How to Solve a Polynomial with Reciprocal Roots?

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In summary, the conversation discusses finding real numbers with a specific property in a polynomial equation. The solution involves setting a condition for the absolute value of a and solving for a using eighth roots of unity. Ultimately, the only solutions for a are +-2.
  • #1
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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.
 
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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

[tex]r^8 =1=(\frac{-a+-\sqrt(a^2-4)}{2})^8[/tex]

Since we want real solutions, I am assuming that a is a real number, and thus we have [tex]-a+-\sqrt(a^2-4)=2u.[/tex] 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.
 
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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!
 

FAQ: How to Solve a Polynomial with Reciprocal Roots?

What is a polynomial?

A polynomial is a mathematical expression consisting of variables and coefficients, combined using addition, subtraction, and multiplication. The highest exponent of the variable is known as the degree of the polynomial.

How do I solve a polynomial?

To solve a polynomial, you must first simplify the expression by combining like terms. Then, use the appropriate method (factoring, long division, etc.) to find the roots or solutions of the polynomial. These solutions are also known as the x-intercepts or zeros of the polynomial.

What are the common methods for solving polynomials?

The most common methods for solving polynomials include factoring, long division, synthetic division, and the quadratic formula. The method used depends on the degree and complexity of the polynomial.

What are the steps for solving a polynomial using factoring?

The steps for solving a polynomial using factoring are:
1. Identify any common factors among the terms.
2. Use the distributive property to rearrange the terms in descending order.
3. Look for patterns or use the quadratic formula to factor the polynomial.
4. Set each factor equal to zero and solve for the roots.
5. Check the solutions by plugging them back into the original polynomial.

Can I use a calculator to solve a polynomial?

Yes, most scientific and graphing calculators have the ability to solve polynomials. However, it is important to understand the steps and methods for solving polynomials without a calculator to ensure accuracy and understanding of the concept.

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