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.

Question

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

Answer 1

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

Answer 2

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.

Answer 3

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 $