How do you solve #(x+6)/(x^2-5x-24)>=0#?

1 Answer
Oct 23, 2016

Answer:

#f(x)>=0 for x⋳[-6,-3]∪[8,+oo]#

Explanation:

Start by factorising the denominator
#x^2-5x-24=(x+3)(x-8)#
Then
#(x+6)/((x+3)(x-8))>=0#
For this we make a sign table

#x##color(white)(aaaaaaa)##-oo##color(white)(aaaa)##-6##color(white)(aaaa)##-3##color(white)(aaaa)##+8##color(white)(aaaa)##+oo#
#x+6##color(white)(aaaaaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+##color(white)(aaaa)##+#
#x+3##color(white)(aaaaaaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#
#x-8##color(white)(aaaaaaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#
#f(x)##color(white)(aaaaaaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##-##color(white)(aaaa)##+#

As we need a result #>=0#, we keep the intervals where #f(x)# os positive

So the result is #x⋳[-6,-3]∪[8,+oo])#