How to solve cube roots question ?

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    Cube Roots
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Discussion Overview

The discussion centers around methods for solving a cubic equation, specifically the equation x^3 - 100x^2 - 7800x + 16300 = 0. Participants explore various approaches, including numerical methods and historical formulas.

Discussion Character

  • Technical explanation, Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant expresses difficulty in solving the cubic equation and seeks alternatives to trial and error.
  • Another participant suggests that numerical methods can provide approximate solutions.
  • A different participant mentions Cardano's method as a potential solution, noting it may be lengthy.
  • Further, a participant provides a transformation of the cubic equation into a simpler form, detailing the steps to apply Cardan's formula.

Areas of Agreement / Disagreement

Participants present multiple methods for solving the cubic equation, but there is no consensus on which method is preferable or most effective.

Contextual Notes

Some methods discussed may require additional assumptions or steps that are not fully resolved in the conversation.

sunny86
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How to solve cube roots question ?
Example :
x^3 - 100x^2 - 7800x + 16300 = 0

I had think long time but still cannot find the way. Besides trial an error, is there anyway to solve this problem ?
thank you.
 
Mathematics news on Phys.org
There are numerical methods to get approximate solutions ...

For an exact solution you may find useful this

http://mizar.uwb.edu.pl/JFM/Vol12/polyeq_1.html"
 
Last edited by a moderator:
For a given equation [tex]x^3+ax^2+bx+c=0[/tex], we can substitute [tex]x=y-\frac{a}{3}[/tex], which implies [tex]y^3+py+q=0[/tex], where [tex]p=b-\frac{a^2}{3}[/tex], and [tex]q=\frac{2a^3}{17}-\frac{ab}{3}+c[/tex]. After further manipulation, one can retrieve [tex]x=\sqrt[3]{-\frac{q}{2}+\sqrt{(\frac{q}{2})^2+(\frac{p}{3})^3}}+\sqrt[3]{-\frac{q}{2}-\sqrt{(\frac{q}{2})^2+(\frac{p}{3})^3}}[/tex], which represents Cardan's formula.
 
thank guy ~
 

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