How do you solve x^2-6x+9>16 using a sign chart?

Mar 1, 2017

The solution is x in ]-oo,-1[uu]7,+oo[

Explanation:

Let's rewrite the inequality and factorise

${x}^{2} - 6 x + 9 - 16 > 0$

${x}^{2} - 6 x - 7 > 0$

$\left(x - 7\right) \left(x + 1\right) > 0$

Let $f \left(x\right) = \left(x - 7\right) \left(x + 1\right)$

We build the sign chart

$\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}$$- 1$$\textcolor{w h i t e}{a a a a}$$7$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x + 1$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$x - 7$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a 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 a}$$+$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

Therefore,

$f \left(x\right) > 0$, when x in ]-oo,-1[uu]7,+oo[