Let a,b,c be in an arithmetic progression.

Question

Let $a,b,c$ be in an arithmetic progression. Let the centroid of the triangle with vertices $(a,c),(2,b),(a,b)$ be $\left(\dfrac{10}3,\dfrac73\right)$. If $\alpha,\beta$ are the roots of $ax^2+bx+1=0$, then the value of $\alpha^2+\beta^2-\alpha\beta$ is

Accepted Answer

Since $a,b,c$ are in AP, $2b=a+c$. Centroid conditions give $\dfrac{2a+2}3=\dfrac{10}3 \implies a=4$, and $\dfrac{c+2b}3=\dfrac73 \implies c+a+c=7 \implies 2c=3 \implies c=\tfrac32$, so $2b=4+\tfrac32 \implies b=\tfrac{11}4$.

For $ax^2+bx+1=0$: $\alpha+\beta=-\dfrac ba=-\dfrac{11}{16}$, $\alpha\beta=\dfrac1a=\dfrac14$.

$$\alpha^2+\beta^2-\alpha\beta=(\alpha+\beta)^2-3\alpha\beta=\frac{121}{256}-\frac34=\frac{121-192}{256}=-\frac{71}{256}.$$

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