Minimum value of $a$ for Quadratic Polynomial

[anemone]

The quadratic polynomial $ax^2+bx+c$ has two distinct roots $p$ and $q$, with $a,\,b,\,c$ are positive integers and with $p>0$ and $q<1$.

Find the minimum possible value of $a$.

 

[kaliprasad]

The quadratic polynomial $ax^2+bx+c$ has two distinct roots $p$ and $q$, with $a,\,b,\,c$ are positive integers and with $p>0$ and $q<1$.

Find the minimum possible value of $a$.

Spoiler

if a,b ,c all are positive then it cannot have a positive root as per Descartes' rules of sign so no solution

 

[anemone]

Spoiler

if a,b ,c all are positive then it cannot have a positive root as per Descartes' rules of sign so no solution

Ops...typo! It was a typo! (Tmi)

The problem should read as:

The quadratic polynomial $ax^2-bx+c$ has two distinct roots $p$ and $q$, with $a,\,b,\,c$ are positive integers and with $p>0$ and $q<1$.

Find the minimum possible value of $a$.

Sorry for my careless mistake and thanks for catching my error.

 

[kaliprasad]

Spoiler

as b is -ve and c is positive so both roots are positive and hence one root is between 0 and 1 and another anything
by looking at rational roots
so let one root be $\frac{r}{s}$ ( s > 1) and another d/e we need to minimize se (s = 2 and e = 1 shall do)

we get $(2x-1)(x-1) = 2x^2- 3x +1 = 0$

so 2 is the lowest a

checking for irrational we have $( 1- \frac{1}{sqrt(2)})$ and its conjugate we get 2
so lowest is 2

 

[anemone]

You're right, kaliprasad and I think this problem has its own flaw since the wording where $p>0$ and $q<1$ isn't that clear because I suspect it might actually want to convey the meaning of $p>0$ and $p\ne 1$.

Thanks for participating, my friend! :D

 

[I like Serena]

The quadratic polynomial $ax^2-bx+c$ has two distinct roots $p$ and $q$, with $a,\,b,\,c$ are positive integers and with $p>0$ and $q<1$.

Find the minimum possible value of $a$.

I suspect I'm missing something... (Thinking)

Spoiler

Suppose we pick $(x-4)(x-\frac 12) = x^2-2x+2$ with $p=4$ and $q=\frac 12$.

[table="class: grid"]
[tr]
[td]quadratic polynomial $ax^2-bx+c$[/td]
[td]$\checkmark$[/td]
[/tr]
[tr]
[td]two distinct roots $p$ and $q$[/td]
[td]$\checkmark$[/td]
[/tr]
[tr]
[td]$a,\,b,\,c$ are positive integers[/td]
[td]$\checkmark$[/td]
[/tr]
[tr]
[td]$p>0$ and $q<1$[/td]
[td]$\checkmark$[/td]
[/tr]
[/table]

That gives me a minimum value $a=1$, which is the smallest positive integer value.
What am I missing? (Wondering)

 

[anemone]

I suspect I'm missing something... (Thinking)

Spoiler

Suppose we pick $(x-4)(x-\frac 12) = x^2-2x+2$ with $p=4$ and $q=\frac 12$.

[TABLE="class: grid"]
[TR]
[TD]quadratic polynomial $ax^2-bx+c$[/TD]
[TD]$\checkmark$[/TD]
[/TR]
[TR]
[TD]two distinct roots $p$ and $q$[/TD]
[TD]$\checkmark$[/TD]
[/TR]
[TR]
[TD]$a,\,b,\,c$ are positive integers[/TD]
[TD]$\checkmark$[/TD]
[/TR]
[TR]
[TD]$p>0$ and $q<1$[/TD]
[TD]$\checkmark$[/TD]
[/TR]
[/TABLE]

That gives me a minimum value $a=1$, which is the smallest positive integer value.
What am I missing? (Wondering)

I think $(x-4)(x-\frac 12) = x^2-\dfrac{9x}{2}+2$ and in that case, we see that $b\ne \text{integer}$.(Thinking)

 

[kaliprasad]

$(x-4)(x-\frac{1}{2}) = x^2 -\frac{9}{2}x + 2$ and b is a fraction

I suspect I'm missing something... (Thinking)

Spoiler

Suppose we pick $(x-4)(x-\frac 12) = x^2-2x+2$ with $p=4$ and $q=\frac 12$.

[table="class: grid"]
[tr]
[td]quadratic polynomial $ax^2-bx+c$[/td]
[td]$\checkmark$[/td]
[/tr]
[tr]
[td]two distinct roots $p$ and $q$[/td]
[td]$\checkmark$[/td]
[/tr]
[tr]
[td]$a,\,b,\,c$ are positive integers[/td]
[td]$\checkmark$[/td]
[/tr]
[tr]
[td]$p>0$ and $q<1$[/td]
[td]$\checkmark$[/td]
[/tr]
[/table]

That gives me a minimum value $a=1$, which is the smallest positive integer value.
What am I missing? (Wondering)

Spoiler

$(x-4)(x-\frac{1}{2}) = x^2 -\frac{9}{2}x + 2$ and b is a fraction

edit: while I was typing anemones post came.

 

[I like Serena]

I think $(x-4)(x-\frac 12) = x^2-\dfrac{9x}{2}+2$ and in that case, we see that $b\ne \text{integer}$.(Thinking)

$(x-4)(x-\frac{1}{2}) = x^2 -\frac{9}{2}x + 2$ and b is a fraction

Spoiler

$(x-4)(x-\frac{1}{2}) = x^2 -\frac{9}{2}x + 2$ and b is a fraction

edit: while I was typing anemones post came.

Oh! (Blush)(Blush)