Say we have a problem with one square root in it; it's easily solvable: $$\sqrt{a}=b$$ $$a=b^2$$ Two: $$\sqrt{a}+\sqrt{b}=c$$ $$a+b+2\sqrt{ab}=c^2$$ And it reduces to a single root problem. Three: $$\sqrt{a}+\sqrt{b}+\sqrt{c}=d$$ $$\sqrt{a}+\sqrt{b}=d-\sqrt{c}$$ $$a+b+2\sqrt{ab}=d^2+c-2d\sqrt{c}$$ And, it's reduces to a double root problem. Four however seems impossible. Why?
There are a few possibilities at the start with how many terms you'll have at either side of the equals sign before you square:
constant = 4 roots --> results in constant = 6 roots
constant+root = 3 roots --> results in constant + root = 3 roots
constant+ 2 roots = 2 roots --> results in constant + 3 roots = root
constant + 3 roots = root --> results in 6 roots = constant
constant + 4 roots = constant --> results in constant = 10 roots
This is why I think it might be impossible (except when $e=0$ of course).
Question: Is it (im)possible? If not is there a maximum amount of roots for which it is possible?
Here is a strategy - suppose the square roots are initially of $a,b,c,d$
First take all the terms involving $\sqrt a$ to one side, and everything else to the other, and square the resulting equation. This leaves a complicated equation with a number of square roots, some of products of terms, but no square roots of $a$ left in any term.
Now do the same with the terms involving $\sqrt b$ (even where multiplied by other square roots) to eliminate that.
After squaring four times (to deal with $\sqrt c$ and $\sqrt d$ as well), you should have no square roots left. Obviously the strategy can be applied to any such expression.