Question

How do solve `x^2+x<12` and write the answer as a inequality and interval notation?

Answer 1

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

Explanation:

The inequation is

`x^2+x-12<0`

Let's factorise the expression

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

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

Now, we can do the sign chart

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

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

`color(white)(aaaa)` `x-3` `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 ] -4,3 [`

or `-4 < x <3`