How do you solve #(x^2-x-2)/(x+3)<0#?

1 Answer
Mar 8, 2017

Answer:

The solution is #x in ]-oo, -3[ uu ]-1,2[#

Explanation:

We solve this inequality with a sign chart

Let's factorise the numerator

#x^2-x-2=(x+1)(x-2)#

The inequality is

#(x^2-x-2)/(x+3)=((x+1)(x-2))/(x+3)<0#

Let #f(x)=((x+1)(x-2))/(x+3)#

The domain of #f(x)# is #D_f(x)=RR-{-3}#

We can build the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaaa)##-3##color(white)(aaaaaa)##-1##color(white)(aaaa)##2##color(white)(aaaaaa)##+oo#

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

#color(white)(aaaa)##x+1##color(white)(aaaa)##-##color(white)(aaaa)##||##color(white)(aaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

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

#color(white)(aaaa)##f(x)##color(white)(aaaaa)##-##color(white)(aaaa)##||##color(white)(aaaa)##+##color(white)(aaaa)##-##color(white)(aaaa)##+#

Therefore,

#f(x)<0# when #x in ]-oo, -3[ uu ]-1,2[#