Should normalisation factor in a QM always be positive? I have a fairly simple question about a normalisation factor. After normalising a wavefunction for a particle in an infinite square well on an interval $-L/2<x<L/2$ I got a quadratic equation for a normalisation factor $A_0$ which has a solution like this: 
$$A_0 = - \frac{8\pi}{6} \pm \sqrt{\frac{8^2\pi^2}{36}+\frac{3}{4}}$$
The sign $\pm$ gives me two options while a normalisation factor can only be one value. So I want to know if there are any criteria on which i could decide which value to choose. Is it possible that normalisation factor should always be a positive value?

I added the procedure I used to derive this eq. :

 A: Simple: $A_0$ is not a common factor to the entire wavefunction. So it's not a normalization factor.
In other words, you have
$$\psi(x) = A_0\sqrt{\frac{2}{L}}\sin\biggl(\frac{\pi}{L}x\biggr) + \frac{1}{2}\sqrt{\frac{2}{L}}\sin\biggl(\frac{2\pi}{L}x\biggr)$$
but if you really want $A_0$ to be a normalizing factor you should have
$$\psi(x) = A_0\Biggl[\sqrt{\frac{2}{L}}\sin\biggl(\frac{\pi}{L}x\biggr) + \frac{1}{2}\sqrt{\frac{2}{L}}\sin\biggl(\frac{2\pi}{L}x\biggr)\Biggr]$$
A: I think you made a mistake.

Using your wavefunction and noting that $$\int^\frac{L}{2}_{-\frac{L}{2}}
\sin\left(\frac{\pi x}{L}\right)\sin\left(\frac{2\pi x}{L}\right) = \frac{4L}{3\pi}$$ we see that
$$1=\int^{L/2}_{-L/2}|\psi(x)|^2 dx = \frac{2}{L}\left(|A_0|^2\frac{L}{2} +\frac{4L}{3\pi}
\left(A_0 + A_0^*\right)+ \frac{1}{4}\frac{L}{2}\right) = \left(|A_0|^2 +\frac{8}{3\pi}
\left(A_0 + A_0^*\right)+ \frac{1}{4}\right)$$
and so we have the constraint
$$\boxed{|A_0|^2 +\frac{8}{3\pi}
\left(A_0 + A_0^*\right)+ \frac{1}{4} = 1}$$
If we let $A_0 = a+ib$, with $a,b$ real, this reduces to
$$a^2+b^2 +\frac{16}{3\pi}a
+ \frac{1}{4} = 1
$$
If we make the choice that $A_0$ is real, that is $A_0 = a$ we find
$$A_0^2 +\frac{16}{3\pi} A_0
+ \frac{1}{4} = 1 ~~\implies ~~ \boxed{A_0 = -\frac{8}{3\pi}\pm\sqrt{\left(\frac{8}{3\pi}\right)^2+\frac{3}{4}}}$$
which is what I am guessing you were trying to get. We see that the most general solution involves both $a$ and $b$ - it is complex. To find a general result, parameterized by $a$ we solve 
$$a^2+b^2 +\frac{16}{3\pi}a
+ \frac{1}{4} = 1~~\implies ~~ b = \pm \sqrt{\frac{3}{4}-a^2-\frac{16}{3\pi}a}$$
and so our general solution is
$$\boxed{A_0 = a \pm i\sqrt{\frac{3}{4}-a^2-\frac{16}{3\pi}a}}
$$
(note we must restrict $a$ to take values for which $\sqrt{\frac{3}{4}-a^2-\frac{16}{3\pi}a}$ is real). So we have a range of possible solutions. I cannot see any way to make a choice of one solution over any other.

Added after comments below: 
It is likely that the question (wherever you found it) has a typo and the interval of interest is actually $[0,L]$. This makes more sense, because the first two energy eigenstates of an infinite square well potential on the interval $[0,L]$ are (up to a normalization) $\sin\left(\frac{\pi x}{L}\right)$ and $\sin\left(\frac{2\pi x}{L}\right)$. Under this scenario, the problem of normalization is much simpler, because these two functions are orthogonal on this interval. This means that we have
$$1 = \int^L_0 dx |\psi(x)|^2 = |A_0|^2 + \frac{1}{4} \implies |A_0| = \frac{\sqrt{3}}{2}$$
so we would have $A_0 = e^{i\theta}\frac{\sqrt{3}}{2}$ for arbitrary, real $\theta$.
