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 $b$
 A: 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 $c = 1$.
A: By inspection, it looks like $y = f(x) = 4x^2 - 4x + 1$ is a candidate.  
Analysis:


*

*$f(0) = c > 0$ because $c \in \mathbb{N}$. 


Thus, the parabola is open upwards because it has to cross two points $0 < x_1, x_2 < 1$ on the $x$ axis and passing the point $(0,c)$ in the first quadrant.


*

*Because the parabola is open upwards, $f(1)$ must be greater than zero:


$$ f(1) = a - b + c > 0 \implies b < a + c. \quad (1)$$


*

*The product of two roots: 


$$x_1\cdot x_2 < 1 \implies c/a < 1 \implies a > c. \quad (2)$$


*

*Finally, to have two real roots:


$$ b^2 - 4ac \ge 0 \implies b^2 \ge 4ac. \quad (3)$$ 
We will tabulate all possible values that satisfy the three conditions above.
A. Case $c = 1:$


*

*According to $(2)$: $\,a = 2, 3, 4, 5, 6, 7, 8, \ldots$


*

*For $a=2:$ 


*

*$(1)$ says $b < 3 \implies b = 1, 2.$

*$(3)$ says $b^2 \ge 8 \implies b = 3, 4, 5, \ldots$

*Thus, no values of $b$ satisfy both conditions.


*For $a=3:$ 


*

*$(1)$ says $b < 4 \implies b = 1, 2, 3.$

*$(3)$ says $b^2 \ge 12 \implies b = 4, 5, 6, \ldots$

*Thus, no values of $b$ satisfy both conditions.


*For $a=4:$ 


*

*$(1)$ says $b < 5 \implies b = 1, 2, 3, 4.$

*$(3)$ says $b^2 \ge 16 \implies b = 4, 5, 6, \ldots$

*Thus, only $b = 4$ satisfies both conditions.

*The first solution is $a=4,\; b =4\; c=1$


*For $a=5:$


*

*$(1)$ says $b < 6 \implies b = 1, 2, 3, 4, 5.$

*$(3)$ says $b^2 \ge 20 \implies b = 5, 6, \ldots$

*Thus, only $b = 5$ satisfies both conditions.

*The second solution is $a=5,\; b =5\; c=1$


*For $a=6:$


*

*$(1)$ says $b < 7 \implies b = 1, 2, 3, 4, 5, 6.$

*$(3)$ says $b^2 \ge 24 \implies b = 5, 6, \ldots$

*Thus, only $b = 6$ satisfies both conditions.

*The second solution is $a=6,\; b =6\; c=1$




In general, we get a set of solutions $a = b = 4, 5, 6, \ldots$ for $c=1$. The resulting parabola is given by:
$$ y = f(x) = ax^2 -ax + 1 = a (x-1/2)^2 + 1 - a/4, \quad a = 4, 5, 6,\ldots $$
B. Case $c = 2:$


*

*According to (2), $a = 3, 4, 5, 6, 7, 8, \ldots$


*

*For $a=3:$ 


*

*$(1)$ says $b < 5 \implies b = 1, 2, 3, 4.$

*$(3)$ says $b^2 \ge 24 \implies b = 5, 6, 7, \ldots$

*Thus, no value of $b$ satisfies both conditions.


*For $a=4:$ 


*

*$(1)$ says $b < 6 \implies b = 1, 2, 3, 4, 5.$

*$(3)$ says $b^2 \ge 32 \implies b = 6, 7, 8, \ldots$

*Thus, no value of $b$ satisfies both conditions.




We can stop here because the next values of $a$ and $b$ to check will exceed the minimum values found in case A.
Thus, the solution with minimum values of $a$ and $b$ is:
$$ y = 4x^2 - 4x +1.$$
