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?

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.
