# Homework Help: Symetric Triple Quad Formula

1. Jun 28, 2014

### Septimra

1. The problem statement, all variables and given/known data

So I am trying to make the triple quad formula: (Q1+Q2-Q3)2 = 4Q1Q2
into the symetric version: (Q1+Q2+Q3)2 = 2(Q12 + Q22 + Q32)

2. Relevant equations

(Q1+Q2-Q3)2 = 4Q1Q2
(Q1+Q2+Q3)2 = 2(Q12 + Q22 + Q32)

3. The attempt at a solution

I get the equation into the form: Q12+Q22+Q32 = 2(Q1Q2+Q1Q3+Q2Q3)

Then any substitutions I do I go back in circles. I attempted to square both sides but no cigar.

2. Jun 28, 2014

### LCKurtz

Use the X2 and X2 icons above the advanced editing box for subscripts and superscripts. That might make your equations readable. Then explain to us what the heck you are talking about, like what the Q's are.

3. Jun 29, 2014

### Septimra

Original Equation
(Q1+Q2-Q3)2 = 4Q1Q2
Recomposed
(Q1+Q2+Q3)2 = 2(Q12 + Q22 + Q32)

Q's stand for quadrances or the area or a square built on the segment that is the square root of the Q.

http://www.parabola.unsw.edu.au/vol43_no1/img145.png [Broken]

Q1 is (A1 to A2)2
Q2 is (A2 to A3)2
Q3 is (A1 to A3)2

How do I get the original formula into the recompose variation?

Last edited by a moderator: May 6, 2017
4. Jun 29, 2014

### thelema418

You can just brute "force" it.
Step 1: Expand the left hand side of the original. Do everything you can to get $2(Q_1^2 + Q_2^2 + Q_3^2)$. But don't worry about what the right hand side looks like.

Step 2: Go back to the original equation. Expand the left hand side again. What do you have to add to both sides of the equation so that you can factor it back to $(Q_1 + Q_2 + Q_3)^2$?

Step 3: Is there a relationship between these two results?

Last edited by a moderator: May 6, 2017
5. Jul 1, 2014

### epenguin

This is algebra so it doesn't matter what they are.

(Even if it is also geometry).

Last edited: Jul 1, 2014
6. Jul 1, 2014

### epenguin

What I would say is one of the 5 questions to ask yourself recommended by Polya in his book "How to solve it" is "have you seen anything like it before?".

And I bet you have:

(A - B)2 = (A + B)2 - 4 AC

Used one way or another in lots of exercises you might recall to prove that something is larger than 0 or something else.

So letting A be (Q1 + Q2) and B be Q3 **

[(Q1 + Q2) - Q3]2 = [(Q1 + Q2) + Q3]2 - 4 (Q1 +Q2)Q3

Now, since you are given [(Q1 + Q2) - Q3]2 = something else (Polya question : are you using all the information you're given? To which I add: have an eye for symmetry) subtracting that relation from the last equation gives you

0 = [(Q1 + Q2) + Q3]2 - 4 (Q1Q2 + Q1Q3 + Q2Q3)

Which is at least a symmetrical relation, even if not yet the one you seek.

The first term is something you are looking for, the second isn't. What else is the unwanted one equal to? Use the general relationship

(Q1 + Q2 + Q3)2 = (Q12 + Q22 + Q32 ) + 2(Q1Q2 + Q1Q3 + Q2Q3)

- eliminate the last term between these two and you have your result.

You will later find this sort of calculation ("symmetric polynomials") a lot used in early university algebra and other subjects, not just in Prof. W's lectures.

I don't say it couldn't be done a bit faster than I did.

** Edit: although you can just straight expand the squared bracket and my argument is really no different from thelma's.

Last edited: Jul 2, 2014
7. Jul 1, 2014

### thelema418

So, you think that if $Q_1$ and $Q_2$ are positive in the original equation, then we should have both positive and negative answers for $Q_3$? Though the poster responded that $Q_3$ is the distance line segment squared.

8. Jul 3, 2014

### Septimra

Its makes so much sense. Thanks to both you guys. But I have one more question. How would someone know the equation is symmetrical? Is there a dead giveaway, or do you just have to try it and see if it works out?

9. Jul 4, 2014

### thelema418

This refers to the concept of a "symmetric polynomial". I don't know if that is part of your pre-calculus course? Usually they teach symmetry across an axis, which is a bit different.

Anyway, a symmetric polynomial means that you can "flip around" the variables. So if you have $x + y + z$ you can change all the x to z and all the y to x and all the z to y. Then you have $z + x + y$. And you can also have $y + z + x$ and $y + x + z$, etc. With the standard rules of addition of numbers, we know all these variations are equal to each other. So, this is a symmetric polynomial.

Consider now the polynomial $x - y$. The "flip around" is $y - x$. For the familiar numbers this is not symmetric: $3 -2 \neq 2 -3$. Yet, this is why we have to know what the variables are. If your polynomials are over the field $\mathbb{Z}_2$, then the situation is symmetric because $1 - 0 = 1$ and $0 - 1 = 1$!!!

10. Jul 4, 2014

### epenguin

The following which I hope others can say better strikes me.

At least if you suspect a symmetry in this formula between 1, 2 and 3 you can get a symmetrical formula, it seems to me always, by applying the symmetry.

E.g here if

(Q1 + Q2 - Q3)2 = 4Q1Q2

Then if there is this symmetry we must also have

(Q1 - Q2 + Q3)2 = 4Q1Q3

and

(-Q1 + Q2 + Q3)2 = 4Q2Q3

Add these all up, expand, and the symmetrical

Q12 + Q22 + Q32 = 2(Q1Q2 + Q1Q3 + Q2Q3)

drops out. Again not the one you wanted, but you now easily find that via a general identity concerning the RHS as before.

This is much better than what we did before because a thinking part is now automatic, we have more of a guide. :thumbs: So good question.

We started with a hypothesis, but it is stronger than that. I mean our three starting forumulae are just one, in which quantities have merely been given different names. If you look at your fig. it may look at first sight your three squares have a hierarchy, are of different nature. But if you push the points through each other you find the hierarchy is not fixed, and Q1 can indifferently express the larger, smallest or middle square...

Surely some didactic mathematician here knows what I am saying and can express it better?

Last edited: Jul 5, 2014
11. Jul 10, 2014

### Septimra

Alright I've been giving it a go. And so far so good. I tried the brute force method and it worked like a charm. I then tried the basic symmetrical law of creating 3 equations, then adding them up to, again, brute force into the symmetrical form desired-- it's just great! I'm learning a lot. Thanks guys.

However I began to get more interested as to how we could manipulate

4(Q1Q2+Q1Q3+Q2Q3) into the other 2 forms by its self. Yes we know it's symmetrical, and it's been proven, but how do we get look like the other 2 variants throughout shear algebraic manipulation alone

I got the one with everything all squared, but the other one appears to be trickier.

12. Jul 10, 2014

### epenguin

I'm glad you find it interesting.

Sorry, what I said in my last post is just unnecessary and the three equations are true but unnecessary, only one is needed, in fact any one of the three gives on expansion and simplification the same result which is symmetrical.

If this has got you interested you are ready for a couple of chapters of an early university level algebra book which will have a chapter or two on symmetric polynomials (expressions in Q1 , Q2,... Qn which are not changed by interchange of any Q's). Applications In theory of polynomial equations, geometry, statistics and I don't know what else.