I am having trouble starting this question.
Given that the sequence $(x_n)$ satisfies the recurrence relation: $$x_{n+1} = a x_n + b x_{n-1},$$ where $n= 1, 2, \dots$ and $a$ and $b$ are given numbers, if the quadratic $r^2-ar-b = 0$ has distinct roots $\alpha$ and $\beta$, show that $x_n = \alpha^n$ and $x_n = \beta^n$ are both solutions of the recurrence relation shown above.
I have looked at this so many times and wherever I start I get nowhere, thanks for any help in advance.
As $\alpha$ is a solution of the mentioned equation, we have: $\alpha^2 = a\alpha + b$. Then, for $n \in \mathbb N$, we have $\alpha^{n-1} \times (\alpha^2) = \alpha^{n-1} \times (a \alpha + b)$, i.e. $\alpha^{n+1} = a \alpha^n + b\alpha^{n-1}$.
Substitute $x_n = \alpha^n$ into the recurrence relation to get
$$ \alpha^{n+1}=a\alpha^n+b\alpha^{n-1}.$$
Dividing by $\alpha^{n-1}$ and rearranging yields
$$ \alpha^{2}-a\alpha-b=0,$$
which is true since $\alpha$ is a root of the characteristic equation.
The same is true for $x_n = \beta^n$.
