How do solve #(x-4)/(x^2+2x)<=0# and write the answer as a inequality and interval notation?

1 Answer
Dec 26, 2016

Answer:

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

Explanation:

Let's rewrite the inequality as

#(x-4)/(x^2+2x)=(x-4)/(x(x+2))#

And

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

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

Now, we can make the sign chart

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

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

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

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

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

Therefore,

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

or, #x<-2# or # 0< x <=4#