Let $p(x)$ be $ax^2 -bx +c =0$ where $a,b,c$ are natural nos. and it has roots lying in the interval $(1,2)$, then find the minimum value of $a,b,c$ individually.
I started writing $p(x)$ as $a(x-\alpha)(x-\beta)$ and made these equations in both the forms:
$\Delta > 0$
$-b/2a$ lies between $1,2$
and $p(1),p(2) >0$
but anyways I cant get the minimum value of $a,b,c$ specifically. What am I missing?
If you multiply out your polynomial to get $ax^2-a(\alpha+\beta)x+a\alpha\beta$ we can see that $a$ can be $1$. It cannot be $0$ if we take "roots" to indicate there are two of them. If $a,\alpha,$ and $\beta$ are all positive, $b \lt 0$ and there is no solution to the problem as stated. If the polynomial is $ax^2-bx+c$ we have $b=a(\alpha+\beta)$ with $\alpha,\beta \gt 1$, so the minimum value of $b$ is $2$. Similarly, $c=a\alpha \beta$ must be at least $2$. These conditions are not independent. In fact $x^2-2x+2=0$ has no real roots at all. We need to increase $b$ to get the parabola to hit the axis at all. I don't find any polynomial that fits the requirement with roots at $1,2$ disallowed by the fact that the interval is open. If you close the interval, $x^2-3x+2=0$ has roots at $1,2$