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

1 Answer
Dec 31, 2016

Answer:

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]}