Bifurcations for 1-dimensional map Consinder the 1-dim map $F(x,\mu)=\mu-\frac{1}{4} x^2$ as $\mu$ increases from $-\infty$ to $5$ and analyse the bifurcations. 
I start the analysis by considering $F^2(x,\mu)=\mu^2-\frac{1}{2}\mu x^2+\frac{1}{16} x^4$
Setting $F^2(x,\mu)=0=1/16(x^2-4\mu)^2$ gives me the following solutions: $+-2\sqrt{\mu}$
They are also fixed points of $F$ We have stability in case of $+2\sqrt{\mu}$ and instability in $-2\sqrt{\mu}$ for $\mu>0$. What else can I say and how can I include the increasing value $\mu$ from $-\infty$ to $5$?
 A: I'm assuming that this map corresponds to a scalar system of the form $\dot{x} = F(x, \mu)$. Then the fixed points of the system are those points for which $\dot{x} = F(x, \mu) = 0$.
As you've correctly identified, the roots of $F$ are $\pm 2\sqrt{\mu}$. Now, since this is a physical system, complex solutions are not meaningful, and $\mu < 0$ does not produce a solution to the equation $F(x, \mu) = 0$; since $\mu < 0$ requires us to take the square root of a negative number, we say that the system has no solution in this case. Then for $\mu < 0$ there is no solution to the equation $\dot{x} = F(x, \mu) = 0$ and for all $\mu \in (-\infty, 0)$ the system has no fixed points. 
At $\mu = 0$, the system has a fixed point at $x = 0$, since $\mu - \frac{1}{4}x^2 = -\frac{1}{4}x^2 = 0$ implies that $x = 0$; since $x = 0$ is the only solution when $\mu = 0$, it is the only fixed point and it turns out to be degenerate in this case. As we increase $\mu$ above $0$, we see that for any value of $\mu > 0$ the system has two solutions, namely the roots $\pm2\sqrt{\mu}$ we found above (and these are "allowed" since they are not complex numbers). The net effect is that as $\mu$ is increased through $0$, the system goes from having no fixed points to having $2$ fixed points and, as a result, we see that a bifurcation occurs. The two fixed points produced have opposite stabilities, with one being stable and the other being unstable (as you correctly identified in your post). For a system of this form, going from $0$ fixed points to $1$ stable fixed point and $1$ unstable fixed point as we vary a single parameter corresponds to a saddle-node bifurcation. 
Geometrically, it can be helpful to picture the equation $F(x, \mu) = \mu - \frac{1}{4}x^2$ as a parabola which opens downward in the $x\mu$ plane (i.e., $x$ is on the horizontal axis and $\mu$ is on the vertical axis). In this setting, the $x$-intercepts give us the solutions to the equation $\dot{x} = F(x, \mu) = 0$ and, as a result, the $x$-intercepts are precisely the values of $x$ for which the system has a fixed point. 
The parameter $\mu$ then slides the parabola up in the $x\mu$ plane as it increases. Negative values of $\mu$ mean that the parabola does not touch the $x$-axis at all (hence there are no real values of $x$ for which $F(x, \mu) = 0$), as you can see in the image below. 
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The value $\mu = 0$ means that the peak of the parabola "bounces" off of the $x$-axis (as you can see in the next image), which is why $x = 0$ is the only solution when $\mu = 0$; this is a double root. 
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When $\mu > 0$, the parabola intersects the $x$-axis in two places, giving two solutions (as is shown in the last image, below). In this geometrical context, the bifurcation occurs when the parabola intersects the $x$-axis. 
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For a very informative and accessible reference on systems like this, I'd recommend "Nonlinear Dynamics and Chaos" by Strogatz. 
