Calculus Examples

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

Step 1

Convert the inequality to an equation.

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

Step 2

Factor using the perfect square rule.

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Step 2.1

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

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

Step 2.2

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 2.3

Rewrite the polynomial.

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

Step 2.4

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 3

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

$x + 3 = 0$

Step 4

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 5

Use each root to create test intervals.

$x < - 3$

$x > - 3$

Step 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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Step 6.1

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

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Step 6.1.1

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

$x = - 6$

Step 6.1.2

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

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

Step 6.1.3

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

True

Step 6.2

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

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Step 6.2.1

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

$x = 0$

Step 6.2.2

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

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

Step 6.2.3

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

True

Step 6.3

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

$x < - 3$ True

$x > - 3$ True

$x < - 3$ True

$x > - 3$ True

Step 7

The solution consists of all of the true intervals.

$x < - 3$ or $x > - 3$

Step 8

Convert the inequality to interval notation.

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

Step 9