How do you solve #(x^2-x-12)/(x^2+4)<=0# using a sign chart?

1 Answer
Jun 30, 2018

Answer:

The solution is # x in [-3,4] #

Explanation:

Factorise the numerator

#x^2-x-12=(x+3)(x-4)#

Therefore,

#(x^2-x-12)/(x^2+4)<=0#

#<=>#, #((x+3)(x-4))/(x^2+4)<=0#

#AA x in RR, (x^2+4)>0#

Let #f(x)=((x+3)(x-4))/(x^2+4)#

Let's build the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaaa)##-3##color(white)(aaaaaaa)##4##color(white)(aaaaa)##+oo#

#color(white)(aaaa)##x+3##color(white)(aaaaaa)##-##color(white)(aa)##0##color(white)(aaa)##+##color(white)(aaaaaa)##+#

#color(white)(aaaa)##x-4##color(white)(aaaaaa)##-##color(white)(aa)####color(white)(aaaa)##-##color(white)(aa)##0##color(white)(aaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaa)##+##color(white)(aa)##0##color(white)(aaa)##-##color(white)(aa)##0##color(white)(aaa)##+#

Therefore,

#f(x)<=0# when # x in [-3,4] #

graph{(x^2-x-12)/(x^2+4) [-10, 10, -5, 5]}