How Does a Cubic Equation Simplify to a Quadratic?

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Discussion Overview

The discussion revolves around the simplification of cubic equations to quadratic equations, exploring methods for achieving this transformation. Participants inquire about examples, relevant topics in mathematics, and recommendations for textbooks.

Discussion Character

  • Exploratory
  • Technical explanation
  • Homework-related

Main Points Raised

  • One participant describes a method for reducing a cubic equation to a quadratic by substituting variables and manipulating the equation.
  • The specific cubic equation discussed is x^3 + 3x + 6 = 0, with a detailed substitution process provided.
  • Another participant expresses interest in understanding the mathematical topic classification of this process, questioning whether it falls under calculus or algebra.
  • A participant shares a personal anecdote about learning this method from an old textbook, though they cannot recall the title.

Areas of Agreement / Disagreement

Participants generally agree that the topic pertains to algebra, but there is no consensus on specific textbook recommendations or the best methods for simplification.

Contextual Notes

Some assumptions about the choice of variables in the substitution method are not fully explored, and there are unresolved details regarding the application of the method to different cubic equations.

Who May Find This Useful

University students studying algebra or those interested in methods for solving polynomial equations may find this discussion relevant.

athrun200
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The auxiliary equation for a cubic was simpler—a quadratic. This “resolvent quadratic” could be solved by the Babylonian method; then the solution of the cubic could be reconstructed by taking a cube root. Why Beauty is Truth P78

Can anyone can give one example how cubic become quadratic?
Or can you recommand me some books? (I am an university student.)
 
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Hi athrun200! :smile:

Here's a method where a cubic equation is reduced to a quadratic equation.


Starting with [itex]x^3+3x+6=0[/itex].

Substitute x=y+z, meaning you have a free choice for either y or z.
So [itex](y+z)^3+3(y+z)+6=0[/itex]

[itex]\Rightarrow (y^3+z^3+3y^2z+3yz^2) + 3(y+z)+6=0[/itex]

[itex]\Rightarrow y^3+z^3+3(yz+1)(y+z)+6=0[/itex]


Choose z such that [itex]yz+1=0[/itex], or [itex]z=-{1 \over y}[/itex]
Then: [itex]y^3 - {1 \over y^3} + 6 = 0[/itex]
[itex]\Rightarrow (y^3)^2 + 6(y^3) - 1 = 0[/itex]


Solve as a quadratic equation and back substituting z gives:
[tex]x=y+z=\sqrt[3]{-3 + \sqrt{10}} - {1 \over \sqrt[3]{-3 + \sqrt{10}}}[/tex]
or
[tex]x=y+z=-\sqrt[3]{3 + \sqrt{10}} + {1 \over \sqrt[3]{3 + \sqrt{10}}}[/tex]

Note that both solutions are the same root.
 
Thx very much!
which topic does it belong to? Calculus? or Algebra?
Are there any textbook about this?
 
Hmm, not sure.
*looking up calculus*
wiki said:
Calculus (Latin, calculus, a small stone used for counting) is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series.
No, not calculus.


*looking up algebra*
wiki said:
Algebra is the branch of mathematics concerning the study of the rules of operations and relations, and the constructions and concepts arising from them, including terms, polynomials, equations and algebraic structures.
Yes, I think it's algebra!


TBH, I learned it from an old book of my father, that he had while studying.
I don't recall what it was called, but it was a thick volume with a purple cover and many yellowed pages... I loved it! :smile:
 

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