Step 1: Find the actual zeroes using the quadratic formula instead of using the sum-and-product shortcut.
Rather than jumping straight to the identity $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}$, let's find $\alpha$ and $\beta$ as real numbers first and substitute them in directly.
Step 2: Identify the coefficients and apply the quadratic formula.
For $p(x) = x^2 - 3x - 1$, we have $a = 1$, $b = -3$, $c = -1$.
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{3 \pm \sqrt{9 + 4}}{2} = \frac{3 \pm \sqrt{13}}{2} \]
So the two zeroes are:
\[ \alpha = \frac{3 + \sqrt{13}}{2}, \qquad \beta = \frac{3 - \sqrt{13}}{2} \]
Step 3: Write $\frac{1}{\alpha} + \frac{1}{\beta}$ as a single fraction.
\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta} \]
We still need $\alpha + \beta$ and $\alpha\beta$, but now we compute them directly from the explicit root values rather than quoting Vieta's relations.
Step 4: Compute $\alpha + \beta$ directly.
\[ \alpha + \beta = \frac{3 + \sqrt{13}}{2} + \frac{3 - \sqrt{13}}{2} = \frac{(3 + \sqrt{13}) + (3 - \sqrt{13})}{2} = \frac{6}{2} = 3 \]
The $\sqrt{13}$ terms cancel out, leaving a clean whole number.
Step 5: Compute $\alpha\beta$ directly.
This is a product of the form $(m+n)(m-n) = m^2 - n^2$ with $m = \frac{3}{2}$ and $n = \frac{\sqrt{13}}{2}$:
\[ \alpha\beta = \left(\frac{3}{2}\right)^2 - \left(\frac{\sqrt{13}}{2}\right)^2 = \frac{9}{4} - \frac{13}{4} = \frac{-4}{4} = -1 \]
Step 6: Substitute back into the fraction.
\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{3}{-1} = -3 \]
Step 7: Final answer.
\[ \boxed{\dfrac{1}{\alpha} + \dfrac{1}{\beta} = -3} \]