So I was wondering how to convert an equation of the form $\sqrt{x_1}+\sqrt{x_2}+\sqrt{x_3}+…\sqrt{x_n}+k=0$ into a polynomial equation based on each $x_i$.
For example if the equation was $$\sqrt{x_1}+\sqrt{x_2}+k=0$$, then subtracting $\sqrt{x_2}$ from each side and squaring yields:$$x_1+k^2+2k\sqrt{x_1}=x_2.$$
This can then be rearranged to: $$2k\sqrt{x_1}=-x_1-k^2+x_2.$$
Squaring both sides yields: $$4k^2x_1=x_1^2+k^4+x_2^2+2k^2x_1-2x_1x_2-2k^2x_2.$$
Rearranging/simplifying yields:
$$x_1^2+x_2^2+k^4-2k^2x_1-2x_1x_2-2k^2x_2 = 0.$$
How can I find an equation of this form given that $n$ is greater than $4$? I am most interested in when $n = 6$.
Best Answer
Well, you may have to do a lot of squaring, and it may not be practical to do it by hand, but here's the theory: let's start with $\sqrt u+\sqrt v+\sqrt w+\sqrt x+\sqrt y+\sqrt z=k$. Square both sides, transfer all the terms without square roots to the right, divide by two, and you get $\sqrt{uv}+\cdots+\sqrt{yz}=k^2+f(u,\dots,z)$ for some polynomial $f$. Square again and move non-roots to the right. On the left, you get a sum with terms of the type $\sqrt{uv}$ and $\sqrt{uvwx}$, on the right some new polynomial $g(u,\dots,z)$. Do it again, on the left you'll have terms of the type $\sqrt{uv}$, $\sqrt{uvwx}$, and $\sqrt{uvwxyz}$, on the right some polynomial $h(u,\dots,z)$.
Keep on doing this. You'll only ever get terms of those three types on the left, and polynomials on the right. Now there are only $15$ different terms of type $\sqrt{uv}$, another $15$ of type $\sqrt{uvwx}$, and just one of type $\sqrt{uvwxyz}$, making $31$ different terms in all. So after you've done the procedure $32$ times, you'll have $32$ linear equations in these $31$ terms, and you can use linear algebra to boil them down to a single equation with no square roots in it, and you win.
I hope you won't expect me to actually carry out this procedure here....