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

Dec 31, 2016

The answer is  x in ] -oo,-1 [ uu ] 0, +oo[

#### Explanation:

Let's factorise the inequality

${x}^{2} + x = x \left(x + 1\right) > 0$

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

We can now establish the sign chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a}$$- 1$$\textcolor{w h i t e}{a a a a a}$$0$$\textcolor{w h i t e}{a 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 a a}$$-$$\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}$$x$$\textcolor{w h i t e}{a a a a a a a a a a}$$-$$\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}$$f \left(x\right)$$\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}$$+$

Therefore,

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

graph{x(x+1) [-3.08, 3.078, -1.54, 1.54]}