<s>[INST] if $x = \sqrt{x+1} + \sqrt{x+2} + \sqrt{x+3}$ then x =? I have got a new question from my friend and it made me nervous:
$$\text{ If }\,x = \sqrt{x+1} + \sqrt{x+2} + \sqrt{x+3}, \text{ then }\,x = \;?$$
A lot of thank you to all comments.
 [/INST] Since $x$ is the sum of three square roots, $x$ is positive. Thus there exists a positive real $y$ such that $x=y^2-2$. 
The equation then becomes
$$y^2-2=\sqrt{y^2-1}+\sqrt{y^2}+\sqrt{y^2+1}$$
$$y^2-y-2=\sqrt{y^2-1}+\sqrt{y^2+1}$$
Squaring gives
$$y^4-2y^3-3y^2+4y+4=2y^2 + 2\sqrt{y^4-1}$$
$$y^4-2y^3-5y^2+4y+4=\sqrt{4y^4-4}$$
Squaring again
$$y^8-4 y^7-6 y^6+28 y^5+17 y^4-56 y^3-24 y^2+32 y+16=4y^4-4$$
$$y^8-4 y^7-6 y^6+28 y^5+13 y^4-56 y^3-24 y^2+32 y+20=0$$
This octic has six real roots, of which five are extraneous. The approximate result is $$y \approx 3.56021$$
Using $x=y^2-2$ this gives $$x \approx 10.67507$$
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