System of three equations and four variables

[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!

 

[jedishrfu]

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?

 

[A_Studen_349q]

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…

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.

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.

What course did this problem come from?

It is an algebra. Just an algebra :(

 

[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.

 

[A_Studen_349q]

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.

 

[jedishrfu]

When I set all four values to zero I get:

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

 

[A_Studen_349q]

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.

 

[jedishrfu]

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.

 

[A_Studen_349q]

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

Nope. Roots must be real and positive.

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 :)

 

[Chestermiller]

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.

 

[A_Studen_349q]

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).

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.

 

[Chestermiller]

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

 

[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

 

[A_Studen_349q]

For chet and ehild: it seems you are wrong. Well, let's enumerate our eqns:
$$\begin{align}<br /> & \sqrt{\frac{x_{1}^{2}x_{4}^{2}}{4}-{{({{x}_{1}}-{{x}_{4}})}^{2}}}-\sqrt{\frac{x_{1}^{2}x_{3}^{2}}{4}-{{({{x}_{1}}-{{x}_{3}})}^{2}}}=\sqrt{\frac{x_{3}^{2}x_{4}^{2}}{4}-{{({{x}_{3}}-{{x}_{4}})}^{2}}}\ \ \ \ \ (1) \\ <br /> & \sqrt{\frac{x_{1}^{2}x_{4}^{2}}{4}-{{({{x}_{1}}-{{x}_{4}})}^{2}}}-\sqrt{\frac{x_{1}^{2}x_{2}^{2}}{4}-{{({{x}_{1}}-{{x}_{2}})}^{2}}}=\sqrt{\frac{x_{2}^{2}x_{4}^{2}}{4}-{{({{x}_{2}}-{{x}_{4}})}^{2}}}\ \ \ \ \ (2) \\ <br /> & \sqrt{\frac{x_{1}^{2}x_{3}^{2}}{4}-{{({{x}_{1}}-{{x}_{3}})}^{2}}}-\sqrt{\frac{x_{1}^{2}x_{2}^{2}}{4}-{{({{x}_{1}}-{{x}_{2}})}^{2}}}=\sqrt{\frac{x_{2}^{2}x_{3}^{2}}{4}-{{({{x}_{2}}-{{x}_{3}})}^{2}}}\ \ \ \ \ (3) \\ <br /> \end{align}$$
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
$$\begin{align}<br /> & f\left( {{x}_{1}},{{x}_{4}} \right)-f\left( {{x}_{1}},{{x}_{3}} \right)=f\left( {{x}_{3}},{{x}_{4}} \right)\ \ \ \ \ ({1}') \\ <br /> & f\left( {{x}_{1}},{{x}_{4}} \right)-f\left( {{x}_{1}},{{x}_{2}} \right)=f\left( {{x}_{2}},{{x}_{4}} \right)\ \ \ \ \ ({2}') \\ <br /> & f\left( {{x}_{1}},{{x}_{3}} \right)-f\left( {{x}_{1}},{{x}_{2}} \right)=f\left( {{x}_{2}},{{x}_{3}} \right)\ \ \ \ \ ({3}') \\ <br /> \end{align}$$

... 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.

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.

 

[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

 

[A_Studen_349q]

$$f(x_2, x_4)-f(x_2,x_3)=f(x_3 ,x_4)$$
You can get similar relation between any pair of the xi-s
ehild

For any? Are you sure?

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.

 

[Ray Vickson]

For chet and ehild: it seems you are wrong. Well, let's enumerate our eqns:
$$\begin{align}<br /> & \sqrt{\frac{x_{1}^{2}x_{4}^{2}}{4}-{{({{x}_{1}}-{{x}_{4}})}^{2}}}-\sqrt{\frac{x_{1}^{2}x_{3}^{2}}{4}-{{({{x}_{1}}-{{x}_{3}})}^{2}}}=\sqrt{\frac{x_{3}^{2}x_{4}^{2}}{4}-{{({{x}_{3}}-{{x}_{4}})}^{2}}}\ \ \ \ \ (1) \\ <br /> & \sqrt{\frac{x_{1}^{2}x_{4}^{2}}{4}-{{({{x}_{1}}-{{x}_{4}})}^{2}}}-\sqrt{\frac{x_{1}^{2}x_{2}^{2}}{4}-{{({{x}_{1}}-{{x}_{2}})}^{2}}}=\sqrt{\frac{x_{2}^{2}x_{4}^{2}}{4}-{{({{x}_{2}}-{{x}_{4}})}^{2}}}\ \ \ \ \ (2) \\ <br /> & \sqrt{\frac{x_{1}^{2}x_{3}^{2}}{4}-{{({{x}_{1}}-{{x}_{3}})}^{2}}}-\sqrt{\frac{x_{1}^{2}x_{2}^{2}}{4}-{{({{x}_{1}}-{{x}_{2}})}^{2}}}=\sqrt{\frac{x_{2}^{2}x_{3}^{2}}{4}-{{({{x}_{2}}-{{x}_{3}})}^{2}}}\ \ \ \ \ (3) \\ <br /> \end{align}$$
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
$$\begin{align}<br /> & f\left( {{x}_{1}},{{x}_{4}} \right)-f\left( {{x}_{1}},{{x}_{3}} \right)=f\left( {{x}_{3}},{{x}_{4}} \right)\ \ \ \ \ ({1}') \\ <br /> & f\left( {{x}_{1}},{{x}_{4}} \right)-f\left( {{x}_{1}},{{x}_{2}} \right)=f\left( {{x}_{2}},{{x}_{4}} \right)\ \ \ \ \ ({2}') \\ <br /> & f\left( {{x}_{1}},{{x}_{3}} \right)-f\left( {{x}_{1}},{{x}_{2}} \right)=f\left( {{x}_{2}},{{x}_{3}} \right)\ \ \ \ \ ({3}') \\ <br /> \end{align}$$

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

 

[A_Studen_349q]

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...

 

[SammyS]

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:

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.

 

[A_Studen_349q]

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.

 

[ehild]

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

 

[jedishrfu]

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) )

 

[gabbagabbahey]

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 $(x_1, x_2, x_3, x_4)=(1, 2, 0, 0)$

 

[ehild]

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

 

[gabbagabbahey]

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!

 

[A_Studen_349q]

So, any new ideas?