$a \in (0,1]$ satisfies $a^{2008} -2a +1 = 0$ and we define $S$ as $S=1+a+a^2+a^3........a^{2007}$. The sum of all possible value(s) of $S$ is?

Question

This is homework.

Let $a \in (0,1]$ satisfies the equation $$a^{2008} -2a +1 = 0$$ and we define $S$ as $$S=1+a+a^2+a^3........a^{2007}$$

The sum of all possible value(s) of $S$ is?

My Attempt

$a=1$ is obviously a solution.Hence one value of $S$ is $2008$.

To find other values of $S$, I need all the other solutions of $a$ lying between $0$ and $1$. When I graphed the function here, the other root was approximately coming out to be $0.5$. But I think that such a problem has to be solved exactly and no approximations are needed. Hence, I am stuck here.

Solutions should preferably not involve a calculator or any computer tool and use college level math only (since this problem was found in a college level book).

Answer 1

Since $1$ is a solution to $a^{2008}−2a+1 = 0$ therefore $a^{2008}−2a+1$ can be divided by $a-1$. If you do so, you'll get $a^{2007}+a^{2006}+...+a^2+a-1$. Since it's trivially a monotonous function and has values of opposite signs at $0$ and $1$, it must have a single real root in the interval $(0;1)$. It is such a number $a$ that $a^{2007}+a^{2006}+...+a^2+a-1 = 0$. Therefore, $a^{2007}+a^{2006}+...+a^2+a+1 = 2$. So you get $S = 2$ for the other root.

The final answer is $2+2008 = 2010$