How to find x such that $\sqrt{x}= \sqrt{a-x} + \sqrt{b-x} + \sqrt{c-x}$? Let $a,b,c \in R$ are given. How to find x such that $\sqrt{x}= \sqrt{a-x} + \sqrt{b-x} + \sqrt{c-x}$? Is there a simple way?
 A: As I wrote in a comment, to get a polynomial in $x$, you must square multiple times. For starting $$(\sqrt{x}- \sqrt{a-x})^2=(\sqrt{b-x} + \sqrt{c-x})^2$$ $$a-2 \sqrt{x} \sqrt{a-x}=2 \sqrt{b-x} \sqrt{c-x}+b+c-2 x$$ $$a-b-c+2 x=2 \sqrt{x} \sqrt{a-x}+2 \sqrt{b-x} \sqrt{c-x}$$ Squaring again $$(a-b-c+2 x)^2=4(\sqrt{x} \sqrt{a-x}+ \sqrt{b-x} \sqrt{c-x})^2$$ Continue the same way, pushing everytime the radicals to the lhs. 
Hoping no mistakes (not sure !), you should end with something like $$80 x^4-64 (a+b+c)x^3 +8  \left(a^2+6 a (b+c)+b^2+14 b c+c^2\right)x^2-64 a b c x+\left(a^2-2 a (b+c)+b^2+6 b
   c+c^2\right)^2=0$$ Now, you have a quartic polynomial ... which can be solced using radicals.
I wish you a very good time !!
By the way, do not forget that squaring introduces extra roots and some ot them need to be discarded later after checks (in your case, if solution exists it must be between $0$ and the minimum of $a,b,c$).
A: This isn't simple, but the "just keep squaring everything" approach doesn't go so badly if you first rearrange to get two square roots on both sides.
$$\sqrt x=\sqrt{a-x}+\sqrt{b-x}+\sqrt{c-x}$$
$$\sqrt x-\sqrt{c-x}=\sqrt{a-x}+\sqrt{b-x}$$
$$x-2\sqrt{x(c-x)}+c-x=a-x+2\sqrt{(a-x)(b-x)}+b-x$$
$$x+\frac{c-a-b}{2}=\sqrt{x(c-x)}+\sqrt{(a-x)(b-x)}$$
$$x^2+(c-a-b)x+\frac{(c-a-b)^2}{4}=x(c-x)+2\sqrt{x(a-x)(b-x)(c-x)}+(a-x)(b-x)$$
$$x^2+\left(\frac{a^2+b^2+c^2}{4}-\frac{ab+bc+ca}{2}\right)=2\sqrt{x(a-x)(b-x)(c-x)}$$
$$x^4+2\left(\frac{a^2+b^2+c^2}{4}-\frac{ab+bc+ca}{2}\right)x^2+\left(\frac{a^2+b^2+c^2}{4}-\frac{ab+bc+ca}{2}\right)^2=4x(a-x)(b-x)(c-x)$$
$$5x^4-4(a+b+c)x^3+\left(\frac{a^2+b^2+c^2}{2}+3(ab+bc+ca)\right)x^2-4abcx+\left(\frac{a^2+b^2+c^2}{4}-\frac{ab+bc+ca}{2}\right)^2=0$$
Is there a nice solution? Trying the example $a=7,b=8,c=9$ gives
$$5x^4-96x^3+670x^2-2016x+2209=0$$
and the exact solutions given by Wolfram Alpha are horrible, which perhaps suggests there is no nice solution.
