Roots of a polynomial

In summary, the problem is asking to find all real numbers a that satisfy the given polynomial equation with roots r and 1/r. One approach is to solve for r and substitute it into the equation, simplifying it into two 10th-order equations for r. Then, using systems of equations, r can be solved for and plugged back into the original equation to find the value of a. Both Krab and JonF suggest this method, with Krab also mentioning the implication that x = r and x = 1/r.
  • #1
Derivative86
26
0
Hey guys can you help me solve this problem:

Find all real numbers [tex]a[/tex] with the property that the polynomial equation [tex]x^{10} + ax + 1 = 0[/tex] has a real solution [tex]r[/tex] such that [tex]1 / r[/tex] is also a solution.

Thank you for helping me :smile:
Sry I have posted this problem once before, but nobody helped me :frown: and I need to know how to solve the problem badly. Please help me :smile:
 
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  • #2
R is the solutions to this equation:

[tex]x^{10} - \frac{x}{x^{-1}+x^9} +1 = 0[/tex]
 
  • #3
Thanks for helping me, you r awesome :smile: But I don't understand how you get there, can u explain?
 
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  • #4
plug 1/r into the equation, multiply the result by r^10. Now you have two 10th-order equations for r. Subtract them. The rest is easy.
 
  • #5
You have two equations in two unknowns, so use systems of equations or some other such thing to solve.

For your conditions to be met both of the following must be true.

x^10 + rx + 1 = 0

x^10 +1/rx + 1 = 0

solve for r in the second equation.

r = -1/(x^-1 + x^9)

plug in r into the first equation, and you get what I got.
 
  • #6
Thanks for helping me Krab, your reply is referring to the orginial thread right?
 
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  • #7
JonF said:
You have two equations in two unknowns, so use systems of equations or some other such thing to solve.

For your conditions to be met both of the following must be true.

x^10 + rx + 1 = 0

x^10 +1/rx + 1 = 0

solve for r in the second equation.

r = -1/(x^-1 + x^9)

plug in r into the first equation, and you get what I got.

First, Thanks for helping me JonF. But you r saying that [tex]r = a[/tex] and [tex]1 / r = a[/tex]. However, the problem said that the polynomial [tex]x^{10} + ax + 1 = 0[/tex] has roots [tex]r[/tex] and [tex]1 / r[/tex]. Which implies that [tex]x = r[/tex] and [tex]x = 1 / r[/tex].
I think Krab is right
 

1. What are the roots of a polynomial?

The roots of a polynomial are the values that make the polynomial equation equal to zero. In other words, they are the x-values where the polynomial intersects the x-axis on a graph.

2. How do you find the roots of a polynomial?

The most common method to find the roots of a polynomial is by factoring. This involves breaking down the polynomial into simpler expressions and setting each factor equal to zero to solve for the roots.

3. Can a polynomial have more than one root?

Yes, a polynomial can have multiple roots. The number of roots a polynomial has is equal to its degree, which is the highest exponent in the equation. For example, a quadratic polynomial can have a maximum of two roots.

4. What is the relationship between the roots of a polynomial and its graph?

The roots of a polynomial are the x-values where the graph of the polynomial intersects the x-axis. This means that the roots can be used to determine the x-intercepts of the graph.

5. Are complex numbers possible roots of a polynomial?

Yes, complex numbers can be roots of a polynomial. Complex numbers are used to represent the solutions to polynomial equations that do not have real number solutions.

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