- Thread starter
- Admin
- #1
- Feb 14, 2012
- 3,967
For some integer $k$, the polynomial $x^3-2011x+k$ has three integer roots $p, q, r$. Evaluate $ |p|+|q|+|r|$.
Welcome to our community
Be a part of something great, join today!
Find |p| + |q| + |r|
- Thread starter anemone
- Start date
mente oscura
Well-known member
- Nov 29, 2013
- 172
Re: Find |p|+|q|+|r|
I do not know if I have interpreted the question well.
Regards.
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]
[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
Re: Find |p|+|q|+|r|
The question asks about sum of absolute values.
The question asks about sum of absolute values.
- Thread starter
- Admin
- #4
- Feb 14, 2012
- 3,967
Re: Find |p|+|q|+|r|
I am sorry, mente oscura because that isn't the correct answer.
- Moderator
- #5
- Feb 7, 2012
- 2,820
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 $|p|+|q|+|r| = 176.$
- 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, $(p, q, r) = (49, 39, -88)$ 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.
- Moderator
- #7
- Feb 7, 2012
- 2,820
Re: Find |p|+|q|+|r|
Yes, I got a sign wrong (you can probably see where).
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?
- Mar 22, 2013
- 573
Re: Find |p|+|q|+|r|
Yep, definitely correct; at least that's what I got.
Yep, definitely correct; at least that's what I got.
- Thread starter
- Admin
- #9
- Feb 14, 2012
- 3,967
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!
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
Re: Find |p|+|q|+|r|
I am very sorry to hear that, hope you get better soon.
I am very sorry to hear that, hope you get better soon.
- Moderator
- #11
- Feb 7, 2012
- 2,820
- Admin
- #12
Re: Find |p|+|q|+|r|
A little birdie told me she has seen her doctor, taken some prescribed medicine, and is feeling much better now.
- Mar 22, 2013
- 573
Re: Find |p|+|q|+|r|
Well, that's good news!
Well, that's good news!
I am not quite convinced that some "little birdie" told you that muchMarkFL said:
Albert
Well-known member
- Jan 25, 2013
- 1,225
Re: Find |p|+|q|+|r|
take care of yourself please ,and hope you will get better soon
sorry to hear it
take care of yourself please ,and hope you will get better soon
- Thread starter
- Admin
- #15
- Feb 14, 2012
- 3,967
Re: Find |p|+|q|+|r|
I'd like to share with you the solution suggested by other:
Thank you so much Opalg for participating, yes, your second attempt is correct, well done, Opalg!
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?
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 $|p|+|q|+|r|=98$.
$p,q,r \ne 0$ 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.
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.