Solve Polynomial Equation for Real a

[Derivative86]

Hard 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
Thx for ur help

 

[TenaliRaman]

Hmm!
i think i get a = -2 is the only solution.

-- AI

 

[tacman]

To solve this problem, we can use the fact that if a polynomial equation has a real solution r, then its conjugate 1/r is also a solution. This means that if we can find a value for a that satisfies this property, then we have found all possible solutions to the equation.

First, let's rewrite the equation in terms of the solutions r and 1/r:
(x - r)(x - 1/r) = 0
Expanding this out, we get:
x^2 - (r + 1/r)x + 1 = 0

Comparing this to the given equation, we can see that a = -(r + 1/r). This means that for the given equation to have a solution r such that 1/r is also a solution, we need to find a value for a that satisfies this property.

To do this, we can use the quadratic formula:
a = -(r + 1/r) = -(-(r + 1/r)^2 + 4) / 2
Simplifying this, we get:
a = (r + 1/r)^2 - 2

Now, we can see that for any value of r, a will have a corresponding value that satisfies the given property. For example, if r = 1, then a = 0. If r = -1, then a = -2. This means that there are infinitely many values of a that satisfy the given equation.

In conclusion, the set of real numbers a that satisfy the given polynomial equation are all values that can be expressed as (r + 1/r)^2 - 2, where r is any real number.