A problem on algebric expansion and indices.

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sankalpmittal
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Homework Statement



If :
p2+q2+4r2 = pq+2qr+2rp

Prove that :
p=q=2r

Homework Equations



I am not sure but I tried using this equation :

(a+b+c)2 = a2+b2+c2+2(ab+bc+ca)

The Attempt at a Solution



p2+q2+4r2 = pq+2qr+2rp
p2+q2+4r2 -pq-2qr-2rp = 0

Now I can't factor. Its not yielding correct answer. Please help.
 
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sankalpmittal said:

Homework Statement



If :
p2+q2+4r2 = pq+2qr+2rp

Prove that :
p=q=2r
Hi sankalpmittal!
It's deceptively easy. (Or do I mean, it's deceptively difficult?) :confused: I find some of these idioms confusing.

Just take out a single first-degree factor; do this a few times. I'll start you off:

p(p-q) + ... = 0
 
The result can also be obtained by writing the equation as a quadratic in r (I think any of the variables will do.). Using the quadratic formula gives a solution for r with √(-(p-q)2 ) which is real only if p=q. The rest follows from there.
 
You can use linear algebra. The quadratic form:
[tex] p^2 + q^2 + 4 r^2 - p q - 2 q r - 2 p r = 0[/tex]
can be written in matrix form:
[tex] (p, q, r) \cdot \left(\begin{array}{ccc}<br /> 1 & -\frac{1}{2} & -1 \\<br /> <br /> -\frac{1}{2} & 1 & - 1 \\<br /> <br /> -1 & -1 & 4<br /> \end{array}\right) \cdot \left( \begin{array}{c}<br /> p \\<br /> <br /> q \\<br /> <br /> r<br /> \end{array}\right) = 0[/tex]
The eigenvalue problem for this matrix is:
[tex] \left\vert \begin{array}{ccc}<br /> 1 - \lambda & -\frac{1}{2} & -1 \\<br /> <br /> -\frac{1}{2} & 1 - \lambda & - 1 \\<br /> <br /> -1 & -1 & 4 - \lambda<br /> \end{array}\right\vert = 0[/tex]

[tex] (1 - \lambda) \left\vert \begin{array}{cc}<br /> 1 - \lambda & -1 \\<br /> -1 & 4 - \lambda<br /> \end{array}\right\vert - \left(-\frac{1}{2}\right) \left\vert \begin{array}{cc}<br /> -\frac{1}{2} & -1 \\<br /> -1 & 4 - \lambda<br /> \end{array}\right\vert + (-1) \left\vert \begin{array}{cc}<br /> -\frac{1}{2} & 1 - \lambda \\<br /> <br /> -1 & -1<br /> \end{array} \right\vert = 0[/tex]
[tex] (1 - \lambda) (3 - 5 \lambda + \lambda^2) + \frac{1}{2} ( -3 + \frac{\lambda}{2} ) - (-\lambda + \frac{3}{2}) = 0[/tex]
[tex] \left. -\lambda^3 + 6 \lambda^2 - 8 \lambda + 3 - \frac{3}{2} + \frac{\lambda}{4} + \lambda - \frac{3}{2} = 0 \right/ \cdot (-4)[/tex]
[tex] 4 \lambda^{3} - 24 \lambda^{2} + 32 \lambda - 12 + 6 - \lambda - 4 \lambda + 6 = 0[/tex]
[tex] 4 \lambda^{3} - 24 \lambda^2 + 27 \lambda = 0[/tex]
[tex] \lambda (4 \lambda^2 - 24 \lambda + 27) = 0[/tex]
 
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[STRIKE]None of those rationals seem to be roots of the cubic equation. Are you sure you haven't made a typo? Or, can you see a typo in my calculations?[/STRIKE] :smile:

EDIT:

I found one error and corrected my pervious post. Now there are more options and the cubic equation is different. Can you check whether some of these rationals are roots?
 
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[STRIKE]I found one rational root is [itex]\lambda = \frac{1}{2}[/itex]. After doing long division of[/STRIKE]

[STRIKE]you get a quadratic equation whose roots you can find using the quadratic formula instead of going through the other cases.[/STRIKE]

EDIT:

Scratch that! I corrected one more error. The resulting characteristic equation is correct and easy to solve.
 
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Multiply the expression by 2:
[tex] 2 p^2 + 2 q^2 + 8 r^2 - 2 p q - 4 q r - 4 p r = 0[/tex]
[tex] (p^2 - 2 p q + q^2) + (p^2 - 4 p r + 4 r^2) + (q^2 - 4 q r + 4 r^2) = 0[/tex]
 
I don't know which method is correct. Everyone is giving different opinions well. I start off from last post.

Hi Dickfore ,

(p-q)2 + (p-2r)2 + (q-2r)2 = 0

Now ?

ok !

(p-q)2 = 0
p=q

(p-2r)2 = 0
p=2r

(q-2r)2 = 0

so , q=2r

p=q=2r
QED

That was simple haha.
Thanks all :)

That does not agree with p=q=2r anyway.

Look at p = q = 2r = 1, which does work.

Hiii SammyS ,
Your relation worked though. How did you get it ? By observation ?
Well by the method which Dickfore suggested ,
p=q=2r=n , where n can be any number.
 
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sankalpmittal said:
I don't know which method is correct. Everyone is giving different opinions well. I start off from last post.
...

Hi SammyS ,
Your relation worked though. How did you get it ? By observation ?
Well by the method which Dickfore suggested ,
p=q=2r=n , where n can be any number.
Well, all the methods work.

I really liked the one by Dickfore.

I had to fiddle around with WolframAlpha to get it to recognize the expression. Then it gave lots of stuff. It wasn't until I messed around with the radical expression that I saw that it gave the solution. Similarly, with Dick's solution, where you have the sum of three squares is equal to zero, the only way to get a real solution is for each of them to be zero.

[itex]4 r^2-2(p+q)r+p^2-p q+q^2 = 0[/itex]

[itex]\displaystyle r=\frac{2(p+q)\pm\sqrt{12} \sqrt{-p^2+2 p q-q^2}}{8}[/itex]

[itex]\displaystyle r=\frac{(p+q)\pm\sqrt{3} \sqrt{-(p-q)^2}}{4}[/itex]

So for a real solution we need p=q. Then r = 2p/4 = p/2 = q/2 .

NascentOxygen's also works. p(p-q) + q(q-2r) + 2r(2r-p) = 0 .
 
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SammyS said:
If p-q = 0 , q-2r = 0 and p-2r = 0, then the expression is zero. But it's not so obvious that this is the only solution.

This method of NascentOxygen is also good.

p(p-q) + q(q-2r) + 2r(2r-p) = 0 .

Here p(p-q) , q(q-2r) , 2r(2r-p) are zero , but its not obvious so ,

let

p(p-q) = 1

so p2 -pq -1 = 0

q(q-2r) = 1
q2 - 2rq - 1 = 0

2r(2r-p) = -2

2r2 - pr +1 = 0

I think solving and combining these equation can help the case.
Well thanks all for taking effort !

Yet to try :
p(p-q) = 1
Thinking the other way
p=1
or p-q = 1

q(q-2r) = 1
q=1
q-2r = 1

2r(2r-p) = -2
2r = -2
2r-p = 1

Here we can obviously get : p=q=2r = 1 or we can try simultaneously solving equations p-q = 1 , q-2r = 1 , 2r-p = 1.

Well we can use other combination in p(p-q) as p=2 and p-q = 1/2

Thanks all of you :)
 
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NascentOxygen said:
There's a trivial solution: p=q=r=0

and a non-trivial solution: p=q=2r

though the trivial solution is contained within the non-trivial.

Well... I still don't get it.
I see no reason why there couldn't be more solutions.
And the problem asks to proof that p=q=2r is the only solution.
 
I like Serena said:
Well... I still don't get it.
I see no reason why there couldn't be more solutions.
And the problem asks to proof that p=q=2r is the only solution.

Hiii ILS !


This method of NascentOxygen is also good.

I Method : is a trivial solution

p=q=r=0

II Method is given below :

p(p-q) + q(q-2r) + 2r(2r-p) = 0 .

Here p(p-q) , q(q-2r) , 2r(2r-p) are zero , but its not obvious so .

Then p = q = 2q = 0

III Method

p(p-q) = 1

so p2 -pq -1 = 0

q(q-2r) = 1
q2 - 2rq - 1 = 0

2r(2r-p) = -2

2r2 - pr +1 = 0

I think solving and combining these equation can help the case.
Well thanks all for taking effort !

IV Method :

In p(p-q) + q(q-2r) + 2r(2r-p) = 0 .

Yet to try :
p(p-q) = 1
Thinking the other way
p=1
p-q = 1

q(q-2r) = 1
q=1
q-2r = 1

2r(2r-p) = -2
2r = 1
2r-p = -2

Since 1+1-2=0

Here we can obviously get : p=q=2r = 1 or we can try simultaneously solving equations p-q = 1 , q-2r = 1 , 2r-p = -2.

Well we can use other combination in p(p-q) as p=2 and p-q = 1/2

p-q = 1 , q-2r = 1 , 2r-p = -2.

Now try solving these equations simultaneously , you will get a non trivial solution of p=q=2r

q=2r+1
p= 2r+2

So 2r+2 - 2r - 1 = 1
1=1 Hmm This question is irritating me now !
Hmm

?
:confused:
 
To be clear, Dickfore's method is a (the?) proper solution:

(p-q)2 + (p-2r)2 + (q-2r)2 = 0

Since a square is always at least zero, each of the 3 terms must be zero.
From there it follows that p=q=2r exactly represents all possible solutions.
 
sankalpmittal said:
I Method is given below :

p(p-q) + q(q-2r) + 2r(2r-p) = 0 .

Here p(p-q) , q(q-2r) , 2r(2r-p) are zero , but its not obvious so .

Then p = q = 2q = 0
I think you have not noticed that there are infinite numerical solutions. The general solution is expressed as: p=q=2r=X where X can be any number you like.
 
NascentOxygen said:
I think you have not noticed that there are infinite numerical solutions. The general solution is expressed as: p=q=2r=X where X can be any number you like.

Suppose p(p-q) = 1 , q(q-2r) = 1 , 2r(2r-p) =-2 ; since 1+1-2 =0

Then how can you get a non trivial solution of p=q=2r ?

Is My trial correct ?:
To get p(p-q) = 1 , p=p-q = 1
To get q(q-2r) = 1 , q=q-2r=1
To get 2r(2r-p) =-2 , either 2r=1 and 2r-p=-2 OR 2r-p=1 and 2r=-2

Considering the first fragment of 2r=1 and 2r-p=-2 ; we get p=q=2r=1

Considering the second fragment of 2r-p=1 and 2r=-2 ; we get
p-q=q-2r=2r-p=1
Solving p-q=q-2r we get 2r+p = 2q .. I
Solving q-2r=2r-p we get 4r-p = q.. II

If we add I and II we get 6r=3q ==> q=2r
From I : Putting q=2r or 2q = 4r we get p=2r
Hence we get p=q=2r=n as a non trivial solution

Am I correct ?

Thanks for the efforts !
 
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I can't fathom what you are trying to do here. But let's look at it:

sankalpmittal said:
Suppose p(p-q) = 1 , q(q-2r) = 1 , 2r(2r-p) =-2 ; since 1+1-2 =0
Why assume these values? Why not p(p-q)=20, q(q-r)=31, 2r(2r-p)=-51

Then how can you get a non trivial solution of p=q=2r ?

Is My trial correct ?:
To get p(p-q) = 1 , p=p-q = 1
If p=p-q then doesn't it follow that this says q=0 ?

To get q(q-2r) = 1 , q=q-2r=1
To get 2r(2r-p) =-2 , either 2r=1 and 2r-p=-2 OR 2r-p=1 and 2r=-2

Considering the first fragment of 2r=1 and 2r-p=-2 ; we get p=q=2r=1
What does it mean to say q=1 when the working above required q=0?

I can't follow what you are doing.
 

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