Let α and β be the roots of x² - 3x + 1 = 0…

Start. $\alpha + \beta = 3$ and $\alpha\beta = 1$.

(a) $\alpha^2 + \beta^2 = 3^2 - 2 \times 1 = 7$.

A rule for the powers. Each root satisfies $t^2 = 3t - 1$, so $t^n = 3t^{n-1} - 1t^{n-2}$. Adding for $\alpha$ and $\beta$: $s_n = 3s_{n-1} - 1s_{n-2}$, with $s_0 = 2$ and $s_1 = 3$.

(b) and (c) $s_2 = 7$, $s_3 = 3 \times 7 - 3 = 18$, $s_4 = 47$, $s_5 = 123$ and $s_6 = 3 \times 123 - 47 = 322$.

(d) $(\alpha - \beta)^2 = 3^2 - 4 \times 1 = 5$, so $|\alpha - \beta| = \sqrt{5} = 2.24$.

(e) Sum: $(\alpha + \beta) + \dfrac{\alpha + \beta}{\alpha\beta} = 3 + 3 = 6$. Product: $\left(\alpha + \dfrac1\beta\right)\left(\beta + \dfrac1\alpha\right) = \alpha\beta + 1 + 1 + \dfrac{1}{\alpha\beta} = 4$. So $p = 6$ and $q = 4$.