Chemistry - Equilibria questions involving ICE tables

[EdTheHead]

In my book for all these equilibria questions involving ice tables when I end up with $$Ka = \frac{(x)(x)}{initial amount - x}$$ they always say that if the Ka is tiny compared to the initial amounts then we can assume x will be too and we can omit as many x's as we like from this equation. This obviously makes solving for x a whole lot easier but I don't get how this works.

First off a weak acid/base example
$$Ka = \frac{(x)(x)}{initial\ amount - x}$$
if I omit 1 x I get $$x = \sqrt{Ka(initial\ amount)}$$
if I omitted 2 x's I'd get $$x = Ka(initial\ amount)$$ a completely different answer. Does this only apply if I have a $$(constant \pm x)$$ then I just omit the plus/minus x?

 

[Borek]

x can be omitted only in sums.

There is a rule of thumb (so called 5% rule) which says that a+x can be approximated by a if x is less than 5% of a. See here.

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