Solve $ x=54+\sqrt{54-\sqrt{x}} $ I tried so much with this equation but nothing seems to work:
$$ x=54+\sqrt{54-\sqrt{x}} $$
I tried squaring but it gives a quadratic equation that I can't solve.
Any ideas please?
 A: $$ x = 54 + \sqrt{54 - \sqrt{x}} \\ k = 54 - \sqrt{x} \\ t = x-54 \\ k+t = x - \sqrt{x}$$
And so:
$$ t = \sqrt{k} \\ t^2 = k \\ t^2 + t = t(t+1) = k+t = x - \sqrt{x} = \sqrt{x}( \sqrt{x} - 1)$$
$$t = \sqrt{x} -1 ~~ \text{and} ~~ t+1 = \sqrt{x} $$
$$ t = x-54 \\ (t+1)^2 = x \\ (x-53)^2 = x$$
Which is a simple quadratic equation, you will get $2$ results, one of them would not be correct because it would make the LHS of your equation negative (after you "move" the 54 to the left), and we know square-roots over the reals give a positive solution.
A: We can put $y=54-\sqrt{x}$ to get a system of equations
\begin{align}
x-54&=\sqrt{y}\\
y-54&=-\sqrt{x}.
\end{align}
Squaring and subtracting from each other (using the fact that $x,y \geq 0$) this simplifies to
$$
(y-x)(108-x-y)=y-x.
$$
Since $y \neq x$ (otherwise the system above would yield $-\sqrt{x}=\sqrt{y}$ and so $x=y=0$, impossible), we have $108-x-y=1$. Plugging that into one of the equations we get $x-54=\sqrt{107-x}$, and so we need to solve the quadratic equation
$$
(x-54)^2=107-x.
$$
That should be a straightforward exercise yielding two real roots only one of which will satisfy $x -54 \geq 0$.
A: With $x=t^2$, $$(t^2-54)^2=54-t$$ is a quartic equation.
Though a CAS tells us the the solutions have a relatively simple expression, I see no better way to reach it than the standard resolution method of the quartic.
