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- Feb 14, 2012

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- Thread starter anemone
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- #1

- Feb 14, 2012

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- Nov 29, 2013

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Hello.

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

[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.

- Mar 22, 2013

- 573

The question asks about sum of absolute values.

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- Feb 14, 2012

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I am sorry,Hello.

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

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

Regards.

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- Feb 7, 2012

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- Mar 22, 2013

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@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.

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- Feb 7, 2012

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

- Mar 22, 2013

- 573

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

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- Feb 14, 2012

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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!

- Mar 22, 2013

- 573

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

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- #11

- Feb 7, 2012

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

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A little birdie told me she has seen her doctor, taken some prescribed medicine, and is feeling much better now.Sorry you're not well. I hope you recover in time for Christmas.

- Mar 22, 2013

- 573

Well, that's good news!

I am not quite convinced that some "little birdie" told you that muchMarkFL said:A little birdie told me ...

- Jan 25, 2013

- 1,225

sorry to hear itHey 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!

take care of yourself please ,and hope you will get better soon

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- Feb 14, 2012

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Thank you so muchYes, I got a sign wrong (you can probably see where).

I'd like to share with you the solution suggested by other:

$p,q,r \ne 0$ 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.

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.sorry to hear it

take care of yourself please ,and hope you will get better soon