# Help with solving a polynomial

1. Oct 8, 2004

### galois427

I need some help on how to solve this question. It asks me to find all real numbers with the property that the polynomial equation x^10 + a*x +1 = 0 has a real solution r such that 1/r is also a solution. I tried plugging in r and 1/r and equating the 2 equations, but that got me nowhere.

2. Oct 16, 2004

### robert Ihnot

If r is a root, so is 1/r, thus r^10+ar+1=0=r^10+ar^9+1. Thus for a not 0, r^8 =1.
This gives: r^2+ar+1=0. Absolute value of a equal or exceeds 2. At a=+-2, we have r=-+1. Since r^8 =1, we can look at

$$r^8 =1=(\frac{-a+-\sqrt(a^2-4)}{2})^8$$

Since we want real solutions, I am assuming that a is a real number, and thus we have $$-a+-\sqrt(a^2-4)=2u.$$ Where u is taken to be one of the eighth roots of unity, but the only ones not complex are +-1. Solving for this we get that a=+-2, as before.

Last edited: Oct 16, 2004