# How do you solve 16-x^2>0 using a sign chart?

May 11, 2017

The solution is $x \in \left(- 4 , + 4\right)$

#### Explanation:

We need

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

We factorise the inequality

$16 - {x}^{2} > 0$

$\left(4 + x\right) \left(4 - x\right) > 0$

Let

$f \left(x\right) = \left(4 + x\right) \left(4 - x\right)$

We can 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}$$- 4$$\textcolor{w h i t e}{a a a a}$$+ 4$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$4 + x$$\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}$$+$

$\textcolor{w h i t e}{a a a a}$$4 - x$$\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}$$-$

$\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 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 $x \in \left(- 4 , + 4\right)$