Question

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

Answer 1

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

Explanation:

The numerator is

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

The denominator is

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

Let

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

Let's build 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+4)(x-4))/((x-5)(x+1)) [-14.24, 14.23, -7.12, 7.12]}

Therefore,

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