Find the values of $m$ in the 2nd degree equation $(3m-2)x^2+2mx+3m=0$ so that it has only one root between -1 and 0

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

-
Please post the question as part of the body of the message. –  Arturo Magidin Dec 18 '10 at 2:43
Ok, sorry. Thanks for fixing it for me. –  Kaeser Dec 18 '10 at 2:53
@Kaeser: I don't understand your "conditions"; you never say what $a$, $S$, or $\Delta$ are. Also, in any case, you don't want the conditions intersected: that would be asking that all of them happen at the same time. You would want the collection of all values of $m$ that make one true; that's the union of the solutions, not the intersection. –  Arturo Magidin Dec 18 '10 at 2:58
@Kaesur: A nitpick: $x_1+x_2/2\neq(x_1+x_2)/2$. Parentheses are important. –  Jonas Meyer Dec 18 '10 at 3:10
@Kaeser: No: $x_1 + x_2/2$ means $x_1 + (x_2/2)$. You need the parenthesis in order for it to mean $(x_1+x_2)/2$. If you write $r+s/t$, then first you do $s/t$, then you add $r$. Division has precedence over sum. –  Arturo Magidin Dec 18 '10 at 3:21

HINT $\rm\ \ 0\ >\ f(-1)\ f(0)\ =\ 3\:m\:(4m-2)\ \iff\ a < m < b\$ for $\rm\ a = \ldots,\ b = \ldots$

-
@Kaeser: To expand a bit on this: if there is one and only one root between $-1$ and $0$ (and not a repeated root), then the quadratic must have opposite signs at $-1$ and at $0$. This is the first condition Bill writes. –  Arturo Magidin Dec 18 '10 at 3:25
I understand that $f(-1).f(0)<0 \implies -1<x_1<0<x_2$ or $x_1<-1<x_2<0$ with $(x_1<x_2)$. The problem is whether is this what the problem is saying. Suppose I didn't tell you the answer $0<m<1/2$. Is the problem badly written? Or is that making up to the problem also be possible this way $f(-1).f(0)>0 \implies m<0 \bigvee m>1/2$, contradictory? Maybe be even the two scenarios intersected $(0<m<1/2)\bigcap(m<0 \bigvee m>1/2)= \emptyset$ being this the correct answer? –  Kaeser Dec 19 '10 at 17:16
@Kaeser: No, the problem is not badly written; it is pretty clear. You want two distinct roots for the quadratic, with one and only one between 0 and 1. Your interpretation of a double root in the interval does not work (that's two roots, which happen to coincide). The ONLY way for one root to be inside and the other outside $[-1,0]$ to happen is if $f(-1)f(0)\lt 0$. If $f(-1)f(0)\gt 0$, then either you have two roots inside the interval, or no roots inside the interval. If that implies "$m\leq 0$ or $m\geq 0.5$", then the negation (what you want) is "$m\gt 0$ and $m\lt 0.5$". –  Arturo Magidin Dec 19 '10 at 20:31
Thanks a lot everyone. I understand everything now. Just a big interpretation error on my behalf. –  Kaeser Dec 19 '10 at 22:34
Is my answer answer and development using the theorems correct? I ask this also because I think I found a question that is incompatible with this kind of solution in this answer. –  Kaeser Dec 21 '10 at 0:15

Some suggestions... If $ax^2 + bx + c$ is any quadratic with $a \neq 0$, it's a parabola (possibly upside down). The center of the parabola is at $x = -{b \over 2a}$, corresponding to $x = {m \over 2 - 3m}$ in your case. So if ${m \over 2 - 3m}$ lies outside the range $(-1,0)$ then the function is monotone on $[-1,0]$. As a result it has exactly one root in $(-1,0)$ if and only if one of $f(0)$ and $f(-1)$ is positive and one is negative.

If $-1 < {m \over 2 - 3m} < 0$ on the other hand then your function is monotone from $-1$ to ${m \over 2 - 3m}$ and also monotone from ${m \over 2 - 3m}$ to $0$. So you just have to look at the signs of $f(-1)$, $f({m \over 2 - 3m})$, and $f(0)$ and consider the various possibilities.

-
should the center be at $-m/(3m-2)$ instead of $m/(2m-3)$? –  Arturo Magidin Dec 18 '10 at 3:20
yeah it's ${m \over 2 - 3m}$, I'll correct it –  Zarrax Dec 18 '10 at 3:27
Oh thanks, I understood everything you said. But as explained in the other answer it was my interpretation error. –  Kaeser Dec 19 '10 at 22:33

Let us define $f( x)$ as $f( x ) = (3m - 2){x^2} + 2mx + 3m.$

Since there must be only one solution of the equatin, therefore $$(D = 0) \Rightarrow 4m^2 - 12m(3m - 2) = 0\quad {\text{ (i)}}$$

Second equation will be $f'\left( x \right) = 0$ because $f\left( x \right)$ has extreme value at the point $( {x,0} ) \Rightarrow 2(3m - 2)x + 2m = 0$ (ii)

If we solve equation (i) we get ${m}_1 = 0$ and $m_2= \frac{3}{4}$

if we substitute those values into equation (ii) we get for ${m_1} \Rightarrow {x_1} = 0$ and for ${m_2} \Rightarrow {x_2} = \frac{3}{5}$

So, only solution which satisfies both conditions of the question is ${m_1} = 0$

-