Question

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

Answer 1

The solution is `x in ]-oo, -3[ uu ]-1,2[`

Explanation:

We solve this inequality with a sign chart

Let's factorise the numerator

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

The inequality is

`(x^2-x-2)/(x+3)=((x+1)(x-2))/(x+3)<0`

Let `f(x)=((x+1)(x-2))/(x+3)`

The domain of `f(x)` is `D_f(x)=RR-{-3}`

We can build the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaaa)` `-3` `color(white)(aaaaaa)` `-1` `color(white)(aaaa)` `2` `color(white)(aaaaaa)` `+oo`

`color(white)(aaaa)` `x+3` `color(white)(aaaa)` `-` `color(white)(aaaa)` `||` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `x+1` `color(white)(aaaa)` `-` `color(white)(aaaa)` `||` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `x-2` `color(white)(aaaa)` `-` `color(white)(aaaa)` `||` `color(white)(aaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `||` `color(white)(aaaa)` `+` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+`

Therefore,

`f(x)<0` when `x in ]-oo, -3[ uu ]-1,2[`