I'm actually reading What Every Computer Scientist Should Know About Floating-Points Arithmetics and I have a little problem with understaing one thing.
Here's a quote:
Write $(b \otimes b) \ominus (4a \otimes c) = (b^{2}(1 + \delta_{1}) - 4ac(1 + \delta_{2}))(1 + \delta_{3})$, where $|\delta_{1}| < \epsilon$. Using $d = b^{2} - 4ac$, this can be rewritten as $(d(1 + \delta_{1}) - 4ac(\delta_{2} - \delta_{1}))(1 + \delta_{3})$. To get an estimate for the size of this error, ignore second order terms in $\delta_{i}$, in which case the absolute error is $d(\delta_{1} + \delta_{3}) - 4ac \delta_{4}$, where $|\delta_{4}| = |\delta_{1} - \delta_{2}| \le 2\epsilon$. Since $d \ll 4ac$, the first term $d(\delta_{1} + \delta_{3})$ can be ignored.
And I really can't figure it out why that first term can be ignored (last sentence). I understand that $d$ is significantly lower than $4ac$, but I have no idea why that makes it possible to ignore.
Thank you for answers!
Nothing in what you have quoted shows $d \ll 4ac$, but that is often the case where roundoff is a problem. I will assume it has been justified. The idea is that all the $\delta$s are about the same size, which is about the machine $\epsilon$, so $\delta_1+\delta_3$ and $\delta_4$ are close in magnitude. That makes the $d(\delta_1+\delta_3)$ term much smaller than the other.