How do you solve #(x+6)/(x^2+6x+8)>=0#?

1 Answer
Jan 11, 2017

Answer:

The answer is #x in [-6, -4[uu ] -2,+oo [ #

Explanation:

Let 's factorise the denominator

#x^2+6x+8=(x+4)(x+2)#

Let #f(x)=(x+6)/(x^2+6x+8)=(x+6)/((x+4)(x+2))#

The domain of #f(x)# is #D_f(x)=RR-{-2,-4}#

We solve the inequality with a sign chart

#color(white)(aaaa)##x##color(white)(aaaaaa)##-oo##color(white)(aaaa)##-6##color(white)(aaaa)##-4##color(white)(aaaaaa)##-2##color(white)(aaaaaa)##+oo#

#color(white)(aaaa)##x+6##color(white)(aaaaaaa)##-##color(white)(aaaa)##+##color(white)(aa)##color(red)(∥)##color(white)(aa)##+##color(white)(aa)##color(red)(∥)##color(white)(aaa)##+#

#color(white)(aaaa)##x+4##color(white)(aaaaaaa)##-##color(white)(aaaa)##-##color(white)(aa)##color(red)(∥)##color(white)(aa)##+##color(white)(aa)##color(red)(∥)##color(white)(aaa)##+#

#color(white)(aaaa)##x+2##color(white)(aaaaaaa)##-##color(white)(aaaa)##-##color(white)(aa)##color(red)(∥)##color(white)(aa)##-##color(white)(aa)##color(red)(∥)##color(white)(aaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaaa)##-##color(white)(aaaa)##+##color(white)(aa)##color(red)(∥)##color(white)(aa)##-##color(white)(aa)##color(red)(∥)##color(white)(aaa)##+#

Therefore,

#f(x)>=0#, when #x in [-6, -4[uu ] -2,+oo [ #