What is the correct answer to this diffferential equation? [Question]
When solving the differential equation:
$$\frac{\mathrm dy}{\mathrm dx} = \sqrt{(y+1)}$$ 
I've found two ways to express $y(x)$:
implicitly: $2\sqrt{(y + 1)} =  x + C$
or directly: $y = (x^2)/4 + (2xC)/4 + (C^2)/4 -1$
Although they look the same, these expressions result in different answers when
applying the  initial condition $y(0) =1$:
When using the implicit expression:
$$2\sqrt{(y(0) + 1)} = 0 + C
\\ 2\sqrt{(1 + 1)} = 0 + C
\\ C = 2\sqrt{2}$$
When using the direct expression:
$$y(0) = 0/4 + 0/4 + (C^2)/4 -1
\\ 2 = (C^2)/4
\\ C = \pm 2\sqrt{2}
\\ \text{since: }
C^2 = 8 \implies C = \pm \sqrt 8 $$
Thus, using the direct expression results in two answers (being a quadratic
equation), when 
using the implicit expression there's only one answer to C.
Which one is the complete answer?
[Additional information]
The solution's manual states that the answer should be:
$C = 2\sqrt 2 $ and the solution to the differential equation with initial condition is
 $2\sqrt{(y + 1)} =  x + 2\sqrt 2$. 
(notice, no minus sign before $2\sqrt 2 $)
However, I do not think this is the complete solution, as shown by the direct
expression, there 
may be another answer to the differential equation:
$C = - 2\sqrt(2)$ and the solution is: $2\sqrt{(y + 1)} =  x - 2\sqrt 2$
I do not know if the implicit expression hides one of the answers or if is there
something 
wrong with my use of the direct expression.

The version with $\sqrt{8}$ is the only correct one. If you use $-\sqrt{8}$, you get that $\sqrt{y+1}$ is negative when $x=0$. But $\sqrt{y+1}$, by definition, is the non-negative number whose square is $y+1$. 
Remark: A familiar related fact is that when we are solving an ordinary equation that involves square roots, squaring often introduces one or more extraneous solutions, that is, non-solutions. 
