What Values of a Allow a Polynomial to Have Reciprocal Roots?

[Derivative86]

Hey guys can you help me solve this problem:

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

Thank you for helping me
Sry I have posted this problem once before, but nobody helped me and I need to know how to solve the problem badly. Please help me

 

[JonF]

R is the solutions to this equation:

$$x^{10} - \frac{x}{x^{-1}+x^9} +1 = 0$$

 

[Derivative86]

Thanks for helping me, you r awesome But I don't understand how you get there, can u explain?

 

[krab]

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.

 

[JonF]

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.

 

[Derivative86]

Thanks for helping me Krab, your reply is referring to the orginial thread right?

 

[Derivative86]

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 $$r = a$$ and $$1 / r = a$$. However, the problem said that the polynomial $$x^{10} + ax + 1 = 0$$ has roots $$r$$ and $$1 / r$$. Which implies that $$x = r$$ and $$x = 1 / r$$.
I think Krab is right