TITLE: solving $12-x=\sqrt{12-\sqrt{x}}$
QUESTION [5 upvotes]: How can I slove: $12-x=\sqrt{12-\sqrt{x}}$?
I tried to put $t=12-\sqrt{x}$
But it got me to polynomial of 4th degree which I don't think its the idea of solving this equation.

REPLY [3 votes]: If $y = 12 - \sqrt{x}$, the equation says 
$x = 12 - \sqrt{y}$.  Note the symmetry.  One possibility is $x=y$ which leads to the equation
$x = (12-x)^2$.  That quadratic has two solutions $x=9$ and $x=16$, but only 
$x=9$ satisfies the original equation.