If α and β are the zeroes of $x^2-6x+3$, the value of α/β + β/α is:

$\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }=\frac{(\alpha +\beta {)}^2-2\alpha \beta }{\alpha \beta }$. With sum 6 and product 3 this is $\frac{36-6}{3}=\frac{30}{3}=10$. 12 forgets the $-2\alpha \beta$ in the numerator; 14 adds $2\alpha \beta$ instead of subtracting it; $\frac{34}{3}$ subtracts 2 rather than $2\alpha \beta =6$.