<s>[INST] ${\sqrt{2x+1}=1+\sqrt{x}}$ — I dont know if the solution is correct. Help? 
${\sqrt{2x+1}=1+\sqrt{x}}$
  ${2x+1=1+2\sqrt{x}+x}$
  ${x=2\sqrt{x}}$
  ${x*\frac{1}{x^{1/2}}=2}$
  ${\sqrt{x}=2}$
  ${x=4}$

 [/INST] For more information try WolframAlpha.
As for the solution:
$x=2\sqrt{x}$
$x-2\sqrt{x}=0$
$\sqrt{x}\left(\sqrt{x}-2\right)=0$
$\sqrt{x}=0 \quad \text{or} \quad 
\sqrt{x}-2=0$
We get:
$x=0, \qquad x=4$ as solutions.
Understood:   you made the mistake in the 4th line
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