1. Given recurrence roots

The numbers $\alpha, \beta$ are roots of $x^2-x-1=0.$ So each root satisfies $r^2=r+1.$ Hence, for any integer $n\ge 0$, $r^{n+2}=r^{n+1}+r^n$ for $r=\alpha,\beta$.

2. Use this for the sequence $a_n$

Given $a_n=p\alpha^n+q\beta^n,$ we get \begin{align*} a_{n+2} &=p\alpha^{n+2}+q\beta^{n+2}\ &=p(\alpha^{n+1}+\alpha^n)+q(\beta^{n+1}+\beta^n)\ &=(p\alpha^{n+1}+q\beta^{n+1})+(p\alpha^n+q\beta^n)\ &=a_{n+1}+a_n. \end{align*} So the sequence satisfies the Fibonacci-type recurrence $a_{n+2}=a_{n+1}+a_n.$

3. Find $a_{12}$

Putting $n=10$, $a_{12}=a_{11}+a_{10}.$ Thus the correct option is:

$\boxed{\text{D: } a_{11}+a_{10}}.$

4. Why the fact about $a+b\sqrt5=0$ is not needed here

The given fact would be useful if we explicitly wrote $\alpha=\frac{1+\sqrt5}{2},\qquad \beta=\frac{1-\sqrt5}{2}$ and compared coefficients of $1$ and $\sqrt5$. But the recurrence follows directly from the quadratic equation, so that fact is unnecessary for this question.