algebra-precalculus
This is the question:
Let $a, b$ are integers, and the equation $ax^2+ bx + 1 = 0$ has two distinct positive roots both of which are less than 1. Find the least value of $a$.
You can solve using AM-GM by making $ax^2+ bx + 1 = 0$ a function of F(x), but I am certain this is not the fastest way. Any faster solutions?
The discriminant of any monic polynomial is the product $\prod_{i \neq j} (x_i - x_j)^2$ of the squares of the differences of the roots (in an algebraic closure, e.g. $C$). Cf. the Wikipedia article on this. Consequently, if the roots are all real and distinct, this must be positive.
(If the polynomial is not monic, the factor $a_0^{2n-2}$ is thrown in, for $a_0$ the leading coefficient and $n$ the degree; this is positive for a real polynomial.)
There is a generic way to solve this classical problem:
Find conditions on the coefficients of a quadratic polynomial $p(x)=ax^2+ bx+c$ so that it has two real roots between $x_0$ and $x_1$.
In the present case, all this translates to: $$ \begin{cases} b^2>4c\\ b+c>-1,\quad b-c<1\\-2<b<2 \end{cases} $$
Graphical representation of the solutions as a domain of the plane:
Best Answer
There are a few things we can read off the expression immediately. For simplicity, set $f(x) = ax^2 - bx + 1$ (yes, I swapped a sign, it makes things cleaner below). We get
Here we note that $0<b\leq a$ together with $\frac{b}4\cdot b > 4a$ means we must have $\frac{b}4 > 1$, and therefore $b>4$ and thus also $a>4$. So we try the first possible case after that, which is $a = b = 5$. And that works.