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Find |p| + |q| + |r|

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anemone

MHB POTW Director
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Feb 14, 2012
3,755
For some integer $k$, the polynomial $x^3-2011x+k$ has three integer roots $p, q, r$. Evaluate $ |p|+|q|+|r|$.
 

mente oscura

Well-known member
Nov 29, 2013
172
Re: Find |p|+|q|+|r|

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.

[tex](x-p)(x-q)(x-r)=0[/tex]

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

[tex]-k=pqr[/tex]

[tex]pq-pr-qr=2011[/tex]

[tex]p+q+r=0[/tex]


Regards.
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
Re: Find |p|+|q|+|r|

The question asks about sum of absolute values.
 
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anemone

MHB POTW Director
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Feb 14, 2012
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Re: Find |p|+|q|+|r|

Hello.

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



[tex]p+q+r=0[/tex]


Regards.
I am sorry, mente oscura because that isn't the correct answer.
 

Opalg

MHB Oldtimer
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Feb 7, 2012
2,725
Re: Find |p|+|q|+|r|

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 $|p|+|q|+|r| = 176.$
 

mathbalarka

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MHB Math Helper
Mar 22, 2013
573
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.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
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.
(Doh) Yes, I got a sign wrong (you can probably see where).

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?
 

mathbalarka

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MHB Math Helper
Mar 22, 2013
573
Re: Find |p|+|q|+|r|

Yep, definitely correct; at least that's what I got.
 
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anemone

MHB POTW Director
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Feb 14, 2012
3,755
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!
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
Re: Find |p|+|q|+|r|

I am very sorry to hear that, hope you get better soon. :(
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
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 you're not well. I hope you recover in time for Christmas.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Find |p|+|q|+|r|

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)
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
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)
 

Albert

Well-known member
Jan 25, 2013
1,225
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!
sorry to hear it
take care of yourself please ,and hope you will get better soon
 

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,755
Re: Find |p|+|q|+|r|

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

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?
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 any one 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 $|p|+|q|+|r|=98$.



I am very sorry to hear that, hope you get better soon. :(
Sorry you're not well. I hope you recover in time for Christmas.
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.:)