# Homework Help: System of three equations and four variables

1. Jul 18, 2012

### A_Studen_349q

How to solve this system?
\begin{align}
& \sqrt{\frac{x_{1}^{2}x_{4}^{2}}{4}-{{({{x}_{1}}-{{x}_{4}})}^{2}}}-\sqrt{\frac{x_{1}^{2}x_{3}^{2}}{4}-{{({{x}_{1}}-{{x}_{3}})}^{2}}}=\sqrt{\frac{x_{3}^{2}x_{4}^{2}}{4}-{{({{x}_{3}}-{{x}_{4}})}^{2}}} \\
& \sqrt{\frac{x_{1}^{2}x_{4}^{2}}{4}-{{({{x}_{1}}-{{x}_{4}})}^{2}}}-\sqrt{\frac{x_{1}^{2}x_{2}^{2}}{4}-{{({{x}_{1}}-{{x}_{2}})}^{2}}}=\sqrt{\frac{x_{2}^{2}x_{4}^{2}}{4}-{{({{x}_{2}}-{{x}_{4}})}^{2}}} \\
& \sqrt{\frac{x_{1}^{2}x_{3}^{2}}{4}-{{({{x}_{1}}-{{x}_{3}})}^{2}}}-\sqrt{\frac{x_{1}^{2}x_{2}^{2}}{4}-{{({{x}_{1}}-{{x}_{2}})}^{2}}}=\sqrt{\frac{x_{2}^{2}x_{3}^{2}}{4}-{{({{x}_{2}}-{{x}_{3}})}^{2}}} \\
\end{align}
Thanks a bunch!

2. Jul 18, 2012

### Staff: Mentor

Welcome to the physics forum!

Before we can help you, you need to show what you've tried and the steps you've taken to solve this homework problem.

I did notice that you have some repeated terms in the equations that may hint at a solution like the first square-root term of the equation is repeated as the first term of the second equation.

Also have you tried squaring both sides to see if there's any reductions that can be made?

What course did this problem come from?

3. Jul 18, 2012

### A_Studen_349q

I tried my best though failed. First of all I tried to yield one of the variables as a function of other variables but got an equation of high (fourth) degree and stopped. Then I tried to compose (add, divide, etc) these equations to make their form easier. Failed again. Etc., etc…

Well, I have noticed that too though got nothing useful of it.

Yes, of course I tried it though equations became large and the reductions gave no good effect.

It is an algebra. Just an algebra :(

4. Jul 18, 2012

### Staff: Mentor

okay have you tried setting the 4 values to simple numbers like zero or 1. By inspection it seems that one of those might work.

5. Jul 18, 2012

### A_Studen_349q

0 gives nothing (sqare root of negative values).
1 gives nothing good either.

6. Jul 18, 2012

### Staff: Mentor

When I set all four values to zero I get:

sqrt(0) - sqrt(0) = sqrt(0) for all eqns.

7. Jul 19, 2012

### A_Studen_349q

Yes, but (0,0,0,0) is only a one (rather obvious) of the roots I have to find. And it gives nothing for finding all other roots.

8. Jul 19, 2012

### Staff: Mentor

could the other roots be imaginary like try 0+i?

Beyond that I don't know how else to solve these. Perhaps your teacher or other students can give you a hint.

My first attempt would be to square both sides and then see if I can use the existing eqns to sub in to eliminate any sqrt terms.

Also perhaps some of the more senior contributors in this forum like Mark44 or micromass could provide some advice.

Last edited: Jul 19, 2012
9. Jul 19, 2012

### A_Studen_349q

Nope. Roots must be real and positive.

Ok, I'll wait for any ideas :)

10. Jul 19, 2012

### Staff: Mentor

This is a very interesting problem, and one I personally haven't solved yet. But I did notice that if you subtract the second equation from the first, and then add the third, the left hand sides cancel, and you are left with a linear combination of the right hand sides. The right hand side of the resulting equation does not contain x1.

Since there are 3 equations and 4 unknowns, the best you can do is to solve for the ratio of three of the parameters to the forth parameter.

11. Jul 19, 2012

### A_Studen_349q

Yes, but this "linear combination" is nothing but eqn (1). Why? Well, lets enumerate our three equations of the initial system as (1), (2) and (3). We have 3 eqns and 4 vars while (1) feels lack of x2, (2) feels lack of x3, (3) feels lack of x4. The "linear combination" (let's call it (4)) you told about will feel lack of x1. Why do I say (4) is nothing but (1)? Well, we can just redesignate our variables around to turn any of our eqns into (4).

Yes, ... but how to reduce all these to ratios? I tried hard but failed.

12. Jul 19, 2012

### Staff: Mentor

No. I'll write some more soon. I don't have time right now. But I can tell you that, when you do what I said, you will be able to show that x1 = x2 for arbitrary values of x3 and x4.

chet

13. Jul 20, 2012

### ehild

Use the notation

$$f(x,y)=\sqrt{\frac{x^{2}y^{2}}{4}-{{({{x}}-{{y}})}^{2}}}$$.

f(x,y) is symmetric for the interchange of the variables: f(y,x)=f(x,y)

It can be derived from the equations given that f(a,b)-f(a,c)=f(c,b), choosing a, b, c in any way from the variables x1,x2,x3.

Try to show that f(xi,xj)=-f(xj,xi) is also true, so f(xi,xj)=0 for x1,x2,x3,x4.

ehild

14. Jul 20, 2012

### A_Studen_349q

For chet and ehild: it seems you are wrong. Well, let's enumerate our eqns:
\begin{align} & \sqrt{\frac{x_{1}^{2}x_{4}^{2}}{4}-{{({{x}_{1}}-{{x}_{4}})}^{2}}}-\sqrt{\frac{x_{1}^{2}x_{3}^{2}}{4}-{{({{x}_{1}}-{{x}_{3}})}^{2}}}=\sqrt{\frac{x_{3}^{2}x_{4}^{2}}{4}-{{({{x}_{3}}-{{x}_{4}})}^{2}}}\ \ \ \ \ (1) \\ & \sqrt{\frac{x_{1}^{2}x_{4}^{2}}{4}-{{({{x}_{1}}-{{x}_{4}})}^{2}}}-\sqrt{\frac{x_{1}^{2}x_{2}^{2}}{4}-{{({{x}_{1}}-{{x}_{2}})}^{2}}}=\sqrt{\frac{x_{2}^{2}x_{4}^{2}}{4}-{{({{x}_{2}}-{{x}_{4}})}^{2}}}\ \ \ \ \ (2) \\ & \sqrt{\frac{x_{1}^{2}x_{3}^{2}}{4}-{{({{x}_{1}}-{{x}_{3}})}^{2}}}-\sqrt{\frac{x_{1}^{2}x_{2}^{2}}{4}-{{({{x}_{1}}-{{x}_{2}})}^{2}}}=\sqrt{\frac{x_{2}^{2}x_{3}^{2}}{4}-{{({{x}_{2}}-{{x}_{3}})}^{2}}}\ \ \ \ \ (3) \\ \end{align}
and introduce function $$f\left( {{x}_{i}},{{x}_{j}} \right)=\sqrt{\frac{x_{i}^{2}x_{j}^{2}}{4}-{{({{x}_{i}}-{{x}_{j}})}^{2}}}$$
with obvious property
$$f\left( {{x}_{i}},{{x}_{j}} \right)=f\left( {{x}_{j}},{{x}_{i}} \right)\ \ \ \ \ (*)$$
then our system will have a form of
\begin{align} & f\left( {{x}_{1}},{{x}_{4}} \right)-f\left( {{x}_{1}},{{x}_{3}} \right)=f\left( {{x}_{3}},{{x}_{4}} \right)\ \ \ \ \ ({1}') \\ & f\left( {{x}_{1}},{{x}_{4}} \right)-f\left( {{x}_{1}},{{x}_{2}} \right)=f\left( {{x}_{2}},{{x}_{4}} \right)\ \ \ \ \ ({2}') \\ & f\left( {{x}_{1}},{{x}_{3}} \right)-f\left( {{x}_{1}},{{x}_{2}} \right)=f\left( {{x}_{2}},{{x}_{3}} \right)\ \ \ \ \ ({3}') \\ \end{align}
No, chet, that can't be true. Why? Suppose x1=x2. Then (3) x1=x2=0 while we said that zero roots are obvious and our goal is to find real positive roots x1,x2,x3,x4.

f(xi,xj)=0 gives only zero root (x1,x2,x3,x4)=(0,0,0,0) while our goal is to find real positive roots x1,x2,x3,x4.

15. Jul 20, 2012

### ehild

Well, not all goals can be achieved.

Subtracting 2' from 1' and comparing it with 3':

$$f(x_1,x_2)-f(x_1,x_3)=f(x_3 ,x_4)-f(x_2,x_4)=-f(x_2, x_3)$$

Rearranging : $$f(x_2, x_4)-f(x_2,x_3)=f(x_3 ,x_4)$$
You can get similar relation between any pair of the xi-s:
Let be a, b, c any different xi-s.

f(a,b)-f(a,c)=f(c,b),

but also

f(b,c)-f(b,a)=f(a,c), or

f(b,a)+f(a,c)=f(b,c)
As f(a,b)=f(b,a) and f(b,c)=f(c,b),

f(a,b)+f(a,c)=f(b,c),

that is f(a,c)=-f(a,c)

I hope my derivation is correct...

ehild

16. Jul 20, 2012

### A_Studen_349q

For any? Are you sure?

Tha is f(a,c)=0? Hm... can you show it at least for one pair of variables? Well, you prooved $$f(x_2, x_4)-f(x_2,x_3)=f(x_3 ,x_4)$$ but (imho) it cannot be combied with any of (1')-(3') to achieve f(a,c)=0 for any pair (a,c) of variables.

17. Jul 20, 2012

### Ray Vickson

How do you know that goal is achievable?

RGV

18. Jul 20, 2012

### A_Studen_349q

Well, first, I know that "to solve a system of equations" = "find all its roots or prove that there are no roots", second, teacher told that this system has non-zero roots, third, one of my classmates tried to use some numerical methods and said he obtained positive values.
So, I think that there must be non-zero roots (may be positive).
In any case, I have to solve this problem. So far, I don't know the solution.
Waiting for ideas...

19. Jul 20, 2012

### SammyS

Staff Emeritus
ehild has obtained that 4th equation, the one that doesn't have x1 in it:
Although that gives a fourth equation, it's not independent of the other three.

20. Jul 20, 2012

### A_Studen_349q

Agrees.

21. Jul 20, 2012

### ehild

The fourth equation has been derived from the first three and the symmetry relation f(a,b)=f(b,a).
You can have three equations of the same type as the original ones either with x2, x3 or x4 as the "leading variable", by combining the original equations and the symmetry condition f(a,b)=f(b,a).

For example, the first two equations can be rewritten as
f(x4,x1)-f(x4,x3)=f(x3,x1)
f(x4,x1)-f(x4,x2)=f(x2,x1)
From these, it follows the third equation with x4:
f(x4,x2)-f(x4,x3)=f(x3,x2).

ehild

22. Jul 20, 2012

### Staff: Mentor

another thing to note is that you can factor the the square root term although that may or may not help:

multiply both side by 2 and pull inside to eliminate the 1/4 factor

then for a given sqrt term notice the x^2 - y^2 pattern to yield (x-y)*(x+y)

sqrt( xi^2 * xj^2 - 4 * (xi - xj)^2 ) = sqrt ( ( xi*xj - 2 * (xi -xj)) * ( xi*xj + 2 * (xi -xj) )

23. Jul 21, 2012

### gabbagabbahey

I really don't see how you can claim this. As a specific example, how can one show that $f(x_1, x_2)-f(x_1, x_3)=f(x_3, x_2)$?

Still, $f(x_i, x_j)=0$ is certainly a permissible solution, but leads to only to trivial solutions, even when restricted to $i \neq j$, and I see no reason that it is the only solution.

Edit: Not all solutions of $f(x_i, x_j)=0$ are trivial; for example $(x_1, x_2, x_3, x_4)=(1, 2, 0, 0)$

Last edited: Jul 21, 2012
24. Jul 22, 2012

### ehild

I can not prove directly that f(a,b)-f(a,c) = f(c,b) true for any three from x1, x2, x3, x4. It is a "lemma", based on the symmetry of the function f(a,b)=f(b,a). But it might be false. If it is true then f(xi,xj)=0 for i,j {1,2,3,4} i≠j.

x3=x4=0 means negative values under square roots, so the equations are not defined over the real numbers. Plugging in your values, you get √(-1) on both sides, but it can be either i or -i. You can not say that (1, 2, 0, 0) is a solution.

ehild

Last edited: Jul 22, 2012
25. Jul 22, 2012

### gabbagabbahey

I agree that the symmetry relation allows you to derive another equation, but I don't see it leading to f(a,b)-f(a,c) = f(c,b) for all permutations (where a≠b≠c). I think exactly half the permutations satisfy this equation, and the other half satisfy f(a,b)-f(a,c) = -f(c,b).

Yes, that was a silly error on my part!