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$f(x)=(a-b)x^2+(b-c)x+(c-a)$

$g(x)=(b-a)x^2+(a+c-2b)x+(b-c)$

given that $a<b $ and $2a^2+b^2+ac<3ab+bc$

Find common root of $f(x)=0 $ and $ g(x)=0$ . Prove $f(x)=0$ has one root in between roots of $g(x)=0$

My Work

First I took roots of $f(x)=0$ as $\alpha,\beta$ and $g(x)=0$ as $\alpha,\gamma$

By substituting $\alpha$ to both equations I was able to find the common root i.e. $x=1$. But I don't know how to proceed. Can you please help me? Thank you!

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A different approach to the first part: If $f(x)=0$ and $g(x)=0$, then also $$0=f(x)+g(x)=(a-b)x+(b-a) $$ and this linear equation has exactly one root $x=1$ (where we use $a\ne b$!). So if they have a common root then it must be $1$. One verifies that $1$ is indeed a root of both.

Once we know one root of a quadratic, we can find the other by dividing out the corresponding linear factor. Or, recall that the product of the roots of $Ax^2+Bx+C$ (with $A\ne 0$) is $\frac CA$ (and the sum of the roots is $-\frac BA$). Knowing that one root is $=1$, the product of the roots is simply the other root. So the other root of $f$ is $\frac {c-a}{a-b}$ and the other root of $g$ is $\frac{b-c}{b-a}$. So the desired result is that $\frac{a-c}{b-a}$ is between $\frac{b-a}{b-a}$ and $\frac{b-c}{b-a}$, or equivalently, $a-c$ is between $b-a$ and $b-c$. From $a<b$, we find $a-c<b-c$, hence what we still need is $b-a<a-c$, or equivalently $$\tag1 b+c<2a.$$ Unfortunaltely,$(1)$ is not a consequence of the given inequalitites. In fact, if $b>a$, then $2a^2+b^2+ac<3ab+bc$ holds for all $c\gg0$, wheras $(1)$ certainly does not. In concreto, we can use this to construct a counterexample: Let $a=2$, $b=4$, $c=1$. Then $a<b$ and $2a^2+b^2+ac<3ab+bc$, but we have $$ \begin{align}f(x)=-2x^2+3x-1&\text{ with roots } \{\tfrac12,1\},\\ g(x)=2x^2-5x+3&\text{ with roots } \{1,\tfrac32\}.\end{align}$$

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