Q:

Finding The Radix of A Quadratic Equation

I have found previous solutions to finding the radix of a quadratic equation, where both of the provided roots return the same radix or base. However, unless I am some type of arithmetic error of which I am not aware, currently I am obtaining two different bases neither of which satisfy my required conditions.
The problem statement and all variables are as follows:
Given x^2 - 11x + 29 = 0 with the solutions x = 4 and x = 8, find the radix of the equation.
I found with x = 4 -> radix = 21/2. This provides the correct root of x = 4, but the second root is not quite equal to the desired value of 8.
Similarly with x = 8 -> radix = 65/6, which again gives the one desired root this time of 8, and a second root which is less than four. 
I'm wondering if perhaps I am to combine these two radixes in some manner, or if I have made an arithmetic error as I have tried this problem over and over, each time obtaining the same two radixes. 

A:

There is no solution to this problem. To see why, lets assume that there is a base b that satisfies the conditions. The polynomial could be written as
x^2 - (b + 1) x + (2b + 9)

On the other hand, if the roots are 4 and 8 we would have that the polynomial should be the same as
(x - 4)(x - 8) = x^2 - 12x + 32

But this can only happen if both polynomials have the very same coefficients:
-(b + 1) = -12
2b + 9 = 32

Multiply the first eq. by 2 and sum it to the second:
7 = -24 + 32

which is absurd. Therefore there is not such a base b.

