Minimum value of $a$ for Quadratic Polynomial

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Discussion Overview

The discussion revolves around finding the minimum possible value of the coefficient $a$ in the quadratic polynomial $ax^2 - bx + c$, given that it has two distinct roots $p$ and $q$, where $p > 0$ and $q < 1$. The participants explore the implications of the conditions on the roots and the coefficients being positive integers.

Discussion Character

  • Exploratory
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants point out a potential flaw in the problem's wording regarding the conditions on the roots, suggesting it might imply $p \ne 1$ instead of $q < 1$.
  • One participant expresses uncertainty about the integer nature of $b$, noting that the expression derived from the roots leads to $b$ being a fraction.
  • Another participant reiterates the polynomial form and the conditions, emphasizing the need to find the minimum value of $a$ under the stated constraints.
  • There is a mention of a specific polynomial $(x-4)(x-\frac{1}{2})$ and its implications for the coefficients, which raises questions about the integer requirement for $b$.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the clarity of the problem statement or the implications of the roots on the coefficients. Multiple competing views regarding the interpretation of the conditions exist.

Contextual Notes

The discussion highlights uncertainties regarding the definitions and implications of the roots and coefficients, particularly the integer nature of $b$ and the conditions on $p$ and $q$.

anemone
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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 $a$.
 
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anemone said:
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 $a$.

if a,b ,c all are positive then it cannot have a positive root as per Descartes' rules of sign so no solution
 
kaliprasad said:
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.
 
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
 
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
 
anemone said:
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)

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 like Serena said:
I suspect I'm missing something... (Thinking)

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)
 
$(x-4)(x-\frac{1}{2}) = x^2 -\frac{9}{2}x + 2$ and b is a fraction
I like Serena said:
I suspect I'm missing something... (Thinking)

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)

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

edit: while I was typing anemones post came.
 
anemone said:
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 said:
$(x-4)(x-\frac{1}{2}) = x^2 -\frac{9}{2}x + 2$ and b is a fraction

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

edit: while I was typing anemones post came.

Oh! (Blush)(Blush)
 

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