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 $n = 6$.

  • 2
    $\begingroup$ Expand a product with $(k\pm\sqrt{x_1}\pm\sqrt{x_2})$ factors with all possible sign combinations. $\endgroup$ – Somos May 6 '19 at 23:52
  • $\begingroup$ Ok that's what I thought @RossMillikan. Thanks for confirming. $\endgroup$ – automaticallyGenerated May 7 '19 at 0:12
  • $\begingroup$ @Ross, but if you keep on squaring, you eventually keep getting the same set of square roots, and when you have enough equations, you can eliminate them all. $\endgroup$ – Gerry Myerson May 7 '19 at 0:19
  • $\begingroup$ @GerryMyerson can u show an example of this? I didn't realize I could cancel parts out when I was working it out by hand. $\endgroup$ – automaticallyGenerated May 7 '19 at 0:28
  • 3
    $\begingroup$ Look at mathpages.com/home/kmath111/kmath111.htm $\endgroup$ – Piquito May 7 '19 at 1:05

COMMENT (A replay to the last comment of the O.P.).-Your problem is actually finding the minimum polynomial of $ -k $ or $ k $ (which, in principle, is an irrational of degree $2 ^ n$) and then the link I gave you offers you the solution you want.

But you post it in a way that could lead to think that it is not. I explain why the minimum polynomial of $-k$ solves the problem. The easiest example is for two radicals and it is enough to understand what I want to say.

You can easily find out the minimal polynomial of $x=\sqrt a+\sqrt b$ which is $$x^4-2(a+b)x^2+(b-a)^2=0$$ and certainly $-k=\sqrt a+\sqrt b$ is a root of it.

Well, in this polynomial you can note that the coefficients are rational functions of $a$ and $b$ then you can pose your problem the way you do in your post by replacing $ a $ for $ x_1 $, $ b $ for $ x_2 $ and $ x $ for $ -k $ or $ k $. This gives the result $$ k ^ 4-2 (x_1 + x_2) k ^ 2 + (x_2-x_1) ^ 2 = 0 $$ what obviously answers your problem for the case $ n = 2 $

  • $\begingroup$ I see my silly misunderstanding - I didn't realize that I could plug in $k$ for $x$. Thanks for the help $\endgroup$ – automaticallyGenerated May 7 '19 at 19:46
  • $\begingroup$ You are welcome. $\endgroup$ – Piquito May 7 '19 at 19:56

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....


Just for fun, the polynomial drawn from $\sqrt a+\sqrt b+\sqrt c+k=0$, using the method by Somos:


Then next challenge is to determine the number of terms as a function of the number of variables.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.