Consider a polynomial of the form $ax^2 - bx + c$ with integer coefficients, such that it has two distinct zeros in the open interval $0 < x < 1.$ Find the least positive integer value of $a$ for which such a polynomial exists.
Open interval? What's that? And how would I approach this problem? Solution are greatly appreciated.
Answer
The open interval $0 < x < 1$ is the set of all real numbers $x$ that are (strictly) greater than $0$ and (strictly) less than $1$. So the question is saying there are two distinct zeros, and both are strictly between $0$ and $1$.
Let $f(x)$ be your polynomial. The minimum of $f(x)$ is at $x = b/(2a)$, which must be between $0$ and $1$. The value of $f(x)$ there is $c - b^2/(4a)$, which must be $< 0$. The values of $f(x)$ at $x=0$ and $x=1$ are
$c$ and $a-b+c$, which must be $>0$.
It seems to me that the least possible positive integer $a$ is $5$, for which
$b=5$, $c=1$ works, for example.
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