Find the values of $m$ in the 2nd degree equation $mx^2-2(m-1)x-m-1=0$ so that it has one root between $-1$ and $2$ Find the values of $m$ in the 2nd degree equation $mx^2-2(m-1)x-m-1=0$ so that it has only one root between $-1$ and $2$.
Like in this almost identical question there are two ways to solve this, one is acknowledging that $f(-1)f(2) \lt 0$, the other applying the theorems for the two possible scenarios $x_1 \lt -1 \lt x_2 \lt 2$ and $-1 \lt x_1 \lt 2 \lt x_2$.
Doing $f(-1)f(2)<0$, we have:
$
\begin{align*}
\left(m(-1)^2-2(m-1)(-1)-m-1\right)\left(m2^2-2(m-1)2-m-1\right)&\lt0\\\
(2m-3)(-m+3) & \lt 0 \\\
-2m^2+9m-9 & \lt 0 \\\
 m \lt \frac{3}{2} \quad \vee \quad m \gt 3
\end{align*}
$
The thing is that this answer is wrong since the correct one is $m \lt \frac{3}{2} \quad \wedge \quad m \neq 0 \quad \vee \quad m \gt 3$. How do I get to know and not only verify that $m \neq 0$ what did I do wrong?
Using the theorems in the scenarios I.$ \quad x_1 \lt -1 \lt x_2 \lt 2$ and  II.$ \quad -1 \lt x_1 \lt 2 \lt x_2$, we have:
I.
$$x_1 \lt -1 \lt x_2 \lt 2
\implies \Bigl(af(-1) \lt 0\Bigr) \wedge  \Bigl(af(2) \gt 0\Bigr)  \wedge \Bigl(\Delta \gt 0 \Bigr)\wedge \Bigl( \frac{S}{2} \lt 2\Bigr).$$
II.
$$
-1 \lt x_1 \lt 2 \lt x_2 \implies
\Bigl( af(-1) \gt 0\Bigr)\wedge\Bigl( af(2) \lt 0\Bigr) \wedge \Bigl(\Delta \gt 0 \Bigr) \wedge\Bigl(\frac{S}{2} \gt -1 \Bigr)$$
So both I and II have the condition $\Delta \gt 0$ in common.
\begin{align*}
\Delta \gt 0 &\implies [-2(m-1)]^2-4m(-m-1) \gt 0\\\
&\implies 8m^2-4m+4 \gt 0 \\\
\end{align*}
The problem is that the $\Delta$ from the equation $8m^2-4m+4=0$ is negative. $\Delta = b^2-4ac = (-4)^2-4(8)(4) \lt 0 \implies \nexists m$ such that $8m^2-4m+4=0
\implies \nexists m$ such that $[-2(m-1)]^2-4m(-m-1) \gt 0$. So doesn't matter what the others conditions says in I (or in II), intersecting them all will give an empty set. And therefore I $\cup$ II will give $\emptyset$.
So I am doing it wrong both ways. Any thoughts on that?
 A: I'm guessing that the reason for $m\neq 0$ is that if $m=0$, then your original equation,
$$mx^2-2(m-1)x - m - 1=0$$
becomes a linear equation, namely $2x - 1 = 0$. Presumably, they are asking that the equation actually have two roots, of which one and only one lies in the desired interval. While it is true that the equation $2x-1=0$ has one and only one solution between $-1$ and $2$ (namely, $x=\frac{1}{2}$), it has no solutions outside the interval, which is probably assumed in the statement.
Note that you did far too much work in the first method. From $(2m-3)(-m+3)\lt 0$, you are set: the two roots are $m=\frac{3}{2}$ and $m=3$; that means that $(2m-3)(m-3)$ takes negative values exactly between them; since you have this product times $-1$ (given the factor $(-m+3)$ instead of $(m-3)$), the negative values for $(2m-3)(-m+3) = -(2m-3)(m-3)$ occur exactly outside that interval: on $(-\infty,\frac{3}{2})$ and on $(3,\infty)$. 
As for the second one, you are not thinking clearly. You want the discriminant positive, yes. The discriminant is $8m^2-4m+4$, yes. And the discriminant of this is negative, yes. And that means is that $8m^2-4m+4$ is never zero, yes. But you don't want it to be zero! You want it to be positive. You are looking for the values where it is zero so you can see where it may change from positive to negative. The fact that it is never zero tells you that it never changes from positive to negative or from negative to positive, so that means that is either always positive or always negative. Plugging in $m=0$ shows you that it is in fact always positive, so that the condition $\Delta \gt 0$ is always satsfied. It is the exact opposite of what you concluded: it's not that the condition is never satisfied, it is always satisfied, so you just need to check the other condition; $\Delta\gt 0$ will always hold.
A: HINT $\ $ The case $\rm\: m = 0\:$ should not be excluded since then $\rm\: f = 2\ x - 1\ $ which has only 1 root (between -1 and 2). If, alternatively, you require that there be another root outside the interval then you need to determine the case(s) where the prior argument breaks down due to the quadratic degenerating to a lower degree polynomial. Here that is precisely when the leading coefficient vanishes, i.e. $\rm\: m = 0\:$. So the argument really is no more difficult than the simple proof I gave in the original thread.
