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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?

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Your formatting really makes things hard to read and follow. You don't put arrows on both previous and next line, just on one of them. And don't use . for multiplication! It's too much like a decimal point. Either use \times or \cdot (the latter will produced a raised dot) – Arturo Magidin Jan 3 '11 at 19:48
I am going to correct it now. Thanks. Right, I'll use \cdot or \times next time. – Kaeser Jan 3 '11 at 19:50
I would like someone to address the theorems solutions, please. – Kaeser Jan 3 '11 at 20:03
There are no "theorems solutions" here. There is your solution to the problem, and I did address it. I explained what your error was: it lies in assuming that because the discriminant of $\Delta$ is never zero, this means that $\Delta$ itself is never positive. That is not the case. – Arturo Magidin Jan 3 '11 at 22:11

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.

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@Arturo I see now why $m=0$ can't possible be a solution, since linear equations doesn't have roots at all. Yet thinking... – Kaeser Jan 3 '11 at 20:20
@Kaeser Linear equations do have roots. Well, one root, anyway. The linear equation $mx + b = 0$ has its root at $x = -\frac{b}{m}$. I think Arturo was saying that the case when $m = 0$ violates the spirit of the problem, since the original question had to do with a quadratic equation and $m = 0$ makes the equation no longer quadratic. – Alex Basson Jan 3 '11 at 20:56
Is it that $\Delta \gt 0$ is always satisfied because $8m^2-4m+4$ has its $a \gt 0$ and $\Delta \lt 0$, such that all $m$'s make it positive? – Kaeser Jan 3 '11 at 21:53
@Kaeser: When you have a quadratic with negative discriminant, it is either always positive or always negative; in particular, if when evaluated at $0$ it is positive, then it is always positive, and if when evaluated at $0$ it is negative, then it will always be negative. In this case, the quadratic $8m^2-4m+4$ has negative discriminant, and at $0$ the value is $4$, which is positive. Since it can never cross the $x$-axis (it is never equal to $0$, because its discriminant is negative), and it is positive in at least one place, it is always positive. So positive for all values of $m$. – Arturo Magidin Jan 3 '11 at 22:13
@Kaeser: I'll add that you see some of the major disadvantages of your notation. You are using, in your next to last comment, $\Delta$ to mean two different things. The first time you use it, it refers to $8m^2-4m+4$; the second, it refers to $(-4)^2-4(4)(8)$. This is likely at least partly to blame for your confusion. You should never use the same symbol to mean two different things at the same time. – Arturo Magidin Jan 3 '11 at 22:40

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.

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@Bill: It all depends on the exact phrasing. If they want one root in the interval, and the other one outside, then $m=0$ is excluded because it fails the second condition. – Arturo Magidin Jan 3 '11 at 19:50
@Arturo: But, if that's what's desired, then it is obvious what to do in this degenerate case - as I remarked. – Bill Dubuque Jan 3 '11 at 20:02
So $f(a)f(b) \lt 0$ is never fully sufficient to find out $m$? Does it means I need always plug $m=0$ additionally, and then, it will be enough? Thanks for the answers. – Kaeser Jan 3 '11 at 20:03
No: my point is that if you want the equation to have TWO roots (as opposed to just one) you need to make sure the equation remains a quadratic. In this case, the leading coefficient is $m$: if $m=0$, then the leading coefficient is $0$, and the equation is not a quadratic, but a linear equation. It's not that $0$ is magical in some way, it's that this is what makes the leading coefficient vanish in this case. – Arturo Magidin Jan 3 '11 at 20:04
@Bill: Yes, I agree; I'm just guessing why they are excluding $m=0$. We are not seeing the original wording of the problem, just the report of what the answer "is supposed to be", and the report of what the problem "is". Of course it's easy to see what to do in the degenerate case; this has never stopped people who write precalculus drill from doing otherwise. – Arturo Magidin Jan 3 '11 at 20:06

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