How do you solve #(x-4)/(x^2+2x)<=0# using a sign chart?

1 Answer
Dec 14, 2016

Answer:

The answer is #x in ] -oo,-2 [ uu ] 0,4] #

Explanation:

The denominator is

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

Let #f(x)=(x-4)/(x(x+2))#

The domain of #f(x)# is #D_f(x)=RR-{0,-2} #

Now, we can do the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-2##color(white)(aaaa)##0##color(white)(aaaa)##4##color(white)(aaaa)##+oo#

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

#color(white)(aaaa)##x##color(white)(aaaaaaaaa)##-##color(white)(aa)##∥##color(white)()##-##color(white)()##∥##color(white)(a)##+##color(white)(aaa)##+#

#color(white)(aaaa)##x-4##color(white)(aaaaa)##-##color(white)(aa)##∥##color(white)()##-##color(white)()##∥##color(white)(a)##-##color(white)(aaa)##+#

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

So,
#f(x)<=0#, when #x in ] -oo,-2 [ uu ] 0,4] #