<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] You are mostly correct but there are two conditions you didn't take into account which you should have.
$\sqrt{2x + 1}=1+\sqrt{x}$  Note, this implies $2x + 1 \ge 0$ i.e $x \ge -\frac 12$.
$2x +1 = 1 + 2\sqrt{x} + x$  Now, we have "lost" that assumption.  It is possible that we will end up with some extraneous answers where $x < -\frac 12$.  As it turns out that isn't an issue and it doesn't happen but it could have. (As we still have $\sqrt{x}$ that implies $x \ge 0$ so $x < -\frac 12$ is impossible).
$x = 2\sqrt{x}$
$x/x^{1/2} = 2$ Here you divided by $x^{1/2}$ in the assumption $x^{1/2} \ne 0$.  You can not make that assumption.  You must consider the possibility that $x^{1/2}$.
So say: Case 1: If $\sqrt{x} = 0$, then $x= 0$ and we have $0 = 2\sqrt{0}$ which consistent so $x = 0$ is a possible answer.
But if $\sqrt{x} \ne 0$ then
$x/x^{1/2} = 2$
$x^{1/2} = 2$
$x =4$ 
So $x=4$ is the only other  possible solution .  So $x = 0$ or $x = 4$.
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