Can you find roots of roots?

$\sqrt { x\quad +\quad 2\sqrt { x\quad +\quad 2\sqrt { x\quad +\quad 2\sqrt { 3x } } } } =\quad x\\ We\quad also\quad know\quad that\quad \\ x\quad =\quad \sqrt { { x }^{ 2 } } \quad =\quad \sqrt { x\quad +\quad (x\quad -\quad 1)x } \\ =\sqrt { x\quad +\quad (x\quad -\quad 1)\sqrt { x\quad +\quad (x\quad -\quad 1)x } } \\ =\sqrt { x\quad +\quad (x\quad -\quad 1)\sqrt { x\quad +\quad (x\quad -\quad 1)\sqrt { x\quad +\quad (x\quad -\quad 1)x } } } \\ =\sqrt { x\quad +\quad (x\quad -\quad 1)\sqrt { x\quad +\quad (x\quad -\quad 1)\sqrt { x\quad +\quad (x\quad -\quad 1)\sqrt { { x }^{ 2 } } } } } \quad \\ Now\quad we\quad compare\quad this\quad with\quad are\quad question\\ We\quad get\quad \\ \sqrt { 3x } \quad =\quad \sqrt { { x }^{ 2 } } \\ 3x\quad =\quad { x }^{ 2 }\\ x\quad =\quad 3\quad \leftarrow ANS$ Quite Easily Done

Write 3x as x+2x. Now, if we assume this as an Infinite Loop, x+2x seems to be a result of x+2 root ( ...... ).

So this root ( .......... ) = x Therefore, root ( x + 2x ) = x, i.e. x^2 = 3x or, x = 3.

The last term 3x can be written as x+2x...now we can put the value of x, that is left side of the equation for 2(left side of the equation)....We notice that this can be repeated infinite times.....thus we can write ✓( x + 2×)=x Solving we get x=0,3....thus the non zero ans is 3

that was pure luck :P really i had no idea it would be 3

squaring both side we get x^2=3x but x not equal to zero so x=3

Why can't zero be the solution? Wasted one try in that ;)