An intriguing system of equation

  • MHB
  • Thread starter anemone
  • Start date
  • Tags
    System
In summary: In this case, we have$\dfrac{2z}{x}=\dfrac{x^2+9k^2}{x^2+k^2}\ge \dfrac{-3xk}{2xk}=-1$$2z\ge -1x$$\therefore z<x$This concludes the solution to the problem.
  • #1
anemone
Gold Member
MHB
POTW Director
3,883
115
Hi MHB,

I've recently come across a system of equation problem that I could not solve it fully, I don't know how to prove that system, the resulted polynomial has only three real roots.

Problem:

Solve for all real solutions in terms of $k$ for the system below:

$x(x^2+9k^2)=2z(x^2+k^2)$

$y(y^2+9k^2)=2x(y^2+k^2)$

$z(z^2+9k^2)=2y(z^2+k^2)$

It's no hard to see that when $x=y=z$, then we have the solutions:

$(x,\,y,\,z)=(-\sqrt{7}a,\,-\sqrt{7}a,\,-\sqrt{7}a)\stackrel{\text{or}}{=}(0,\,0,\,0)\stackrel{\text{or}}{=}(\sqrt{7}a,\,\sqrt{7}a,\,\sqrt{7}a)$

But I don't know what other conclusions that I could draw when I considered the condition where $x\ne y \ne z$ that could lead me to solve for this problem.

Wolfram has confirmed that those are the only real solutions the the given system, and I tried, using the knowledge that I borrowed from inequality, geometry, function, etc and nope, nothing helped. And here I am, hoping to gain some useful advice from the members to solve this system successfully.

Any help would be much appreciated.

Thanks.
 
Mathematics news on Phys.org
  • #2
This is just a hunch but the cyclic nature of the system along with the only solution being $x = y = z$ suggests that $x < y$ implies $y < x$ and vice versa, and same for $(y, z)$ and $(z, x)$ (or the other way around) so that the only solution can be $x = y = z$. I haven't tried, but if you haven't tried that it seems like it may be fruitful.
 
  • #3
Thanks Bacterius for your reply and yes, I think I was able to complete the solution based on your suggestion and here is my work:

First, if we rewrite the first given equation as $\dfrac{2z}{x}=\dfrac{x^2+9k^2}{x^2+k^2}$, we see that $\dfrac{2z}{x}>0$ for all real $x$ and $k$.

This means both $x$ and $z$ are either both positive or both negative. By the same token, we also have $x$ and $y$ are both positive or negative, and so are $y$ and $z$.

Since $x^2,\,9k^2,\,k^2$ are all positive and if we apply AM-GM inequality on those, we see that

$x^2+9k^2\ge 2(\sqrt{9x^2k^2})=2(3xk)=6xk$ and $x^2+k^2\ge 2(\sqrt{x^2k^2})=2(xk)=2xk$

Now, if we consider the case where $x$ and $z$ are both positive, we see that

$\dfrac{2z}{x}=\dfrac{x^2+9k^2}{x^2+k^2}\ge \dfrac{6xk}{2xk}=3$

$2z\ge 3x$

$\therefore z>x$

Similarly, from $\dfrac{2x}{y}=\dfrac{y^2+9k^2}{y^2+k^2}$, we have $x>y$ and this suggests $z>y$ but from $\dfrac{2y}{z}=\dfrac{z^2+9k^2}{z^2+k^2}$, we have $y>z$. This leads to a contradiction.

The other case for which all $x,\,y,\,z$ are negative works the same way and so, there is only one condition in the given system of equation that we need to consider, and that is when $x=y=z$.
 

FAQ: An intriguing system of equation

1. What is a system of equations?

A system of equations is a set of two or more equations that contain the same variables. The solution to a system of equations is the set of values that make all of the equations in the system true simultaneously.

2. How do you solve a system of equations?

There are several methods for solving a system of equations, including substitution, elimination, and graphing. The most efficient method will depend on the specific equations in the system.

3. Can a system of equations have more than one solution?

Yes, a system of equations can have one, infinitely many, or no solutions. This will depend on the relationship between the equations and the number of variables in the system.

4. What are real-world applications of systems of equations?

Systems of equations are commonly used in science, engineering, and economics to model and solve real-world problems. For example, they can be used to determine the optimal production levels for a company or the trajectory of a projectile.

5. Is there a limit to the number of equations that can be in a system?

No, there is no limit to the number of equations in a system. However, as the number of equations increases, the complexity of solving the system also increases.

Similar threads

Replies
7
Views
2K
Replies
1
Views
951
Replies
1
Views
2K
Replies
1
Views
1K
Replies
6
Views
949
Replies
6
Views
1K
Back
Top