Evaluate |p|+|q|+|r| for $x^3-2011x+k$

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Discussion Overview

The discussion revolves around evaluating the expression |p|+|q|+|r| for the polynomial $x^3-2011x+k$, where $p$, $q$, and $r$ are integer roots. Participants explore the relationships between the roots and the implications for the value of $k$.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants note the relationships between the roots, specifically that $p+q+r=0$ and $pq+pr+qr=-2011$.
  • One participant proposes a method involving completing the square to derive a relationship between $p$ and $q$, leading to the conclusion that both must be odd numbers ending in $1$ or $9$.
  • A participant initially suggests $(p, q, r) = (49, 39, -88)$, but another participant challenges this by stating it does not yield a valid $k$.
  • After correcting a sign error, a participant presents a revised solution with $(p, q, r) = (10, 39, -49)$, resulting in $|p| + |q| + |r| = 98$ and $k = 19,110$.
  • Several participants express uncertainty about the correctness of earlier claims and engage in clarifying discussions regarding the calculations and assumptions made.
  • There are personal exchanges regarding one participant's health, which are not directly related to the mathematical discussion.

Areas of Agreement / Disagreement

There is no consensus on the initial proposed values of the roots, as participants challenge and refine each other's claims. The discussion includes multiple competing views on the correct values of $p$, $q$, and $r$, and the corresponding value of $k$ remains a point of contention.

Contextual Notes

Participants express uncertainty about the separability of the cubic polynomial over $\mathbb{Z}[x]$ and the implications of the derived relationships. There are unresolved mathematical steps and assumptions regarding the roots.

anemone
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For some integer $k$, the polynomial $x^3-2011x+k$ has three integer roots $p, q, r$. Evaluate $ |p|+|q|+|r|$.
 
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Re: Find |p|+|q|+|r|

anemone said:
For some integer $k$, the polynomial $x^3-2011x+k$ has three integer roots $p, q, r$. Evaluate $ |p|+|q|+|r|$.

Hello.

I do not know if I have interpreted the question well.

(x-p)(x-q)(x-r)=0

x^3-(p+q+r)x^2+(pq-pr-qr)x-pqr=0

-k=pqr

pq-pr-qr=2011

p+q+r=0

Regards.
 
Re: Find |p|+|q|+|r|

The question asks about sum of absolute values.
 
Re: Find |p|+|q|+|r|

mente oscura said:
Hello.

I do not know if I have interpreted the question well.

p+q+r=0

Regards.

I am sorry, mente oscura because that isn't the correct answer.
 
Re: Find |p|+|q|+|r|

[sp]The relations between the roots are $p+q+r=0$ and $pq+pr+qr= -2011$. Writing $r=-(p+q)$ in the second one, we get $pq - (p+q)^2 = -2011$, or p^2-pq+q^2 = 2011. Multiply by $4$ and complete the square: $(2p-q)^2 + 3q^2 = 8044$. A bit of calculation and guesswork leads to the conclusion that $p$ and $q$ must both be odd numbers ending in $1$ or $9$. Then a few minutes with a calculator comes up with the solution $p=49$, $q=39$. Then $r = -88$, and $|p|+|q|+|r| = 176.$[/sp]
 
Re: Find |p|+|q|+|r|

@Opalg, $(p, q, r) = (49, 39, -88)$ does not seem to yield a $k$ such that the cubic is separable over $\mathbb{Z}[x]$, let alone being each of $p, q$ and $r$ a root.
 
Re: Find |p|+|q|+|r|

mathbalarka said:
@Opalg, $(p, q, r) = (49, 39, -88)$ does not seem to yield a $k$ such that the cubic is separable over $\mathbb{Z}[x]$, let alone being each of $p, q$ and $r$ a root.
(Doh) Yes, I got a sign wrong (you can probably see where).

[sp]It should have been $(2p+q)^2 + 3q^2 = 8044$, and the result of that is $p=10$, $q=39$, and therefore $r=-49$, so that $|p| + |q| + |r| = 98$. The value of $k$ is then $-pqr = 19\,110$. Better?[/sp]
 
Re: Find |p|+|q|+|r|

Yep, definitely correct; at least that's what I got.
 
Re: Find |p|+|q|+|r|

Hey MHB,

I will reply to this thread after I gained much more energy, because now I am very sick and ill with symptoms including stomach pain, vomiting, diarrhea and nausea.

Sorry guys!
 
  • #10
Re: Find |p|+|q|+|r|

I am very sorry to hear that, hope you get better soon. :(
 
  • #11
Re: Find |p|+|q|+|r|

anemone said:
Hey MHB,

I will reply to this thread after I gained much more energy, because now I am very sick and ill with symptoms including stomach pain, vomiting, diarrhea and nausea.
Sorry you're not well. I hope you recover in time for Christmas.
 
  • #12
Re: Find |p|+|q|+|r|

Opalg said:
Sorry you're not well. I hope you recover in time for Christmas.

A little birdie told me she has seen her doctor, taken some prescribed medicine, and is feeling much better now. (Clapping)
 
  • #13
Re: Find |p|+|q|+|r|

Well, that's good news!

MarkFL said:
A little birdie told me ...

I am not quite convinced that some "little birdie" told you that much (Nerd)
 
  • #14
Re: Find |p|+|q|+|r|

anemone said:
Hey MHB,

I will reply to this thread after I gained much more energy, because now I am very sick and ill with symptoms including stomach pain, vomiting, diarrhea and nausea.

Sorry guys!
sorry to hear it
take care of yourself please ,and hope you will get better soon
 
  • #15
Re: Find |p|+|q|+|r|

Opalg said:
(Doh) Yes, I got a sign wrong (you can probably see where).

[sp]It should have been $(2p+q)^2 + 3q^2 = 8044$, and the result of that is $p=10$, $q=39$, and therefore $r=-49$, so that $|p| + |q| + |r| = 98$. The value of $k$ is then $-pqr = 19\,110$. Better?[/sp]

Thank you so much Opalg for participating, yes, your second attempt is correct, well done, Opalg!:)

I'd like to share with you the solution suggested by other:
With Vieta's formula we have $p+q+r=0$ and $pq+qr+pr=-2011$.

$p,q,r \ne 0$ since anyone being zero will make the other $2\pm\sqrt{2011}$.

$\therefore a=-(b+c)$

WLOG, let $|p|\ge |q|\ge |r|$.

If $p>0$, then $q,r<0$ and if $p<0$, then $q,r>0$.

$pq+qr+pr=-2011=p(q+r)+qr=-p^2+qr$

$p^2=2011+qr$

We know that $q, r$ have the same sign. So $|p|\ge 45$. ($44^2<2011$ and $45^2=2025$)

Also, $qr$ maximize when $q=r$ if we fixed $p$. Hence, $2011=p^2-qr>\dfrac{3p^2}{4}$.

So $p^2<\dfrac{4(2011)}{3}=2681+\dfrac{1}{3}$ but $52^2=2704$ so we have $|p|\ge 51$

Now we have limited $p$ to $45 \le |p| \le 51$.

A little calculation leads us to the case where $|p|=49$, $|q|=39$ and $|r|=10$ works. Hence $|p|+|q|+|r|=98$.
mathbalarka said:
I am very sorry to hear that, hope you get better soon. :(

Opalg said:
Sorry you're not well. I hope you recover in time for Christmas.

Albert said:
sorry to hear it
take care of yourself please ,and hope you will get better soon

Thank you so much for yours concern, as Mark has already mentioned, I felt much better after taking the medicines, but I am not feeling completely okay yet, because I vomited last midnight but I am positively that I will be okay after finishing all of the prescribed medicines.:)
 

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