Question

How do you solve `2v^2+7v<4` using a sign chart?

Answer 1

The answer is `v in ] -4,1/2 [`

Explanation:

Let's rewrite the inequality as

`2v^2+7v-4<0`

Let `f(v)=2v^2+7v-4`

The domain of `f(v)` is `D_f(v)=RR`

Let's factorise the expression

`2v^2+7v-4=(2v-1)(v+4)`

Now we establish the sign chart

`color(white)(aaaa)` `v` `color(white)(aaaaa)` `-oo` `color(white)(aaaaa)` `-4` `color(white)(aaaaa)` `1/2` `color(white)(aaaaa)` `+oo`

`color(white)(aaaa)` `v+4` `color(white)(aaaaaaa)` `-` `color(white)(aaaaa)` `+` `color(white)(aaaaa)` `+`

`color(white)(aaaa)` `2v-1` `color(white)(aaaaaa)` `-` `color(white)(aaaaa)` `-` `color(white)(aaaaa)` `+`

`color(white)(aaaa)` `f(v)` `color(white)(aaaaaaaa)` `+` `color(white)(aaaaa)` `-` `color(white)(aaaaa)` `+`

Therefore,

`f(v)<0` when `v in ] -4,1/2 [`