Let the two zeroes be $\alpha$ and $\dfrac{1}{\alpha}$ (reciprocal of each other).
Using the relation: Product of zeroes $= \dfrac{\text{Constant term}}{\text{Coefficient of } x^2}$
$$\alpha \times \frac{1}{\alpha} = \frac{-(k-2)}{6}$$
$$1 = \frac{-(k-2)}{6}$$
$$6 = -(k-2)$$
$$6 = -k + 2$$
$$k = -4$$
Source: Chapter 2, Section 2.3 — Relationship between Zeroes and Coefficients of a Polynomial
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The key insight is that if one zero is the reciprocal of the other, their product = 1. Then apply the formula $\alpha\beta = \dfrac{c}{a}$, where here $c = -(k-2)$ and $a = 6$. Set the product equal to 1 and solve for $k$. Examiners award 1 mark for correctly setting up the product condition and 1 mark for the correct value of $k$.