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

 A: The fourth step is not correct. You may rather write
$$
x=2\sqrt{x}
$$ $$
x-2\sqrt{x}=0
$$$$
\sqrt{x}\left(\sqrt{x}-2\right)=0
$$
$$
\sqrt{x}=0 \quad \text{or} \quad 
\sqrt{x}-2=0
$$ giving easily 
$$x=0, \qquad x=4
$$ as solutions.
A: it must be $$x\geq 0$$ after squaring we get
$$x=2\sqrt{x}$$
squaring again we obtain $$x(x-4)=0$$ thus we get $$x=0$$ or $$x=4$$ which are indeed solutions.
A: When you divide by $\sqrt{x}$, you are assuming something. This assumption discards another valid solution.
A: 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 $x = 4$.
A: You divided  $x^{\frac12}$ without checking whether $x=0$ is a solution.  When using division to solve an equation, you get only solutions where the divisor isn't zero, so be sure to separately check cases where the divisor is zero.
A: 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
A: In addition to the omission of the solution $0,$ there's an important point here that I think some of the other answers are glossing over.  Your method does not show that $0$ and $4$ are solutions — what it shows is that no number other than $0$ or $4$ is a solution.  You still need to check that $0$ and $4$ actually work.  (In this case, they do, but in some similar problems, you can get spurious solutions in addition to the actual solutions.)
Here's a complete argument:
If $${\sqrt{2x+1}=1+\sqrt{x}},$$ then deduce $$x=2\sqrt{x}$$ just as you did. Square both sides to get $$x^2=4x.$$ It follows that either (a) $x=0$ or (b) you can divide both sides by $x$ to get $x=4.$
At this point, all we have shown is that if $x$ is a solution, then $x$ must equal either $0$ or $4.$ So $0$ and $4$ are the only two possibilities for a solution.  You still need to check out each one of these possible solutions individually:
$${\sqrt{2\cdot 0+1}=1=1+\sqrt{0}}$$
and
$${\sqrt{2\cdot 4+1}=3=1+\sqrt{4}},$$
so in fact both $0$ and $4$ are solutions (and so they're the only solutions, since we already know that any solution has to be either $0$ or $4).$
