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

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

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

given that $$a and $$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!

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 $$a, we find $$a-c, hence what we still need is $$b-a, or equivalently $$\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 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}