Solving Cubic Equations - General Method

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Discussion Overview

The discussion revolves around methods for solving cubic equations of the form ax3 + bx2 + cx + d = 0. Participants explore both general approaches and specific techniques, including the reduction of cubic equations and the implications of multiple solutions.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant requests a general method for solving cubic equations.
  • Another participant shares a link to a resource that outlines the solution process for cubic equations.
  • A detailed explanation is provided regarding the manipulation of the cubic equation, including the transformation to a reduced cubic form and the use of a quadratic equation derived from it.
  • There is a discussion about the nature of the solutions, specifically whether the method yields one or three values for x, with a mention of complex solutions and the potential for multiple distinct roots.

Areas of Agreement / Disagreement

Participants express uncertainty about the correctness of the equations presented, and there is a question regarding the number of solutions provided by the method discussed. The discussion remains unresolved regarding the implications of the derived solutions.

Contextual Notes

Some participants express a lack of ability to verify the equations presented, indicating potential limitations in their understanding or the complexity of the material discussed.

TheDestroyer
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Hi guyz,

How can I solve the equation with the form :

ax^3 + bx^2 + cx + d = 0

I want the general way to solve allllllllll cubic equations..

Thanks
 
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The post above is very good but here it is in a nutshell:

We know that (a- b)3= a3-3a2b+ 3ab2- b3 and that 3ab(a-b)= 3a2b- 3ab so:

(a-b)3+ 3ab(a-b)= a3- b3
(the middle terms cancel).

That is: if we pick any two numbers a, b and let x= a- b, m= 3ab and n= a3- b3, then x3+ mx= n.

What about the other way? If we are given m and n, can we solve for a and b (and so find x)?

Yes, we can. From m= 3ab, we have b= m/(3a). Put that into
a3- b3= n and we have a3- m3/(33a3)= n

Multiply both sides of the equation by a3 and we have the (6th degree!) equation
a6- (m/3)3= naa3.

But if we let u= a3, this reduces to a quadratic equation for u: u2- nu- (m/3)3= 0.

We can solve that by the quadratic formula:
u= a3= (n +/- √(n2+ 4(m/3)3))/2= (n/2)+/- √((n/2)2+ (m/3)3).

Since a3- b3= n,
b3= a3- n
= (-n/2)+/- √((n/2)2+ (m/3)3).

Solving for a and b,

a= (((n/2)+/- √((n/2)2+ (m/3)3))1/3
b= ((-(n/2)+/- √((n/2)2+ (m/3)3))1/3

and, finally, x= a-b.


Notice that the equation was x3+ mx- n= 0 which has no x2. This is the "reduced" cubic.

To solve a general cubic, x3+ ax2+ bx+ c= 0,
"Shift" the variable: replace x by y+ y0 in this equation and then choose y0 to make the coefficient of y2 equal to 0.
 
Thanks Guyz,

Thanks Guyz, I hope these equations are true, because i can't check them right now :)

I'm very thankful :):)
 
?

Hey Guyz, Does the equation "imhereyeah" posted solves with 3 values of x? or 1 only?
 


Originally posted by TheDestroyer
Hey Guyz, Does the equation "imhereyeah" posted solves with 3 values of x? or 1 only?

Yes. Note that the "a" and "b" in that reply are derived from cube roots. So "a" and "b" each have 3 possible (complex) values giving a total of 9 combinations in the solution (though at most 3 will be distinct).
 
Last edited:

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