## System of three equations and four variables

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!
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 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?

 Quote by jedishrfu 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 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…

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

 Quote by jedishrfu Also have you tried squaring both sides to see if there's any reductions that can be made?
Yes, of course I tried it though equations became large and the reductions gave no good effect.

 Quote by jedishrfu What course did this problem come from?
It is an algebra. Just an algebra :(

## System of three equations and four variables

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.

 Quote by jedishrfu 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.
0 gives nothing (sqare root of negative values).
1 gives nothing good either.
 When I set all four values to zero I get: sqrt(0) - sqrt(0) = sqrt(0) for all eqns.

 Quote by jedishrfu When I set all four values to zero I get: sqrt(0) - sqrt(0) = sqrt(0) for all eqns.
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.
 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.

 Quote by jedishrfu could the other roots be imaginary like try 0+i?
Nope. Roots must be real and positive.

 Quote by jedishrfu Also perhaps some of the more senior contributors in this forum like Mark44 or micromass could provide some advice.
Ok, I'll wait for any ideas :)

Recognitions:
Gold Member
 Quote by 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!
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.

 Quote by Chestermiller 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
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).

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

Recognitions:
Gold Member
 Quote by 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.
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
 Recognitions: Homework Help 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

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}
 Quote by Chestermiller ... x1 = x2 for arbitrary values of x3 and x4. chet
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.

 Quote by ehild f(xi,xj)=0 for x1,x2,x3,x4. ehild
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.
 Recognitions: Homework Help 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

 Quote by ehild $$f(x_2, x_4)-f(x_2,x_3)=f(x_3 ,x_4)$$ You can get similar relation between any pair of the xi-s ehild
For any? Are you sure?

 Quote by ehild f(a,c)=-f(a,c) ehild
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.

Recognitions:
Homework Help
 Quote by 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.
How do you know that goal is achievable?

RGV

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