Question

How do you solve `((x+7)(x-3))/(x-5)^2>0`?

Answer 1

The answer is `x in ] -oo,-7 [ uu ] 3,5 [uu ] 5,+oo [`

Explanation:

Let `f(x)=((x+7)(x-3))/(x-5)^2`

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

`AA x in D_f(x)`, the denominator is `>0`

We can now do the sign chart

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

`color(white)(aaaa)` `x+7` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaa)` `+` `color(white)()` `∥` `color(white)(aa)` `+`

`color(white)(aaaa)` `x-3` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaa)` `+` `color(white)()` `∥` `color(white)(aa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaa)` `+` `color(white)(aaaa)` `-` `color(white)(aaa)` `+` `color(white)()` `∥` `color(white)(aa)` `+`

Therefore,

`f(x)>0` when `x in ] -oo,-7 [ uu ] 3,5 [uu ] 5,+oo [`

graph{((x+7)(x-3))/(x-5)^2 [-23.16, 34.55, -11.88, 16.99]}