Question

How do solve `(x^2-x-12)/(x^2+4)<=0` and write the answer as a inequality and interval notation?

Answer 1

The answer is `x in [-3, 4 ]` or `-3<=x<=4`

Explanation:

Let's factorise the numerator

`x^2-x-12=(x+3)(x-4)`

Let `f(x)=(x^2-x-12)/(x^2+4)`

The domain of `f(x)` is `RR`

The denominator is `>0`

We can build the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaaa)` `-oo` `color(white)(aaaa)` `-3` `color(white)(aaaa)` `4` `color(white)(aaaaa)` `+oo`

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

`color(white)(aaaa)` `x-4` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaaa)` `+` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+`

Therefore,

`f(x)<=0` when `x in [-3, 4 ]`

`-3<=x<=4`