I am looking at the following problem:

> 
> Find all values of $m$ for which the equation $(3m-2)x^2 + 2mx + 3m = 0$ shas only one root between $-1$ and $0$.
> 
> 
> 

I don't know what only one root between -1 and 0 actually means. I count these possibilities:

1. $-1<x\_1=x\_2<0$
2. $x\_1<-1<x\_2<0$
3. $-1<x\_1<0<x\_2$

Let $a=3m-2$, $\Delta$ be the discriminant, and $S/2$ the $x$-coordinate of the vertex ($S/2 = -m/(3m-2)$) and also the arithmetic mean value of the roots $(x\_1+x\_2)/2$. 

Conditions to solve 1 alone would be:
$af(-1)<0$,
$af(0)<0$ and
$\Delta \geq 0$

Conditions to solve 2 alone would be:
$af(-1)<0$,
$af(0)>0$,
$\Delta>0$ and
$S/2<0$

Conditions to solve 3 alone would be:

$af(-1)>0$,
$af(0)<0$,
$\Delta>0$, and
$S/2>-1$

Then it could also mean that it is all conditions intersected $ 1 \cap 2 \cap 3$
or just $2 \cap 3$.

What am I doing wrong? the correct answer for this problem according to the book is $0<m<1/2$

PS.: Although this question was already answered correctly I would like to see how to use the theorems that I brought in, and the book as well, to solve this. Solving questions any way or another is always welcomed and helpful, what should certainly be done. Yet the purpose in the book is not a general math challenge, it's intended to teach the theorems and aplly them. So if I can, let me try it also:

I will put the theorems here because no one seems to know what theorems I am refering to:

If $f(x)=ax^2+bx+c$ presents two real roots $x\_1 \leq x\_2$ and $ \alpha $ is the real number that will be compared to $x\_1$ and $x\_2$, we have:

* $af(\alpha)<0 \implies x\_1 < \alpha <x\_2$.
* $af(\alpha)=0 \implies \alpha$ is one of the roots.
* If $af(\alpha)>0$ and $\Delta \geq 0$, then

(a) $\alpha <x\_1 \leq x\_2$ if $\alpha < S/2$;

(b) $x\_1 \leq x\_2 < \alpha$ if $\alpha > S/2$

There is two situations in this problem: $-1<x\_1<0<x\_2$ and $x\_1<-1<x\_2<0$.

**I.**
$$-1<x\_1<0<x\_2 \implies af(-1)>0 \wedge af(0)<0 \wedge \Delta > 0 \wedge S/2 > -1.$$
\begin{align\*}
af(-1)>0 &\implies (3m-2)[(3m-2)(-1)^2+2m(-1)+3m]>0\\
 &\implies (3m-2)(4m-2)>0 \implies 12m^2-14m+4>0\\
 &\implies m<1/2\text{ or }m>2/3.\\
af(0)<0 &\implies (3m-2)[(3m-2)0^2+2m0+3m]<0 \\
&\implies 9m^2-6m\lt 0 \implies 0\lt m\lt 2/3.\\
\Delta \gt 0 &\implies (2m)^2-4(3m-2)3m\gt 0 \implies 4m^2-12m(3m-2)\gt 0\\
 &\implies 4m^2-36m^2+24\gt 0 \implies -32m^2+24\gt 0 \implies 0 \lt m \lt 3/4.\\
S/2\gt -1 &\implies -b/2a\gt -1 \implies -m/(3m-2) \gt -1\\
 &\implies (-m+3m-2)/(3m-2) \gt 0 \implies (2m-2)/(3m-2) \gt 0\\
 &\implies m \lt 2/3 \text{ or } m \gt 1.
\end{align\*}

$$(af(-1) \gt 0) \wedge (af(0) \lt 0) \wedge (\Delta \gt 0) \wedge (S/2 \gt -1) \implies 0 \lt m \lt 1/2.$$

**II.**
$$x\_1<-1<x\_2<0 \implies af(-1)<0 \wedge af(0)>0 \wedge \Delta > 0 \wedge S/2<0.$$
\begin{align\*}
af(-1)<0 &\implies 1/2 \lt m \lt 2/3.\\
af(0)>0 &\implies m \lt 0\text{ or }m \gt 2/3.\\
\Delta \gt 0 &\implies 0 \lt m \lt 3/4.\\
S/2 \lt 0 &\implies -b/2a<0 \implies -m/(3m-2)<0 \\
&\implies m \lt 0 \text{ or }m \gt 2/3.
\end{align\*}

$(af(-1)<0 \wedge af(0)>0 \wedge \Delta \gt 0 \wedge S/2 \lt 0) = \emptyset$

As both set of solutions are possible answers, non-exclusory, I think the answer is to unite the two sets: $I \cup II = (0 \lt m \lt 1/2) \cup \emptyset = 0 \lt m \lt 1/2$.

Final answer $0 \lt m \lt 1/2$.