Convert to Set Notation x^2+36>12x

$x^{2} + 36 > 12 x$

Solve $x^{2} + 36 > 12 x$ .

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Subtract $12 x$ from both sides of the inequality.

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

Convert the inequality to an equation.

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

Factor using the perfect square rule.

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Rearrange terms.

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

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

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

Check the middle term by multiplying $2 a b$ and compare this result with the middle term in the original expression.

$2 a b = 2 \cdot x \cdot - 6$

Simplify.

$2 a b = - 12 x$

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

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

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

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

Use each root to create test intervals.

$x < 6$

$x > 6$

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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

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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.

${(0)}^{2} + 36 > 12 (0)$

The left side $36$ is greater 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.

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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.

${(8)}^{2} + 36 > 12 (8)$

The left side $100$ is greater than the right side $96$ , which means that the given statement is always true.

True

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

$x < 6$ True

$x > 6$ True

$x < 6$ True

$x > 6$ True

The solution consists of all of the true intervals.

$x < 6$ or $x > 6$

Use the inequality $x < 6 or x > 6$ to build the set notation.

${x | x < 6 or x > 6}$

Convert to Set Notation x^2+36>12x

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