Question

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

Answer 1

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

Explanation:

Let's factorise the inequality

`x^2+x=x(x+1)>0`

Let `f(x)=x(x+1)`

We can now establish the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaaa)` `-oo` `color(white)(aaaaa)` `-1` `color(white)(aaaaa)` `0` `color(white)(aaaaa)` `+oo`

`color(white)(aaaa)` `x+1` `color(white)(aaaaaaa)` `-` `color(white)(aaaaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `x` `color(white)(aaaaaaaaaa)` `-` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaaaa)` `+` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `+`

Therefore,

`f(x)>0`, when `x in ] -oo,-1 [ uu ] 0, +oo[`

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