# How do you solve x^2+6x+9>=0 using a sign chart?

Dec 23, 2016

The answer is $x \in \mathbb{R}$

#### Explanation:

We use

${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$

Factorise the expression

${x}^{2} + 6 x + 9 = {\left(x + 3\right)}^{2}$

Let $f \left(x\right) = {\left(x + 3\right)}^{2}$

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R}$

and

$\forall x \in \mathbb{R}$, $f \left(x\right) \ge 0$

So,

$x \in \mathbb{R}$

The sign chart is simple

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- 3$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x + 3$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a}$$0$$\textcolor{w h i t e}{a a}$$+$

$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a}$$+$$\textcolor{w h i t e}{a a a}$$0$$\textcolor{w h i t e}{a a}$$+$

Therefore,

$f \left(x\right) \ge 0$ when $x \in \mathbb{R}$