<s>[INST] 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..
 [/INST] From Vieta's:
$$2<x_1+x_2=\frac ba<4 \Rightarrow 2a<b<4a\\
1<x_1x_2=\frac ca<4 \Rightarrow a<c<4a\\
0<x_2-x_1<1 \Rightarrow 0<\frac{\sqrt{b^2-4ac}}{a}<1 \Rightarrow 0<b^2-4ac<a^2$$
Also:
$$f(1)=a-b+c>0 \Rightarrow c>b-a\\
f(2)=4a-2b+c>0 \Rightarrow c>2b-4a$$
Cases: $a=1$
$$2<b<4 \Rightarrow b=3\\
1<c<4\Rightarrow c=2,3\\
 0<9-4c<1 \Rightarrow \emptyset$$
$a=2$:
$$4<b<8\Rightarrow b=5,6,7\\
2<c<8 \Rightarrow 3,4,5,6,7\\
0<b^2-8c<4 \ \text{and} \ c>b-a \ \text{and} \ c>2b-4a \Rightarrow \emptyset$$
So, by method of elimination, the answer is (if there is a single correct answer) $a=5$.

Let's consider $a=5$ as well:
$$10<b<20 \Rightarrow b=11,12,...,19\\
2<c<20 \Rightarrow c=3,4,...,19\\
0<b^2-20c<25 \ \text{and} \ c>b-5 \ \text{and} \ c>2b-20 \Rightarrow \\
(b,c)=(15,11),(18,15) \Rightarrow \\
5x^2-15x+11=0 \Rightarrow x_{1,2}\approx 1.28; 1.72 \ \color{green}{\checkmark}\\
5x^2-18x+15=0 \Rightarrow x_{1,2}\approx 1.31; 2.29 \not\in (1,2).$$
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