Question

How do you solve `(2y^2+3y-20)/(y^3-y^2)>0`?

Answer 1

The answer is `y in [-4, 0[ uu] 0,1 [uu [5/2, +oo[`

Explanation:

Factorise the numerator and denominator

`2y^2+3y-20=(2y-5)(y+4)`

`y^3-y^2=y^2(y-1)`

So,

`(2y^2+3y-20)/(y^3-y^2)=((2y-5)(y+4))/(y^2(y-1))`

Let `y=((2y-5)(y+4))/(y^2(y-1))`

The domain of `y` is `D_y=RR-{0,1}`

`y^2>0 AA y in D_y`

Let's do a sign chart

`color(white)(aaaa)` `y` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-4` `color(white)(aaaaaaaa)` `0` `color(white)(aaaa)` `1` `color(white)(aaaaa)` `5/2` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `y+4` `color(white)(aaaa)` `-` `color(white)(aaaaa)` `+` `color(white)(aaaa)` `color(red)(∥)` `color(white)()` `+` `color(red)(∥)` `color(white)(a)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `y-1` `color(white)(aaaa)` `-` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `color(red)(∥)` `color(white)()` `-` `color(red)(∥)` `color(white)(a)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `2y-5` `color(white)(aaa)` `-` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `color(red)(∥)` `color(white)()` `-` `color(red)(∥)` `color(white)(a)` `-` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `f(y)` `color(white)(aaaaa)` `-` `color(white)(aaaaa)` `+` `color(white)(aaaa)` `color(red)(∥)` `color(white)()` `+` `color(red)(∥)` `color(white)(a)` `-` `color(white)(aaaa)` `+`

Therefore,

`f(y)>0`, when `y in [-4, 0[ uu] 0,1 [uu [5/2, +oo[`