How to Solve a Polynomial with Reciprocal Roots?

[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.

 

[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.

 

[M98Ranger]

To solve this question, we need to use the properties of polynomials and their roots. First, let's recall that for a polynomial equation of degree n, there can be at most n distinct real roots. In this case, we have a polynomial of degree 10, so it can have at most 10 real roots.

Now, let's consider the given condition that 1/r is also a solution. This means that if r is a root of the polynomial, then 1/r must also be a root. This can be represented mathematically as:

x^10 + a*x + 1 = 0

Substituting x = r, we get:

r^10 + a*r + 1 = 0

And substituting x = 1/r, we get:

(1/r)^10 + a*(1/r) + 1 = 0

Simplifying, we get:

1 + a/r^9 + 1 = 0

a/r^9 = -2

a = -2*r^9

Now, we can substitute this value of a in the original polynomial equation:

x^10 + (-2*r^9)*x + 1 = 0

This is a polynomial of degree 10, and we know that it has at most 10 real roots. Therefore, we can conclude that the only possible values of r that satisfy the given condition are the roots of this new polynomial.

To find these roots, we can use techniques such as synthetic division, factoring, or the rational root theorem. Once we have found the roots, we can check if they satisfy the given condition by plugging them in the original equation and checking if 1/r is also a solution.

I hope this explanation helps you in solving this polynomial equation. Remember to always use the properties of polynomials and their roots to your advantage. Good luck!