How do you solve #x^2+x-6<0# using a sign chart?

1 Answer
Jan 15, 2017

Answer:

The answer is #x in ] -3,2 [#

Explanation:

Let's factorise the expression,

#x^2+x-6=(x-2)(x+3)#

and let #f(x)=x^2+x-6#

Now we can do the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-3##color(white)(aaaaa)##2##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x+3##color(white)(aaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##x-2##color(white)(aaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##+##color(white)(aaaa)##-##color(white)(aaaa)##+#

Therefore,

#f(x)<0#, when #x in ] -3,2 [ #