Question

How do you solve `3(2w^2-5)<w` using a sign chart?

Answer 1

The answer is `w in]-3/2, 5/3 [`

Explanation:

Let's rewrite the inequality

`6w^2-w-6<0`

Let's factorise

`(3w-5)(2w+3)<0`

Let `f(w)=(3w-5)(2w+3)`

Now, we can build the sign chart

`color(white)(aaaa)` `w` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-3/2` `color(white)(aaaa)` `5/3` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `2w+3` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `3w-5` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+`

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

Therefore,

`f(w)<0` when `w in]-3/2, 5/3 [`