how can one solve for $x$, $x =\sqrt{2+\sqrt{2+\sqrt{2\cdots }}}$ 
Possible Duplicate:
Limit of the nested radical $\sqrt{7+\sqrt{7+\sqrt{7+\cdots}}}$ 

how can one solve for $x$, $x =\sqrt[]{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2\cdots }}}}}}$
we know, if $x=\sqrt[]{2+\sqrt{2}}$, then, $x^2=2+\sqrt{2}$
now, if $x=\sqrt[]{2+\sqrt{2}}$, then, $(x-\sqrt{2})(x+\sqrt{2})=\sqrt{2}$

put $x = \sqrt{2+x} 
 \implies x^2 = 2+x \implies x^2 -x-2=0$
Now just solve the quadratic equation.
