Question

How do you solve the inequality `x^2-3x-18>0`?

Answer 1

The answer is `( -oo < x<-3)` `uu` `(6< x<+oo )`

Explanation:

We factorise the expression `x^2-3x-18=(x+3)(x-6)`
then we do a sign chart
`color(white)(aaaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-3` `color(white)(aaaa)` `6` `color(white)(aaa)` `+oo`
`color(white)(aaaaa)` `x+3` `color(white)(aaaaa)` `-` `color(white)(aa)` `0` `color(white)(aa)` `+` `color(white)(aa)` `+`
`color(white)(aaaaa)` `x-6` `color(white)(aaaaa)` `-` `color(white)(aaaaa)` `-` `color(white)(aa)` `+`
`color(white)(a)` `(x-6((x+3)` `color(white)(aa)` `+` `color(white)(aaaaa)` `-` `color(white)(aa)` `+`

So `x^2-3x-18>0` when

`-oo < x<-3` and `6< x<+oo`