Question

How do you solve `(x+6)/(x^2-5x-24)>=0`?

Answer 1

`f(x)>=0 for x⋳[-6,-3]∪[8,+oo]`

Explanation:

Start by factorising the denominator
`x^2-5x-24=(x+3)(x-8)`
Then
`(x+6)/((x+3)(x-8))>=0`
For this we make a sign table

`x` `color(white)(aaaaaaa)` `-oo` `color(white)(aaaa)` `-6` `color(white)(aaaa)` `-3` `color(white)(aaaa)` `+8` `color(white)(aaaa)` `+oo`
`x+6` `color(white)(aaaaaaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+`
`x+3` `color(white)(aaaaaaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+`
`x-8` `color(white)(aaaaaaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+`
`f(x)` `color(white)(aaaaaaaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+`

As we need a result `>=0`, we keep the intervals where `f(x)` os positive

So the result is `x⋳[-6,-3]∪[8,+oo])`