# How do you solve x^2-3x-4>=0 using a sign chart?

Nov 1, 2016

The answer is $- \infty < x \le - 1$ and $4 \le x < + \infty$

#### Explanation:

Start by factorising
${x}^{2} - 3 x - 4 = \left(x + 1\right) \left(x - 4\right)$
let $y = \left(x + 1\right) \left(x - 4\right)$
So we can make the sign chart
$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- 1$$\textcolor{w h i t e}{a a a a}$$4$$\textcolor{w h i t e}{a a a a}$$+ \infty$
$x + 1$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$
$x - 4$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$
$y$$\textcolor{w h i t e}{a a a a a a a a}$$+$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

So ${x}^{2} - 3 x - 4 \ge 0$
when $- \infty < x \le - 1$ and when $4 \le x < + \infty$