Consider the quadratic equation $ax^2-bx+c=0, a,b,c \in N. $ If the given equation has two distinct real root...

Question

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..

Answer 1

Firstly we can make some simple remarks about the problem:

1. let $ax^2-bx+c = f(x)$
2. $a,b,c \in \Bbb{N} \Rightarrow a,b,c \ge 0$
3. there are $2$ distinct roots, so the parabola is not a degenerate one $\Rightarrow a \ge 1$
4. $x_{1,2} \in ]1,2[ \Rightarrow x_1 \cdot x_2 > 1\cdot 1 = 1 \Rightarrow \frac{c}{a} > 1 \Rightarrow a < c \Rightarrow c \ge 2$
5. by applying Rolle's Theorem in $]1,2[$ we get $e \in ]1,2[ | f'(e) = 0 = 2\cdot a\cdot e-b = 0$
6. solutions $a=-1,-5$ can be excluded
7. 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 $$a=5, b=15, c=11 \Rightarrow x_1 \approx 1.28, x_1 \approx 1.72$$

Answer 2

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 $\min(a) = 5$