The sum of reciprocals of zeros of the quadratic polynomial acx^2 + (b^2 - ac)x + 1 is -

For $P(x) = acx^2 + (b^2-ac)x + 1$, if the zeros are $\alpha$ and $\beta$, then $\alpha + \beta = -\frac{b^2-ac}{ac} = \frac{ac-b^2}{ac}$ and $\alpha\beta = \frac{1}{ac}$. Thus, the sum of reciprocals is $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta} = ac - b^2$.