Consider the quadratic equation $ax^2-bx+c=0, a,b,c \in N. $ If the given equation has two distinct real root... Problem :
Consider the quadratic equation $~ax^2-bx+c=0, \quad a,b,c \in N. ~$ If the given equation has two distinct real roots belonging to the interval $~(1,2)~ $ then the minimum possible values of $~a~$ is
$(i) \quad -1 $
$(ii)\qquad 5 $
$(iii)~~\quad 2 $
$(iv)\quad -5 $
$(v) \qquad1 $
My approach :
We know the condition that two roots will between the two numbers viz. $(1,2)$
$(1) \quad f(1) >0 ; \qquad $
$(2)\quad f(2)>0\qquad$
$(3) \quad 1 < \frac{-b}{2a} <2\qquad $
$(4) \quad D \geq 0$
By using the above I got the following :
$(1) \quad a-b+c >0$
$(2)\quad 4a-2b+c >0$
$(3)\quad 1 < \frac{b}{2a} <2$
$(4) \quad b^2-4ac \geq 0$
Please guide further how to get the answer given in above five options. Thanks..

Let $\alpha,\beta$ be two distinct roots of $ax^2-bx+c=0,$ Where $a,b,c\in \mathbb{N}$
So here $1<\alpha,\beta <2,$ So here $\displaystyle \alpha+\beta = \frac{b}{a}$ and $\displaystyle \alpha \cdot \beta = \frac{c}{a}$
Now Using $\bf{A.M\geq G.M}$ For $0<\alpha, 1-\alpha,\beta, 1-\beta<1$
So $$\frac{1-\alpha+\alpha}{2}\geq \sqrt{\alpha(1-\alpha)}\Rightarrow \alpha (1-\alpha )\leq  \frac{1}{4}$$
Similarly $$\beta(1-\beta) \leq  \frac{1}{4}$$
So we get $$\alpha \beta (1-\alpha)(1-\beta)< \frac{1}{16}$$
So $$\alpha \beta \left[1-(\alpha +\beta)+\alpha\beta\right]< \frac{1}{16}$$
So $$16c\left(a-b+c\right)<a^2$$
Here $a,b,c\in \mathbb{N}.$ So $\min(c(a-b+c)) = 1$
So we get $a^2>16\Rightarrow a>4$
So we get $\min(a) = 5$
