Solving a Tricky Nonlinear Equation System: A Quest for Closed Form Solutions

In summary, the conversation discusses the search for a closed form solution to a system of equations involving x and y. It is determined that the resulting quartic equation has two real and a pair of complex-conjugate solutions, and that no nice integers or rationals can be found. The use of Wolfram Alpha is suggested for solving the equation. A simpler method for solving the problem is also mentioned.
  • #1
n7imo
7
2
I'm trying to find a closed form (an algebraic solution) for the following system:

x² - y² = 5
x + y = xy

It's a bit tricky but I manage to end up with the quartic equation:
x^4 - 2x^3 + 5x^2 -10x + 5 =0
And this is where I get stuck looking for a closed form root.
Any suggestion would be appreciated
 
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  • #2
trial and error of synthetic division.
 
  • #3
n7imo said:
I'm trying to find a closed form (an algebraic solution) for the following system:

x² - y² = 5
x + y = xy

It's a bit tricky but I manage to end up with the quartic equation:
x^4 - 2x^3 + 5x^2 -10x + 5 =0
And this is where I get stuck looking for a closed form root.
Any suggestion would be appreciated
You won't find any integer or rational solutions. The general solution to a fourth degree equation is pretty daunting.

https://en.wikipedia.org/wiki/Quartic_function

This particular equation has two real and a pair of complex-conjugate roots.

BTW, I checked your algebra in reducing your system of equations to one equation in x. I think you have some mistakes there, since I don't obtain your particular quartic equation.

In any event, the resulting quartic still has two real and a pair of complex-conjugate solutions, none of which are nice integers or rationals.

I used Wolfram Alpha to solve for the roots. It's much easier than anything else.
 
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  • #4
SteamKing said:
You won't find any integer or rational solutions. The general solution to a fourth degree equation is pretty daunting.

https://en.wikipedia.org/wiki/Quartic_function

This particular equation has two real and a pair of complex-conjugate roots.

BTW, I checked your algebra in reducing your system of equations to one equation in x. I think you have some mistakes there, since I don't obtain your particular quartic equation.

In any event, the resulting quartic still has two real and a pair of complex-conjugate solutions, none of which are nice integers or rationals.

I used Wolfram Alpha to solve for the roots. It's much easier than anything else.

Indeed, the right resulting quartic equation is x^4 - 2x^3 - 5x^2 -10x - 5 =0. I used Cardano and Lagrange method to find the real roots, but their form is very ugly.
Actually I got this equation while trying to solve a simple geometrical problem. I'll post it today on a new thread, I'm interested in finding a simpler method to solving it since mine leads to a quartic equation.

Thanks for the contribution.
 

1. What is a nonlinear equation system?

A nonlinear equation system is a set of mathematical equations that cannot be expressed as a linear combination of variables. This means that the equations involve powers, products, or quotients of variables, making them more complex to solve compared to linear equations.

2. How is a nonlinear equation system different from a linear equation system?

A linear equation system can be solved by finding the values of its variables that satisfy all the equations, while a nonlinear equation system may have multiple solutions or no solution at all. Nonlinear equations also have a curved graph, unlike linear equations which have a straight line graph.

3. What are some real-life applications of nonlinear equation systems?

Nonlinear equation systems are used in various fields such as physics, engineering, economics, and biology to model complex relationships between variables. For example, they can be used to predict the growth of a population, the trajectory of a projectile, or the behavior of a chemical reaction.

4. How do you solve a nonlinear equation system?

There is no one method for solving all types of nonlinear equation systems. Some methods include substitution, elimination, and graphing. In some cases, numerical methods such as Newton's method or the bisection method may be used to approximate the solutions.

5. What are the challenges of working with nonlinear equation systems?

Nonlinear equation systems can be more difficult to solve compared to linear equations because they may have multiple solutions, no solutions, or require advanced mathematical techniques. They also often involve complex calculations, making it challenging to find an exact solution. Additionally, small changes in the equations or initial conditions can significantly affect the outcome, making it challenging to predict the behavior of the system accurately.

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