Can you solve this double cubic algebraic equation?

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Discussion Overview

The discussion revolves around solving a double cubic algebraic equation, specifically focusing on the expression $x+\frac{1}{x}$ and its implications for finding real and complex solutions. Participants explore different methods for solving the equation, including algebraic manipulations and substitutions.

Discussion Character

  • Mathematical reasoning, Technical explanation

Main Points Raised

  • One participant suggests viewing $x+\frac{1}{x}$ as a whole and claims there are four real solutions: $x=\frac{3±\sqrt{5}}{2}$ and $x=\frac{-3±\sqrt{5}}{2}$, along with two complex solutions: $x=±i$.
  • Another participant proposes an alternative method by multiplying both sides by $x^3$ and substituting $y = x^2$, leading to a cubic equation for $y$ that can be solved using the rational root theorem.
  • A participant expresses appreciation for the feedback received on their initial idea.

Areas of Agreement / Disagreement

There are multiple competing views on how to approach solving the equation, with no consensus on a single method being preferred.

Contextual Notes

The discussion does not clarify the assumptions underlying the proposed methods or the specific conditions under which the solutions hold.

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The key idea is to view $x+\frac{1}{x}$ as a whole. There are four real solutions: $x=\frac{3±\sqrt{5}}{2}$, $x=\frac{-3±\sqrt{5}}{2}$ (and two complex solutions: $x=±i$). Here is the explanation:
 
MathTutoringByDrLiang said:
The key idea is to view $x+\frac{1}{x}$ as a whole. There are four real solutions: $x=\frac{3±\sqrt{5}}{2}$, $x=\frac{-3±\sqrt{5}}{2}$ (and two complex solutions: $x=±i$). Here is the explanation:

Nice idea!

Or you could just multiply both sides by $x^3$ and sub in $y = x^2$. The resulting cubic equation for y is easy to solve using the rational root theorem.

-Dan
 
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topsquark said:
Nice idea!

Or you could just multiply both sides by $x^3$ and sub in $y = x^2$. The resulting cubic equation for y is easy to solve using the rational root theorem.

-Dan
Thank you very much for your feedback!

Derek
 

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