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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
  2. $x_1<-1
  3. $-1

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

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
  • $af(\alpha)=0 \implies \alpha$ is one of the roots.
  • If $af(\alpha)>0$ and $\Delta \geq 0$, then

    (a) $\alpha

    (b) $x_1 \leq x_2 < \alpha$ if $\alpha > S/2$

There is two situations in this problem: $-1

I. $$-10 \wedge af(0)<0 \wedge \Delta > 0 \wedge S/2 > -1.$$ \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. $$x_1<-10 \wedge \Delta > 0 \wedge S/2<0.$$ \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 $0 \lt m \lt 1/2$.

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    Please post the question as part of the body of the message.2010-12-18
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    Ok, sorry. Thanks for fixing it for me.2010-12-18
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    @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.2010-12-18
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    @Arturo Magidin: there are theorems for this. $a$ is from $ax^2+bx+c$, $\Delta=b^2-4ac$, $S=x_1+x_2/2=-b/2a$. There are theorems for comparing a real number $\alpha$ and roots of a 2nd degree equation, let me list them: $a.f(\alpha)>0$ if $\alpha$ is to the right or left of the roots $\alpha2010-12-18
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    @Kaesur: A nitpick: $x_1+x_2/2\neq(x_1+x_2)/2$. Parentheses are important.2010-12-18
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    Your condition (1) is wrong, then: for the two roots to be equal, you need the discriminant (your $\Delta$) to be equal to $0$, not merely nonnegative.2010-12-18
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    $S/2=x_1+x_2/2=-b/2a$2010-12-18
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    @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.2010-12-18
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    Correction: $S/2 = (x_1+x_2)/2 = -b/2a$2010-12-18
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    Although $S/2$ is really the x-coordinate of the vertex the book only mentions it as arithmetic mean value $(x_1+x_2)/2$ of the roots. The reason is that using the theorem that states $a.f(\alpha)>0$ and $\Delta \geq 0$ $\implies$ $\alpha$\alpha$ is to the right of the roots or to the left. So comparing it with $S/2$ is intersting since it is between $x_1$ and $x_2$. $x_1 \leq x_2 \implies x_1 \leq \frac{x_1+x_2}{2} \leq x_2 \implies x_1 \leq S/2 \leq x_2$. I am yet trying to understand the answers given. Anyone to address this? – 2010-12-18
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    @Kaeser: Don't use a period for multiplication. Just use juxtaposition (that is, don't write `a.f(1)`, just write `af(1)`. Also, `<` and `>` sometimes confuse the rendered; try using `\lt` and `\gt` instead. I reformatted so that it is easier to read, and I think I caught all those issues, but if I didn't I'm sorry.2010-12-20
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    @Kaeser: You have a sign error in case I; from $4m^2-36m^2+24\gt )$ you get $32m^2+24\gt 0$. That should be $-32m^2+24\gt 0$.2010-12-20
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    Oh, what would I be without you. Thanks, I will fix it all.2010-12-20
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    So I think evertything is right with the latex and typos. Is the development and answer right?2010-12-20

3 Answers 3

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

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    @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.2010-12-18
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    I understand that $f(-1).f(0)<0 \implies -10 \implies m<0 \bigvee m>1/2$, contradictory? Maybe be even the two scenarios intersected $(01/2)= \emptyset$ being this the correct answer?2010-12-19
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    @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* $m\lt 0.5$".2010-12-19
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    Thanks a lot everyone. I understand everything now. Just a big interpretation error on my behalf.2010-12-19
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    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.2010-12-21
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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.

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    should the center be at $-m/(3m-2)$ instead of $m/(2m-3)$?2010-12-18
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    yeah it's ${m \over 2 - 3m}$, I'll correct it2010-12-18
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    Oh thanks, I understood everything you said. But as explained in the other answer it was my interpretation error.2010-12-19
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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 $