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..
 A: Firstly we can make some simple remarks about the problem:


*

*let $ax^2-bx+c = f(x)$

*$a,b,c \in \Bbb{N} \Rightarrow a,b,c \ge 0$

*there are $2$ distinct roots, so the parabola is not a degenerate one $\Rightarrow a \ge 1$

*$x_{1,2} \in ]1,2[ \Rightarrow x_1 \cdot x_2 > 1\cdot 1 = 1 \Rightarrow \frac{c}{a} > 1 \Rightarrow a < c \Rightarrow c \ge 2$

*by applying Rolle's Theorem in $]1,2[$ we get $e \in ]1,2[ | f'(e) = 0 = 2\cdot a\cdot e-b = 0$

*solutions $a=-1,-5$ can be excluded

*obviously $x_{1,2} \in \Bbb{R} \Rightarrow \Delta = b^2-4ac > 0$


By fact #5 we get $$\exists e \in ]1,2[ | f'(e) = 2\cdot a\cdot e-b = 0 \Rightarrow\\a=\frac{b}{2e}\Rightarrow\\max\{a\} = \frac{b}{2}, min\{a\} = \frac{b}{4}\Rightarrow\\\frac{b}{4} < a < \frac{b}{2}$$
Now let's plug in some values


*

*b=0 $\Rightarrow 0<a<0$ which is impossible

*b=1 $\Rightarrow \frac14<a<\frac12$, impossible too

*b=2 $\Rightarrow \frac14<a<\frac12$, another impossible case

*b=3 $\Rightarrow \frac34<a<\frac32 \Rightarrow a = 1$ but $x_{1,2} = \frac{3\pm\sqrt{9-4c}}{2} = \frac32 \pm \frac{\sqrt{9-4c}}{2} \notin ]1,2[$ if $c=2$; obviously $c>2 \Rightarrow \Delta <0$

*b=4 $\Rightarrow 1<a<2 \Rightarrow a \in \Bbb{N}$, impossible; so $a=1$ can be excluded

*b=5 $\Rightarrow \frac54<a<\frac52 \Rightarrow a =2$, yet $x_{1,2} = \frac{5\pm\sqrt{25-8c}}{4} = \frac54 \pm \frac{\sqrt{25-8c}}{4} \notin ]1,2[$ for $c\le3$, while $c>3 \Rightarrow \Delta <0$

*b=6 $\Rightarrow \frac32<a<3 \Rightarrow a =2$, but if you check, $x_{1,2}\notin]1,2[$ for any acceptable value of $c$

*b=7 $\Rightarrow \frac74<a<72 \Rightarrow a =2,3$, and you can check these values can fit

*b=8 $\Rightarrow 2<a<4 \Rightarrow a =2 $ can be discarded and the minimum value of $a$ is $5$; infact if $$a=5, b=15, c=11 \Rightarrow x_1 \approx 1.28, x_1 \approx 1.72$$

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