Problem Statement:-
If the equation $ax^2-bx+c=0$ has two distinct real roots b/w $1$ and $2$ where $a,b,c\in\Bbb{N}$, then show that $a\ge5$ and $b\ge11$.
Attempt at a solution:-
As the roots are real and distinct then $b^2-4ac\gt0\implies b^2\gt4ac$
Also, as the roots lie in b/w $1$ and $2$, hence $a-b+c\gt0$ and $4a-2b+c\gt0$.
As the roots are greater than $1$, so the product of the roots must be greater than $1$, hence $c\gt a$.
Due to the roots of the equation lying $ax^2-bx+c=0$ in b/w $1$ and $2$, hence $$1\lt\dfrac{b}{2a}\lt2\implies 2a\lt b\lt4a$$.
Now, let's arrange the inequalities in an orderly fashion. $$b^2\gt4ac\tag{1}$$ $$b\lt a+c\tag{2}$$ $$2b\lt4a+c\tag{3}$$ $$c\gt a \tag{4}$$ $$2a\lt b\lt4a\tag{5}$$
From the above set of inequalities it can be concluded that $a$ will have the smallest value. So, lets start by assigning $a=1$. Then we get the following set of inequalities for $b$ and $c$. $$b^2\gt4c\\ b\lt1+c\\ 2b\lt4+c\\ c\gt1\\ 2\lt b\lt4$$
On solving all of these inequalities for $b$ and $c$, we get that there are no natural numbers which satisfy $b$ and $c$ simultaneously.
Going on with this process we find that the first set of natural numbers which satisfy the set of inequalities occurs when $a=5$, which are $b=15$ and $c=11$. And from the inequalities it is pretty clear that this is the least value of $b$ as the bound for $b$ in $2a\lt b \lt 4a$ keeps increasing.
My deal with the question:-
The question states that the least value of $b$ has to be proven to be $11$ but at that value combined with $c=11\; a=5$ we get a root to be one too. But doesn't the question states that the roots ought to be in b/w the $1$ and $2$, or did I understand the question wrong.
Also, please do submit your solutions which can teach me something new at high school level.