Graph x^2+2x+1>0

$x^{2} + 2 x + 1 > 0$

Step 1

Convert the inequality to an equation.

$x^{2} + 2 x + 1 = 0$

Step 2

Factor using the perfect square rule.

Rewrite $1$ as $1^{2}$ .

$x^{2} + 2 x + 1^{2} = 0$

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

$2 x = 2 \cdot x \cdot 1$

Rewrite the polynomial.

$x^{2} + 2 \cdot x \cdot 1 + 1^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 1$ .

${(x + 1)}^{2} = 0$

Step 3

Set the $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 4

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 5

Use each root to create test intervals.

$x < - 1$

$x > - 1$

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 $x < - 1$ to see if it makes the inequality true.

Choose a value on the interval $x < - 1$ and see if this value makes the original inequality true.

$x = - 4$

Replace $x$ with $- 4$ in the original inequality.

${(- 4)}^{2} + 2 (- 4) + 1 > 0$

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

True

Test a value on the interval $x > - 1$ to see if it makes the inequality true.

Choose a value on the interval $x > - 1$ and see if this value makes the original inequality true.

$x = 0$

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

${(0)}^{2} + 2 (0) + 1 > 0$

The left side $1$ 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.

$x < - 1$ True

$x > - 1$ True

$x < - 1$ True

$x > - 1$ True

Step 7

The solution consists of all of the true intervals.

$x < - 1$ or $x > - 1$

Step 8

Graph x^2+2x+1>0

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