Efficient Solutions for Fractional Polynomial Equations

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  • Thread starter Thread starter ACLerok
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Discussion Overview

The discussion revolves around finding efficient methods for solving fractional polynomial equations, particularly focusing on quadratic and cubic equations. Participants explore various approaches, including the quadratic formula and Cardano's method, while also expressing concerns about the complexity of these methods.

Discussion Character

  • Technical explanation
  • Homework-related
  • Debate/contested

Main Points Raised

  • One participant inquires about the quickest method to solve a quadratic equation with fractional coefficients, specifically the equation (x^2) - (1/6)x - 1/6 = 0.
  • Another participant suggests using the quadratic formula and notes the difficulty of solving higher-degree polynomials analytically, mentioning methods like Newton's Method for approximations.
  • A different participant provides the general solutions for a quadratic equation in the form x^2 + px + q = 0, applying it to the example given.
  • One participant asks about solving a cubic polynomial, x^3 + 0.5x^2 - 0.25x - 0.125 = 0, and whether they should factor or adjust the equation.
  • Another participant references Cardano's method for solving cubic equations, providing links for further reading.
  • A participant expresses frustration about the complexity of the processes involved in solving cubic equations.
  • One participant suggests finding an initial solution (x=0.5) and then factoring the cubic polynomial, demonstrating a specific approach to solving it.
  • Another participant comments on the difficulty of solving certain problems, implying that some may be insoluble.
  • A participant challenges the effectiveness of the methods discussed, suggesting that simpler methods may be applicable for homework and exam tasks.

Areas of Agreement / Disagreement

Participants express a range of views on the methods for solving polynomial equations, with some advocating for the quadratic formula and others for factoring techniques. There is no consensus on the best approach, and the discussion reflects differing opinions on the complexity and solvability of polynomial equations.

Contextual Notes

Participants highlight the challenges associated with solving higher-degree polynomials and the potential for certain problems to be insoluble. The discussion includes references to specific methods without resolving the effectiveness or applicability of these methods in all contexts.

ACLerok
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Hi, I was just reading up on some material for one of my exams and came across something. I realize this may seem like a trivial, obvious question but:
What is the easiest way to find the solutions of the polynomial (for example):
(x^2) - (1/6)x - 1/6 = 0
Other than guessing and testing for the correct values of x, what is the quickest and most efficient method of solving for x when the coefficients of each term is a fraction rather than a whole number? Thanks in advance.
 
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well you just let your fraction, 1/6, equate to b, as the equation was a quadratic of the form [itex]ax^2 + bx + c = 0[/itex]. Then substitute your values for a, b and c into the quadratic formula, [tex]x=\frac {-b(+-)(b^2 - 4ac)^0.5}{2a}[/tex] If you mean in general, you it is quite hard to solve equations of degree 3 and 4 analytically, and after 5 it can only be approximated using methods such as Newtons Method.
 
Given the equation

[tex]x^2 + px + q = 0[/tex]

the solutions are:

[tex]x_{1} = - \frac{p}{2} - \sqrt{ \left( \frac{p}{2} \right)^2-q }[/tex]

and

[tex]x_{2} = - \frac{p}{2} + \sqrt{ \left( \frac{p}{2} \right)^2-q }[/tex]

So in your example:

p=-1/6 and q=-1/6
 
Last edited:
Thanks a lot!
 
one more thing: if I am trying to find the solutions if the polynomial is of order 3, for example:
x^3 + 0.5x^2 - 0.25x - 0.125 = 0

Do I add 0.125 to both sides then factor an x or do must I do something else? Thanks
 
Wow what a long and involved process :cry:
There is no other process of finding them? oh well
Thanks
 
You can try to find (adjust) one solution ([tex]x=0.5[/tex]). After that you can divide [tex]x^3 + 0.5x^2 - 0.25x - 0.125[/tex] by [tex]x-0.5[/tex].
The result is [tex]x^2+x+0.25[/tex].
So [tex]x^3 + 0.5x^2 - 0.25x - 0.125=(x-0.5)(x^2+x+0.25)=0[/tex]
You already can solve it.
 
Last edited:
ACLerok said:
Wow what a long and involved process :cry:
There is no other process of finding them? oh well
Thanks

Not all problems easily stated are easily solved.

Most happen to be insoluble, I'm afraid.
 
  • #10
arildno
I don't think that ACLerock solve very difficult problems.
It is possible to apply more easy methods when solving home and exam tasks.
(for example the method from my post).
 

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