Find Linear Combination of Coefficients Given Term in Quadratic Root Sequence

Given that $\alpha$ and $\beta$ are the distinct roots of the quadratic equation $x^2 - x - 1 = 0$, we can express them as: $\alpha = \frac{1 + \sqrt{5}}{2}, \quad \beta = \frac{1 - \sqrt{5}}{2}$

Since $\alpha$ and $\beta$ satisfy $x^2 = x + 1$, we can find higher powers using the recurrence relation: $x^3 = x^2 + x = (x + 1) + x = 2x + 1$ $x^4 = 2x^2 + x = 2(x + 1) + x = 3x + 2$

Thus, we have: $\alpha^4 = 3\alpha + 2$ $\beta^4 = 3\beta + 2$

The sequence term $a_4$ is defined as: $a_4 = p\alpha^4 + q\beta^4$

Substituting the expressions for $\alpha^4$ and $\beta^4$: $a_4 = p(3\alpha + 2) + q(3\beta + 2) = 3(p\alpha + q\beta) + 2(p + q)$

Now, evaluate $p\alpha + q\beta$: $p\alpha + q\beta = p\left(\frac{1 + \sqrt{5}}{2}\right) + q\left(\frac{1 - \sqrt{5}}{2}\right) = \left(\frac{p + q}{2}\right) + \left(\frac{p - q}{2}\right)\sqrt{5}$

Substituting this back into $a_4$: $a_4 = 3\left[ \left(\frac{p + q}{2}\right) + \left(\frac{p - q}{2}\right)\sqrt{5} \right] + 2(p + q)$ $a_4 = \frac{7(p + q)}{2} + \frac{3(p - q)}{2}\sqrt{5}$

We are given $a_4 = 28$, so: $\frac{7(p + q)}{2} + \frac{3(p - q)}{2}\sqrt{5} = 28$ $\left( \frac{7(p + q)}{2} - 28 \right) + \left( \frac{3(p - q)}{2} \right)\sqrt{5} = 0$

Since $p, q \in \mathbb{Z}$, both $\left( \frac{7(p + q)}{2} - 28 \right)$ and $\left( \frac{3(p - q)}{2} \right)$ are rational numbers. By the given FACT ($a + b\sqrt{5} = 0 \implies a = 0$ and $b = 0$ for rational numbers $a, b$), we obtain the system of equations:

1. $\frac{3(p - q)}{2} = 0 \implies p = q$
2. $\frac{7(p + q)}{2} - 28 = 0 \implies 7(p + q) = 56 \implies p + q = 8$

Since $p = q$: $2p = 8 \implies p = 4 \quad \text{and} \quad q = 4$

Finally, we find the value of $p + 2q$: $p + 2q = 4 + 2(4) = 12$

Thus, the correct option is D.