# How do you solve (q+4)^2>10q+31 using a sign chart?

##### 1 Answer
Aug 7, 2017

The solution is $q \in \left(- \infty , - 3\right) \cup \left(5 , + \infty\right)$

#### Explanation:

Let's rearrange and factorise the inequality

${\left(q + 4\right)}^{2} > 10 q + 31$

${q}^{2} + 8 q + 16 > 10 q + 31$

${q}^{2} + 8 q + 16 - 10 q - 31 > 0$

${q}^{2} - 2 q - 15 > 0$

$\left(q + 3\right) \left(q - 5\right) > 0$

Let $f \left(q\right) = \left(q + 3\right) \left(q - 5\right)$

We can build the sign chart

$\textcolor{w h i t e}{a a a a}$$q$$\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}$$5$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$q + 3$$\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}$$q - 5$$\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}$$f \left(q\right)$$\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,

$f \left(q\right) > 0$ when $q \in \left(- \infty , - 3\right) \cup \left(5 , + \infty\right)$

graph{(x+4)^2-10x-31 [-27.53, 23.8, -17.2, 8.47]}