Suppose the quadratic polynomial p(x)=ax^2+bx | vedclass

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(C) Given $p(x)=ax^2+bx+c$ with $a, b, c > 0$ and $b-a=c-b$,which implies $2b=a+c$,so $a, b, c$ are in arithmetic progression.Let $\alpha, \beta$ be t

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$7$

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(C) Given $p(x)=ax^2+bx+c$ with $a, b, c > 0$ and $b-a=c-b$,which implies $2b=a+c$,so $a, b, c$ are in arithmetic progression.Let $\alpha, \beta$ be the integer roots of $p(x)=0$. By Vieta's formulas,$\alpha+\beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.Since $b = \frac{a+c}{2}$,we have $\alpha+\beta = -\frac{a+c}{2a} = -\frac{1}{2} - \frac{c}{2a}$.Also,$\alpha\beta = \frac{c}{a}$. Substituting $\frac{c}{a} = \alpha\beta$ into the sum equation:$\alpha+\beta = -\frac{1}{2} - \frac{\alpha\beta}{2}$ $\Rightarrow 2(\alpha+\beta) = -1 - \alpha\beta$ $\Rightarrow \alpha\beta + 2\alpha + 2\beta = -1$.Adding $4$ to both sides to factorize:$(\alpha+2)(\beta+2) = 3$.Since $\alpha, \beta$ are integers,the possible pairs for $(\alpha+2, \beta+2)$ are $(1, 3), (3, 1), (-1, -3), (-3, -1)$.Case $1$: $(\alpha+2, \beta+2) = (1, 3) \Rightarrow \alpha=-1, \beta=1$. Then $\alpha+\beta+\alpha\beta = -1+1+(-1)(1) = -1$ (Not in range).Case $2$: $(\alpha+2, \beta+2) = (-1, -3) \Rightarrow \alpha=-3, \beta=-5$. Then $\alpha+\beta+\alpha\beta = -3-5+(-3)(-5) = -8+15 = 7$.Since $7$ is in the range $0 \leq \alpha+\beta+\alpha\beta \leq 8$,the possible value is $7$.

Suppose the quadratic polynomial $p(x)=ax^2+bx+c$ has positive coefficients $a, b, c$ such that $b-a=c-b$. If $p(x)=0$ has integer roots $\alpha$ and $\beta$,then what could be the possible value of $\alpha+\beta+\alpha\beta$ if $0 \leq \alpha+\beta+\alpha\beta \leq 8$?

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