System of three equations and four variables

  • Thread starter Thread starter A_Studen_349q
  • Start date Start date
  • Tags Tags
    System Variables
AI Thread Summary
The discussion revolves around solving a system of three equations with four variables, which is challenging due to the complexity of the equations involving square roots. Participants suggest various approaches, including squaring both sides and looking for patterns in the equations, but many attempts lead to high-degree equations or unhelpful results. The equations exhibit repeated terms, hinting at potential simplifications, yet these have not yielded clear solutions. There is a consensus that with three equations and four variables, the best outcome may be to express one variable in terms of the others. The goal remains to find real positive roots, as indicated by the problem's context.
A_Studen_349q
Messages
14
Reaction score
0
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!
 
Physics news on Phys.org
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?
 
jedishrfu said:
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…

jedishrfu said:
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.

jedishrfu said:
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.

jedishrfu said:
What course did this problem come from?
It is an algebra. Just an algebra :(
 
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.
 
jedishrfu said:
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.
 
jedishrfu said:
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 into 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:
jedishrfu said:
could the other roots be imaginary like try 0+i?
Nope. Roots must be real and positive.

jedishrfu said:
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 :)
 
  • #10
A_Studen_349q said:
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.
 
  • #11
Chestermiller said:
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, let's 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).

Chestermiller said:
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.
 
  • #12
A_Studen_349q said:
Yes, but this "linear combination" is nothing but eqn (1). Why? Well, let's 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
 
  • #13
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
For chet and ehild: it seems you are wrong. Well, let's enumerate our eqns:
\begin{align}<br /> &amp; \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) \\ <br /> &amp; \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) \\ <br /> &amp; \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) \\ <br /> \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}<br /> &amp; f\left( {{x}_{1}},{{x}_{4}} \right)-f\left( {{x}_{1}},{{x}_{3}} \right)=f\left( {{x}_{3}},{{x}_{4}} \right)\ \ \ \ \ ({1}&#039;) \\ <br /> &amp; f\left( {{x}_{1}},{{x}_{4}} \right)-f\left( {{x}_{1}},{{x}_{2}} \right)=f\left( {{x}_{2}},{{x}_{4}} \right)\ \ \ \ \ ({2}&#039;) \\ <br /> &amp; f\left( {{x}_{1}},{{x}_{3}} \right)-f\left( {{x}_{1}},{{x}_{2}} \right)=f\left( {{x}_{2}},{{x}_{3}} \right)\ \ \ \ \ ({3}&#039;) \\ <br /> \end{align}
Chestermiller said:
... 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.

ehild said:
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.
 
  • #15
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
ehild said:
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?

ehild said:
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.
 
  • #17
A_Studen_349q said:
For chet and ehild: it seems you are wrong. Well, let's enumerate our eqns:
\begin{align}<br /> &amp; \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) \\ <br /> &amp; \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) \\ <br /> &amp; \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) \\ <br /> \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}<br /> &amp; f\left( {{x}_{1}},{{x}_{4}} \right)-f\left( {{x}_{1}},{{x}_{3}} \right)=f\left( {{x}_{3}},{{x}_{4}} \right)\ \ \ \ \ ({1}&#039;) \\ <br /> &amp; f\left( {{x}_{1}},{{x}_{4}} \right)-f\left( {{x}_{1}},{{x}_{2}} \right)=f\left( {{x}_{2}},{{x}_{4}} \right)\ \ \ \ \ ({2}&#039;) \\ <br /> &amp; f\left( {{x}_{1}},{{x}_{3}} \right)-f\left( {{x}_{1}},{{x}_{2}} \right)=f\left( {{x}_{2}},{{x}_{3}} \right)\ \ \ \ \ ({3}&#039;) \\ <br /> \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
 
  • #18
Ray Vickson said:
How do you know that goal is achievable?
RGV
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
A_Studen_349q said:
Yes, but this "linear combination" is nothing but eqn (1). Why? Well, let's 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 re-designate our variables around to turn any of our eqns into (4).
...
ehild has obtained that 4th equation, the one that doesn't have x1 in it:
ehild said:
Well, not all goals can be achieved.

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

\dots\ =f(x_3 ,x_4)-f(x_2,x_4)=-f(x_2, x_3)

...

ehild

Although that gives a fourth equation, it's not independent of the other three.
 
  • #20
SammyS said:
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.
Agrees.
 
  • #21
SammyS said:
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.

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
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
ehild said:
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)

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:
  • #24
gabbagabbahey said:
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)?

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.
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)

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:
  • #25
ehild said:
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). If it is true then f(xi,xj)=0 for i,j {1,2,3,4} i≠j.

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).

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.

Yes, that was a silly error on my part!
 
  • #26
So, any new ideas?
 

Similar threads

Solve the problem that involves iteration
Replies
21
Views
2K
Obtain the two incongruent solutions modulo 210 of the system
Replies
7
Views
3K
What was the least number of coins that could have been stolen?
Replies
3
Views
2K
Solve ##x\equiv 5\pmod{11},x\equiv 14\pmod{29},x\equiv 15\pmod{31}##
Replies
3
Views
3K
Finding Integer with Chinese Remainder Theorem
Replies
3
Views
2K
How to obtain three consecutive integers?
Replies
3
Views
1K
Solve the simultaneous linear congruences
Replies
4
Views
2K
Solve ##x\equiv 1\pmod {3},x\equiv 2\pmod {5},x\equiv 3\pmod 7 ##
Replies
1
Views
2K
How should I show that this root is given approximately by this?
Replies
2
Views
2K
Prove that ## \sum_{n\leq x}\frac{\phi(n)}{\sqrt{n}}=Cx^{3/2}+.... ##.
Replies
4
Views
2K

Hot Threads

Recent Insights

Back
Top