Question

How do you solve `2x^2+3x>0` using a sign chart?

Answer 1

The answer is `x in ]-oo, -3/2[ uu ]0, +oo[`

Explanation:

Let `f(x)=2x^2+3x=x(2x+3)`

We now, build the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaaaa)` `-oo` `color(white)(aaaa)` `-3/2` `color(white)(aaaa)` `0` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `2x+3` `color(white)(aaaaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `x` `color(white)(aaaaaaaaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+`

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

Therefore,

`f(x)>0` when `x in ]-oo, -3/2[ uu ]0, +oo[`