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Let the polynomial $f(x) = ax^2 – bx + c $ (where $a$, $b$ & $c$ are positive integers). If $f(p) = f(q) = 0$, where $ 0 < p < q < 1$, then find the minimum possible value of $a$.

The vertex is $-\frac{-b}{2a}=\frac{b}{2a}>0$ and lies between $0$ & $1$.

$f(0)>0$ and also $f(1)>0$, hence $c>0$ and $a-b+c>0$, also $b^2-4ac>0$. Even after proceeding up to these steps I am not able to find the minimum value of $a$.

halrankard
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Samar Imam Zaidi
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2 Answers2

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Let the quadratic be $a(x-u)^2-v$, where $u=(p+q)/2$ and $-v=f(u)\lt0$.
Then $f(0)=au^2-v\ge1$ so $au^2\gt1$ and likewise $a(1-u)^2\gt1$. Either $u\le\frac12$ or $1-u\le\frac12$ so $\frac a4\gt1$, and $a\ge5$.
An example is $f(x)=5x^2-5x+1$

Empy2
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You have $ax^2-bx+c = a(x-p)(x-q)=ax^2-a(p+q)x+apq$. So $a(p+q)=b\geq 1$ and $apq=c\geq 1$. This implies $a\geq 1/(p+q)$ and $a\geq 1/pq$. But your constraints on $pq$ force $1/pq>1/(p+q)$. So really the bound is:

$$ a\geq 1/pq. $$

Without more information there is no improvement.

For example let $p=1/3$ and $q=1/2$. Then You can take $a=6$, $b=5$, and $c=1$.

halrankard
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  • Combined with the nice answer by @Empy2 you get $a\geq\max{1/pq,5}$. They have an example where $pq=1/5$. – halrankard Jul 05 '20 at 16:19
  • Would the downvoter like to explain with constructive criticism? My answer provides a potentially better lower bound on $a$ in terms of the roots $p$ and $q$. The other answer gives the smallest "absolute minimum", so makes sense as the accepted answer. But my answer adds a different perspective which could be useful. – halrankard Jul 05 '20 at 17:54