How do you solve #(x^2+2x-15)/(x^2+7x)>0#?

1 Answer
Nov 30, 2016

Answer:

The answer is #x in ] -oo,-7 [uu ] -5, 0[ uu ]3, oo[#

Explanation:

Let #f(x)=(x^2+2x-15)/(x^2+7x)#

#=((x-3)(x+5))/(x(x+7))#

The domain of #f(x)# is #D_f(x)=RR-{0,-7} #

To solve the equation #f(x)>0#, we need a sign chart.

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-7##color(white)(aaaa)##-5##color(white)(aaaa)##0##color(white)(aaaa)##3##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x+7##color(white)(aaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+##color(white)(aaaa)##+##color(white)(aaa)##+#

#color(white)(aaaa)##x+5##color(white)(aaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+##color(white)(aaa)##+#

#color(white)(aaaa)##x##color(white)(aaaaaaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+##color(white)(aaa)##+#

#color(white)(aaaa)##x-3##color(white)(aaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##-##color(white)(aaa)##+#

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

So,

#f(x)>0#

when #x in ] -oo,-7 [uu ] -5, 0[ uu ]3, oo[#

graph{(x^2+2x-15)/(x^2+7x) [-16.02, 16.01, -8.01, 8.01]}