Answer
No solution
Work Step by Step
$\sqrt{1+\sqrt{x+\sqrt{2x+1}}}=\sqrt{5+\sqrt{x}}$ (Take the square)
$1+\sqrt{x+\sqrt{2x+1}}=5+\sqrt{x}$
$\sqrt{x+\sqrt{2x+1}}=\sqrt{x}+4$ (Take the square)
$x+\sqrt{2x+1}=x+8\sqrt{x}+16$
$\sqrt{2x+1}=8\sqrt{x}+16$ (Take the square)
$2x+1=64x+256\sqrt{x}+256$
$62x+256\sqrt{x}+255=0$
$62\sqrt{x}^2+256\sqrt{x}+255=0$
Let $\sqrt{x_1}$ and $\sqrt{x_2}$ be the roots of the last expression.
Then,
(i) $\sqrt{x_1}+\sqrt{x_2}=-\frac{b}{a}=-\frac{256}{62}<0$
(ii) $\sqrt{x_1}\cdot \sqrt{x_2}=\frac{c}{a}=\frac{255}{62}>0$
It tells us that $\sqrt{x_1}<0$ and $\sqrt{x_2}<0$.
However, there are no real numbers $x_1$ and $x_2$ satisfying it.
So, the equation has no real solution.
