Roots of Quadratic lies in $(0, 1)$ Given that the Quadratic equation $f(x)=ax^2-bx+c=0$ has two roots in $(0 \: 1)$
if $a,b,c \in \mathbb{N}$ Find Minimum values of $a$ and $b$
Since $a$ is Natural number graph of parabola will be open upwards.
Now $f(0) \gt 0$ and $f(1) \gt 0$ so we get
$c \gt 0$ and $a+c \gt b$ and since roots are in $(0 \: 1)$ product of roots also lies in $(0 \: 1)$ 
we get
$$\frac{c}{a} \lt 1$$ or $$a \gt c$$
Also Discriminant $$b^2 \gt 4ac \gt 4c^2$$
now how can we find minimum values of $a$ and $b$

For $f(x) = ax^2 - bx + c$ , we get $f^{'}(x) = 2ax - b$.
Thus the minima is at $\displaystyle f^{'}(x) = 0 \iff 2ax-b = 0 \iff x = {b\over 2a}$. 
Therefore a root must be in $[0, b/2a]$ and one in $[b/2a, 1]$.
And since $f(0) > 0$ therefore $f(b/2a) < 0$
therefore $f(b/2a)  < f(0) \implies c < b^2/4a$
Similarly, $\displaystyle f(1) > f(b/2a) \implies a-b +c > {-b^2\over4a} + c \implies b < 2a $
Since there is no zero at $1$ and $f(1) > 0$ we get $a + c > b$. 
So we need $3$ three numbers that satisfy, 


*

*$c < b^2/4a$

*$b < 2a$

*$a + c > b$


The smallest number $z \in \mathbb{N}$ that satisfy $z > b$ is $b+1$ therefore on comparing this with $(3)$ we get $a = b$ and $c = 1$(*).   
Plugging $a = b$ and $c = 1$ in $(1)$ we get $a = b = 5$ and $c = 1$.

(*) :- If $a = 1$ and $c = b$ then from $(1)$ we get $4 < b$ and from $(2)$ we will get $b < 2$ which is not possible, thus $a = b$ and $c = 1$.
