How do you solve #(x-2)/(2x+5)<0# using a sign chart?

1 Answer
Dec 27, 2016

Answer:

The answer is #x in ] -5/2, 2[#

Explanation:

Let #f(x)=(x-2)/(2x+5)#

The domain of #f(x)# is #D_f(x)=RR-{-5/2}#

Now, we can do the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-5/2##color(white)(aaaaa)##2##color(white)(aaaa)##+oo#

#color(white)(aaaa)##2x+5##color(white)(aaaa)##-##color(white)(aaa)##∣∣##color(white)(aa)##+##color(white)(aaa)##+#

#color(white)(aaaa)##x-2##color(white)(aaaaa)##-##color(white)(aaa)##∣∣##color(white)(aa)##-##color(white)(aaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##+##color(white)(aaa)##∣∣##color(white)(aa)##-##color(white)(aaa)##+#

Therefore,

#f(x)<0#, when #x in ] -5/2, 2[#