the inequality to interval notation.

$(- \infty, 2) \cup (6, \infty)$

Step 11

Convert to Interval Notation x^2-8x+12>0

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