Solving Higher Degree Polynomial For Real Solution(s).

In summary, there are four real solutions to the equation (x^2-9x-1)^{10}+99x^{10}=10x^9(x^2-1). However, only two of them can be explicitly determined. Graphing is not a suitable method to find the solutions, and the real roots are between $-\infty$ and $+\infty$.
  • #1
anemone
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Find real solution(s) to the equation \(\displaystyle (x^2-9x-1)^{10}+99x^{10}=10x^9(x^2-1)\)
 
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  • #2
anemone said:
Find real solution(s) to the equation \(\displaystyle (x^2-9x-1)^{10}+99x^{10}=10x^9(x^2-1)\)
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.
 
  • #3
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 \(\displaystyle (x^2-9x-1)^{10}-10x^9(x^2-9x-1)+9x^{10}=0\)

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

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

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

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.
 
  • #4
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.
 
  • #5
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!:eek::p:)
 
  • #6
anemone said:
Find real solution(s) to the equation \(\displaystyle (x^2-9x-1)^{10}+99x^{10}=10x^9(x^2-1)\)
Graph it! Hey, I'm a Physicist. (Beer) (Physics is always better with beer.)

-Dan
 
  • #7
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)
 
  • #8
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 $\chi$ $\sigma$
 

FAQ: Solving Higher Degree Polynomial For Real Solution(s).

What is a higher degree polynomial?

A higher degree polynomial is a polynomial function with the highest power of the variable being greater than 2. Examples include quadratic, cubic, and quartic functions.

What is the process for solving a higher degree polynomial?

The process for solving a higher degree polynomial involves factoring the polynomial, setting each factor equal to zero, and solving for the variable. This can be done using various methods such as the quadratic formula, synthetic division, or the rational root theorem.

How do I know if a higher degree polynomial has real solutions?

A higher degree polynomial will have real solutions if the discriminant (b^2 - 4ac) is greater than or equal to zero. If the discriminant is less than zero, the polynomial will have complex solutions.

What is the difference between real and complex solutions?

Real solutions are solutions in which the variable is a real number, meaning it can be plotted on a number line. Complex solutions involve imaginary numbers and cannot be plotted on a number line.

Can a higher degree polynomial have more than one real solution?

Yes, a higher degree polynomial can have multiple real solutions. The number of real solutions will be equal to the degree of the polynomial, but some solutions may be repeated or have a multiplicity greater than one.

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