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

[anemone]

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

 

[mente oscura]

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.

Spoiler

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

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

$$-k=pqr$$

$$pq-pr-qr=2011$$

$$p+q+r=0$$

Regards.

 

[mathbalarka]

Re: Find |p|+|q|+|r|

The question asks about sum of absolute values.

 

[anemone]

Re: Find |p|+|q|+|r|

Hello.

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

Spoiler

$$p+q+r=0$$

Regards.

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

 

[Opalg]

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]

 

[mathbalarka]

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]

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).

[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]

 

[mathbalarka]

Re: Find |p|+|q|+|r|

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

 

[anemone]

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]

Re: Find |p|+|q|+|r|

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

 

[Opalg]

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]

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]

Re: Find |p|+|q|+|r|

Well, that's good news!

A little birdie told me ...

I am not quite convinced that some "little birdie" told you that much (Nerd)

 

[Albert1]

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]

Re: Find |p|+|q|+|r|

(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:

Spoiler

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 $|p|+|q|+|r|=98$.

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.:)