Convert to Set Notation x^2-11x+18>0

$x^{2} - 11 x + 18 > 0$

Step 1

Solve $x < 2 or x > 9$ .

Convert the inequality to an equation.

$x^{2} - 11 x + 18 = 0$

Factor $x^{2} - 11 x + 18$ 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 $18$ and whose sum is $- 11$ .

$- 9, - 2$

Write the factored form using these integers.

$(x - 9) (x - 2) = 0$

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

$x - 9 = 0$

$x - 2 = 0$

Set $x - 9$ equal to $0$ and solve for $x$ .

Set $x - 9$ equal to $0$ .

$x - 9 = 0$

Add $9$ to both sides of the equation.

$x = 9$

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.

$x = 2$

The final solution is all the values that make $(x - 9) (x - 2) = 0$ true.

$x = 9, 2$

Use each root to create test intervals.

$x < 2$

$2 < x < 9$

$x > 9$

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.

${(0)}^{2} - 11 \cdot 0 + 18 > 0$

The left side $18$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $2 < x < 9$ to see if it makes the inequality true.

Choose a value on the interval $2 < x < 9$ and see if this value makes the original inequality true.

$x = 6$

Replace $x$ with $6$ in the original inequality.

${(6)}^{2} - 11 \cdot 6 + 18 > 0$

The left side $- 12$ is not greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > 9$ to see if it makes the inequality true.

Choose a value on the interval $x > 9$ and see if this value makes the original inequality true.

$x = 12$

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

${(12)}^{2} - 11 \cdot 12 + 18 > 0$

The left side $30$ 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

$2 < x < 9$ False

$x > 9$ True

$x < 2$ True

$2 < x < 9$ False

$x > 9$ True

The solution consists of all of the true intervals.

$x < 2$ or $x > 9$

Step 2

Use the inequality $x < 2 or x > 9$ to build the set notation.

${x | x < 2 or x > 9}$

Step 3

Convert to Set Notation x^2-11x+18>0

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