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The quadratic polynomial $ax^2+bx+c$ has positive coefficients $a,b,c$ in A.P. in the given order. If it has integer roots $\alpha,\beta,$ find $\alpha+\beta+\alpha \beta.$

I tried with Vieta's theorem and putting $b=\frac{a+c}{2}$ to get $\alpha+\beta+\alpha \beta=\frac{b}{a}-1=\frac{c-a}{2a}$ but couldn't arrive at a solution.

P.S. The question had the following options given of which one and only one is the correct answer (if they are of any help)-$3,5,7,14$.

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    $\begingroup$ @Nilknarf Thanks for your interest. See the edit. $\endgroup$ – tatan Jul 11 '17 at 14:43
  • $\begingroup$ Hint. You haven't used the fact that the coefficients are in A.P. Show that implies $b/a - 1 = d$, the common difference in the A.P. Since the answer can't be unique (you could just multiply through by any positive integer) the P.S. is very relevant. Perhaps now trial and error? $\endgroup$ – Ethan Bolker Jul 11 '17 at 15:02
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The roots have sum $-\frac{b}{a}$ and product $\frac{c}{a}$, and these sum to $\frac{c-b}{a}=\frac{b}{a}-1$. Comparing this to the sum of the roots, $(\alpha +2)(\beta +2)=3$. As the roots are integers, $\alpha, \,\beta $ are $-1,\,1$ or $-5,\,-3$ in some order. These both give a suitable quadratic, with the desired quantity either $-1$ or $7$.

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  • $\begingroup$ I've already arrived at this. What to do now? $\endgroup$ – tatan Jul 11 '17 at 14:56
  • $\begingroup$ @tatan I've added some more details. $\endgroup$ – J.G. Jul 11 '17 at 15:01
  • $\begingroup$ Thanks. This was really helpful. I lked the intuition that you fatorized it into $(\alpha +2)(\beta+2) $ to get a prime on the R.H.S. $\endgroup$ – tatan Jul 11 '17 at 15:04
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Unpacking J.G.'s answer for the masses:

$ax^2+bx+c$

$$\text{if} \quad (a,b,c) \quad \text{are in "AP", then} \quad $$ $$\begin{align}\begin{cases} a&=a \\ b&=a+d \\ c&=a+2d \end{cases} \ \ &\underbrace{\implies}_{\text{b is the average of a and c}} \ \ \left[\frac{a+c}{2} =\frac{2a+2d}{2}=a+d=b\right] \\ &\qquad \quad \implies \qquad \quad \ \left[c=2b-a\right]\end{align}$$

Vieta: if $\alpha$ and $\beta$ are the roots $\implies \frac{-b}{a}=\alpha + \beta$ and $\frac{c}{a}=\frac{2b-a}{a}=\alpha \beta $

$$\begin{align} \implies \alpha+\beta+\alpha \beta&=\frac{c-b}{a} \\ &=\frac{b-a}{a} \\ &=\frac{b}{a}-1 \\ &=-(\alpha+\beta)-1 \\ \alpha \beta +2(\alpha+\beta)&=-1 \\ \alpha \beta +2(\alpha +\beta)+4&=3 \\ (\alpha+2)(\beta+2)&=3=d_1 \cdot d_2 \\ \text{such that}\quad (d_1,d_2) &\in \{(1,3),(3,1),(-1,-3),(-3,-1) \\ &\begin{cases} \alpha=d_1-2 \\ \beta=d_2-2 \end{cases} \\ \text{so} \quad (\alpha,\beta)&\in\{(-1,1),(1,-1),(-3,-5),(-5,-3) \\ \implies \alpha + \beta + \alpha \beta&=-1 \quad \lor \quad 7 \end{align}$$

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