How do you solve the inequality #(x^2-16)/(x^2-4x-5)=0#?

Question

How do you solve the inequality #(x^2-16)/(x^2-4x-5)>=0#?

Benjamin Achtenberg · Accepted Answer

The solution is #x in (-oo,-4] uu(-1,4] uu (5,+oo)#

The number one is

#x^2-16=(x-4)(x+4)#

The numerator is

#x^2-4x-5=(x-5)(x+1)#

Let

#f(x)=((x-4)(x+4))/((x-5)(x+1))#

Let's construct a sign chart.

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaaa)##-4##color(white)(aaaa)##-1##color(white)(aaaaa)##4##color(white)(aaaaa)##5##color(white)(aaaaaa)##+oo###

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

#color(white)(aaaa)##x+1##color(white)(aaaaa)##-##color(white)(aaa)####color(white)(aaaa)##-##color(white)(a)##||##color(white)(a)##+##color(white)(aaaa)##+##color(white)(aaaa)##+#

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

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

#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##+##color(white)(aaa)##0##color(white)(aaa)##-##color(white)(a)##||##color(white)(a)##+##color(white)(a)##0##color(white)(aa)##-##color(white)(a)##||##color(white)(aa)##+#

graph{((x+1)(x-4))/((x-5)(x+1)) [-17.12, -7.12, -7.12]}

Consequently,

#f(x)>=0# when #x in (-oo,-4] uu(-1,4] uu (5,+oo)#

How do you solve the inequality #(x^2-16)/(x^2-4x-5)&gt;=0#?

How do you solve the inequality #(x^2-16)/(x^2-4x-5)>=0#?