Solving Higher Degree Polynomial For Real Solution(s).

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Discussion Overview

The discussion revolves around finding real solutions to the polynomial equation $$(x^2-9x-1)^{10}+99x^{10}=10x^9(x^2-1)$$. Participants explore various approaches to solving the equation, including hints and graphical methods, while addressing the existence of real solutions.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants suggest that there are two real solutions, while others express uncertainty about the total number of real solutions.
  • One participant acknowledges a mistake in their earlier assessment regarding the number of real solutions, stating there are no additional solutions.
  • Another participant proposes graphing the equation as a method to visualize the solutions, though they note that graphing may not be the most effective approach in this case.
  • There is a light-hearted exchange about the enjoyment of physics and mathematics, with references to beer, which adds a casual tone to the discussion.

Areas of Agreement / Disagreement

Participants express differing views on the number of real solutions, with some asserting there are two solutions and others retracting earlier claims about additional solutions. The discussion remains unresolved regarding the exact count of real solutions.

Contextual Notes

Participants mention hints and approaches without providing explicit details, leaving some assumptions and steps in the reasoning process unclear. The effectiveness of graphing as a solution method is also debated.

anemone
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Find real solution(s) to the equation $$(x^2-9x-1)^{10}+99x^{10}=10x^9(x^2-1)$$
 
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anemone said:
Find real solution(s) to the equation $$(x^2-9x-1)^{10}+99x^{10}=10x^9(x^2-1)$$
Hint 1:
Divide both sides by $x^{10}$.
Hint 2:
Let $y = x-\frac1x$.
Hint 3:
What happens when $y=10$?
That will give you two real solutions. There are two others, but I do not think that they can be explicitly determined.
 
Hi Opalg,

I thank you for taking the time to share your helpful hints in this challenge problem.

My approach is different from yours and I wish to share it here too.

Hint 1:
We first rewrite the equation as $$(x^2-9x-1)^{10}-10x^9(x^2-9x-1)+9x^{10}=0$$

Hint 2:
Assume that $$(x^2-9x-1)^{10}+9x^{10}=10x^9(x^2-9x-1)$$ is true.

Hint 3:
We need to show that $$(x^2-9x-1)^{10}=x^{10}$$ and $$x^9(x^2-9x-1)=x^{10}$$ are true by obtaining two equivalent equation, i.e. $$x^2-10x-1=0$$ in order to prove our assumption is true.

Hint 4:
The original problem is solved if we solve for x for $$x^2-10x-1=0$$

Opalg said:
That will give you two real solutions. There are two others, but I do not think that they can be explicitly determined.

I failed to realize that there are two other real solutions to this problem. Would you mind to teach me how to tell how many real solutions there are for this particular problem?

Thanks in advance.
 
anemone said:
I failed to realize that there are two other real solutions to this problem. Would you mind to teach me how to tell how many real solutions there are for this particular problem?
That was my silly mistake. There are no other real solutions. Sorry about that.
 
Opalg said:
That was my silly mistake. There are no other real solutions. Sorry about that.

I see...hey, don't worry about that! :)

You are and always will be one of my favorite mathematicians at this site!:o:p:)
 
anemone said:
Find real solution(s) to the equation $$(x^2-9x-1)^{10}+99x^{10}=10x^9(x^2-1)$$
Graph it! Hey, I'm a Physicist. (Beer) (Physics is always better with beer.)

-Dan
 
topsquark said:
Graph it! Hey, I'm a Physicist. (Beer) (Physics is always better with beer.)

-Dan

Indeed. I am now confident the real roots are between $-\infty$ and $+\infty$, more or less (Smoking)
 
topsquark said:
Graph it! Hey, I'm a Physicist. (Beer) (Physics is always better with beer.)

-Dan

Unfortunately in this case graphing in not a comfortable way to arrive to the solution (Wasntme)...

http://www.123homepage.it/u/i69151835._szw380h285_.jpg.jfif

All is allways better with beer! (Beer)...Kind regards $\chi$ $\sigma$
 

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