Suppose the quadratic polynomial $$p(x)=ax^2 + bx + c$$ has positive coefficients $a, b, c$ such that these are in AP in the given order. If $m$ and $n$ are the integer zeros of the polynomial then what is the value of $m+n+mn$? I have tried quadratic formula but ain't getting the answer.


Since $m$ and $n$ are roots of the polynomial we have \begin{align*} ax^2+bx+c&=a\left(x-m\right)\left(x-n\right)\\ &=ax^2-a(m+n)x+amn \end{align*} Then, $$m+n=-\frac ba\quad\text{and}\quad mn=\frac ca$$ So, since $a,b$ and $c$ are in AP we get $$m+n+mn=-\frac ba+\frac ca=\frac{c-b}a=\frac{b+b-a-b}a=\frac{b-a}a=\frac ba-1$$ and $\frac ba =-(m+n)$, then $$m+n+mn=-(m+n)-1\quad \implies \quad mn+2(m+n)+4=3$$ So $$(m+2)(n+2)=3$$ Now, since $3$ is a prime we have $(m,n)\in\left\{(-1,1),(1,-1),(-3,-5),(-5,-3)\right\}$. Notice that $(m,n)=(-1,1)$ or $(m,n)=(1,-1)$ implies $b=0$, but we know that $a,\,b$ and $c$ are positive. So, $(m,n)=(-3,-5)$ or $(-5,-3)$, and then $$\boxed{m+n+mn=-8+15=7}$$

  • $\begingroup$ The answer is an integer $\endgroup$ – Arunabh Feb 8 '18 at 5:20
  • $\begingroup$ Or more specifically the answer is a whole number $\endgroup$ – Arunabh Feb 8 '18 at 5:26
  • $\begingroup$ How can u take the common difference as b for the given AP? $\endgroup$ – Arunabh Feb 8 '18 at 6:12
  • $\begingroup$ Arunabh I don't take the common difference as $b$. I put $$c=b+\underbrace{b-a}_{\text{common diff.}}$$ $\endgroup$ – Ángel Mario Gallegos Feb 8 '18 at 6:20
  • $\begingroup$ Just one more doubt how did u obtain the values (m,n)∈{(−1,1),(1,−1),(−3,−5),(−5,−3)} $\endgroup$ – Arunabh Feb 8 '18 at 9:24

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