Question

Find the real roots of the equation
$\sqrt{x+2\sqrt{x+2\sqrt{x+...+2\sqrt{x+2\sqrt{3x}}}}}=x$
find the sum of solutions .

Answer 1

Rewrite the given equation
$\sqrt{x+2\sqrt{x+2\sqrt{x+...+2\sqrt{x+2\sqrt{x+2x}}}}}=x........\left(1\right)$
on replacing the last letter x on the L.H.S. of the equation (1) by the value of x expressed by (1) we obtain
$x=\sqrt{x+2\sqrt{x+2\sqrt{x+...+\sqrt{x+2x}}}}\left(2nradicalsigns\right)$
Further, let us replace the last letter x by the same expression; again and again yields
$\therefore x=\sqrt{x+2\sqrt{x+2\sqrt{x+...+2\sqrt{x+2x}}}}\left(3nradicalsigns\right)=\sqrt{x+2\sqrt{x+2\sqrt{x+...+2\sqrt{x+2x}}}}=...\left(4nradicalsigns\right)$
we can write
$x=\sqrt{x+2\sqrt{x+2\sqrt{x+....}}}=\underset{N\to \infty }{Lim}\sqrt{x+2\sqrt{x+2\sqrt{x+...+2\sqrt{x+2x}}}}\left(Nradicalsigns\right)$
If follows that
$x=\sqrt{x+2\sqrt{x+2\sqrt{x+....}}}==\sqrt{x+2\left(\sqrt{x+2\sqrt{x+....}}\right)}=\sqrt{x+2x}$
Hence $x^2=x+2x\Rightarrow x^2-3x=0\therefore x=0,3.$