Question

How do you solve `(\frac { 2x + 1} { x - 5} ) ^ { 2} > 0`?

Answer 1

The solution is `x in (-oo, -1/2)uu(-1/2,5)uu(5,+oo)`

Explanation:

Let `f(x)=((2x+1)/(x-5))^2`

The domain of `f(x)` is `x in RR-{5}`

The inequality is `f(x)>0`

Let's build a sign chart

`color(white)(aaaa)` `x` `color(white)(aaaaaaa)` `-oo` `color(white)(aaaaaa)` `-1/2` `color(white)(aaaaaaa)` `5` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `(2x+1)^2` `color(white)(aaaa)` `+` `color(white)(aaaaaa)` `0` `color(white)(aaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `(x-5)^2` `color(white)(aaaaa)` `+` `color(white)(aaaaaa)` ###color(white)(aaaa)#`+` `color(white)(aa)` `||` `color(white)(a)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaaaa)` `+` `color(white)(aaaaaa)` `0` `color(white)(aaaa)` `+` `color(white)(a)` `||` `color(white)(a)` `+`

Therefore,

`f(x)>0` when `x in (-oo, -1/2)uu(-1/2,5)uu(5,+oo)`

graph{(2x+1)^2/(x-5)^2 [-40.24, 41.92, -17.72, 23.4]}