Suppose that $\alpha$ and $\alpha'$ are two negative real numbers. Consider the following two equations
(1) $b^2=b\alpha'+\alpha$
(2) $b^2\alpha'+b\alpha=1$
Find the number the solutions of the two equations with $b\in \mathbb C$.
The equation (2) implies that $b(b\alpha'+\alpha)=1$. And replacing the equation (1) we have that $$b^3=1.$$
This implies that the cube roots of unity are the only possible solutions. Note that $b=1$ is impossible, because $\alpha,\alpha'$ are negative.
The cube roots of unity are they possible solutions or are there more possible solutions?
How are these types of equations solved?
Thanks you all.
Let $w=e^{2\pi i/3}$. Then the cube roots of unity are $1,w,w^2$, and one also has $1+w+w^2=0$. From your computations it follows that $b$ cannot be 1. So it is either $w$ or $w^2$. First we will consider (1). If $b=w$, then $w^2=w\alpha'+\alpha$, i.e., $(\alpha'+1)w+(\alpha+1)=0$, which is only possible if $\alpha'=\alpha=-1$. Similarly, one can see that $b=w^2$ is a solution for (1) if $\alpha'=\alpha=-1$. Exactly the same way one treats (2) by considering the possibilities $b=w,w^2$. If one wants a common solution to (1) and (2), then one only obtains that if $\alpha'=\alpha=-1$, and in that case both $b=w$ and $w^2$ is a solution.