# How do you solve x^2-4x<-3 using a sign chart?

Dec 13, 2016

The answer is x in ] 1,3 [

#### Explanation:

Rewrite the equation as, ${x}^{2} - 4 x + 3 < 0$

Let $f \left(x\right) = {x}^{2} - 4 x + 3 = \left(x - 1\right) \left(x - 3\right)$

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

Now, we can do our 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}$$+ 3$$\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}$$x - 3$$\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}$$f \left(x\right)$$\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}$$+$

So,

$f \left(x\right) < 0$ when  x in ] 1,3 [