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

#### anemone

##### MHB POTW Director
Staff member
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
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.

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

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

$$-k=pqr$$

$$pq-pr-qr=2011$$

$$p+q+r=0$$

Regards.

#### mathbalarka

##### Well-known member
MHB Math Helper
Re: Find |p|+|q|+|r|

#### anemone

##### MHB POTW Director
Staff member
Re: Find |p|+|q|+|r|

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.

#### Opalg

##### MHB Oldtimer
Staff member
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

##### Well-known member
MHB Math Helper
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
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. 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

##### Well-known member
MHB Math Helper
Re: Find |p|+|q|+|r|

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

#### anemone

##### MHB POTW Director
Staff member
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
Re: Find |p|+|q|+|r|

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

##### MHB Oldtimer
Staff member
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

Staff member
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. #### mathbalarka

##### Well-known member
MHB Math Helper
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 #### Albert

##### Well-known member
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
Re: Find |p|+|q|+|r| 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. 