Convert to Interval Notation x^2-12x+36<0

$x^{2} - 12 x + 36 < 0$

Step 1

Convert the inequality to an equation.

$x^{2} - 12 x + 36 = 0$

Step 2

Factor using the perfect square rule.

Rewrite $36$ as $6^{2}$ .

$x^{2} - 12 x + 6^{2} = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$12 x = 2 \cdot x \cdot 6$

Rewrite the polynomial.

$x^{2} - 2 \cdot x \cdot 6 + 6^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 6$ .

${(x - 6)}^{2} = 0$

Step 3

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

$x - 6 = 0$

Step 4

Add $6$ to both sides of the equation.

$x = 6$

Step 5

Use each root to create test intervals.

$x < 6$

$x > 6$

Step 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 $x < 6$ 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.

$x = 0$

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

${(0)}^{2} - 12 \cdot 0 + 36 < 0$

The left side $36$ is not less 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.

${(8)}^{2} - 12 \cdot 8 + 36 < 0$

The left side $4$ is not less 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.

$x < 6$ False

$x > 6$ False

$x < 6$ False

$x > 6$ False

Step 7

Since there are no numbers that fall within the interval, this inequality has no solution.

No solution

Convert to Interval Notation x^2-12x+36<0

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