# How do you solve x^2+4x+4>=9 using a sign chart?

Oct 23, 2016

the answer is $x \in \left(- \infty , - 5\right) \cup \left(1 , + \infty\right)$

#### Explanation:

${x}^{2} + 4 x + 4 \ge 9$
so ${x}^{2} + 4 x - 5 \ge 0$
Factorising $\left(x - 1\right) \left(x + 5\right) \ge 0$
The values we are looking at are $x = 1$ and $x = - 5$

the sign chart is the following

$x$$\textcolor{w h i t e}{a a a a a a a a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a a}$$- 5$$\textcolor{w h i t e}{a a a a a a}$$1$$\textcolor{w h i t e}{a a a a a a}$$+ \infty$
$x - 1$$\textcolor{w h i t e}{a a a a a a a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a a a}$$+$
$x + 5$$\textcolor{w h i t e}{a a a a a a a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a a a}$$+$$\textcolor{w h i t e}{a a a a a}$$+$
$\left(x - 1\right) \left(x + 5\right)$$\textcolor{w h i t e}{a a a a a a}$$+$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a a a}$$+$

From the sign chart, we see that $x \in \left(- \infty , - 5\right)$ for the function to be positive
From the sign chart, we see that $x \in \left(1 , + \infty\right)$ for the function to be positive