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Question

For $0<c<b<a$, let $(a+b-2c)x^2+(b+c-2a)x+(c+a-2b)=0$ and $\alpha
eq1$ be one of its roots. Then, among the two statements:

(I) If $\alpha\in(-1,0)$, then $b$ cannot be the geometric mean of $a$ and $c$.

(II) If $\alpha\in(0,1)$, then $b$ may be the geometric mean of $a$ and $c$.

Accepted Answer

Note that $x=1$ always makes the coefficients sum to $(a+b-2c)+(b+c-2a)+(c+a-2b)=0$, so $x=1$ is always a root; the other root $\alpha$ equals $\dfrac{c+a-2b}{a+b-2c}$ by the product-of-roots relation. Analysis of the sign and range of this expression under the ordering $0<c<b<a$ (together with the condition for $b$ to be a geometric mean, $b^2=ac$) confirms both statements (I) and (II) hold, so the answer is that both (I) and (II) are true.

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