Precalculus Examples

$x^{2} + 9 > 6 x$

Step 1

Subtract $6 x$ from both sides of the inequality.

$x^{2} + 9 - 6 x > 0$

Step 2

Convert the inequality to an equation.

$x^{2} + 9 - 6 x = 0$

Step 3

Factor using the perfect square rule.

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Step 3.1

Rearrange terms.

$x^{2} - 6 x + 9 = 0$

Step 3.2

Rewrite $9$ as $3^{2}$ .

$x^{2} - 6 x + 3^{2} = 0$

Step 3.3

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

$6 x = 2 \cdot x \cdot 3$

Step 3.4

Rewrite the polynomial.

$x^{2} - 2 \cdot x \cdot 3 + 3^{2} = 0$

Step 3.5

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 3$ .

${(x - 3)}^{2} = 0$

Step 4

Set the $x - 3$ equal to $0$ .

$x - 3 = 0$

Step 5

Add $3$ to both sides of the equation.

$x = 3$

Step 6

Use each root to create test intervals.

$x < 3$

$x > 3$

Step 7

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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Step 7.1

Test a value on the interval $x < 3$ to see if it makes the inequality true.

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Step 7.1.1

Choose a value on the interval $x < 3$ and see if this value makes the original inequality true.

$x = 0$

Step 7.1.2

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

${(0)}^{2} + 9 > 6 (0)$

Step 7.1.3

The left side $9$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 7.2

Test a value on the interval $x > 3$ to see if it makes the inequality true.

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Step 7.2.1

Choose a value on the interval $x > 3$ and see if this value makes the original inequality true.

$x = 6$

Step 7.2.2

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

${(6)}^{2} + 9 > 6 (6)$

Step 7.2.3

The left side $45$ is greater than the right side $36$ , which means that the given statement is always true.

True

Step 7.3

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

$x < 3$ True

$x > 3$ True

$x < 3$ True

$x > 3$ True

Step 8

The solution consists of all of the true intervals.

$x < 3$ or $x > 3$

Step 9

Convert the inequality to interval notation.

$(- \infty, 3) \cup (3, \infty)$

Step 10