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

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Answer 1 · 2016-12-31T13:52:53

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

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]}

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