Special relativity: circumventing velocity-addition formula Two spaceships approach an observer from an equal distance and from an opposite direction with an equal speed $v$ in the observer's intertial reference frame $O$.
The speed of a spaceship in the intertial reference frame of the other spaceship $S$ is $0.8c$, what is the speed of one of the spaceships in $O$?
I proceeded as follows:
Let $2l'$ be the distance between the two spaceships in $S$.
In S the two spaceships will collide after a time $t' = \frac{2.5l'}{c}$.
Let $\gamma$ be the squareroot of $1 - v²/c²$. In $O$ the two spaceships will collide when $vt = l$ or $v\gamma t' = \frac{l'}{\gamma}$ ($O$ has to correct for what he perceives as the time dilations and space contractions of the measurements made in $S$)
Substituting we get the equation $x(1-x^2) = 0.4$ with $x = \frac{v}{c}$, if you solve the equation you conclude that this line of reasoning was wrong (but when we replace $0.4$ by $0.375$ we do get the right solution, which is $0.5c$).
Where's the flaw?
 A: 
Where's the flaw?

Here's one:  

Let γ be the squareroot of 1−v²/c²

but, in fact,
$$\gamma_v = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$
But, there is a flaw in your reasoning too so correcting the error in the formula for $\gamma$ will not get you to the correct answer.
Let's work it out with coordinates to see explicitly what's going on.
Let the $x$ coordinates of the two spacecraft, $S_A$ and $S_B$, in the observer's frame $O$ be given by
$$x_A = l - vt$$
$$x_B = -x_A$$
Let the reference frame $O'$ have relative velocity $v$ in $O$ and assume standard configuration.
According to the Lorentz transformations, the $x'$ coordinates of the spacecraft in $O'$ are given by
$$x_A' = \left(\frac{\frac{l}{\gamma_v} - 2vt'}{1+ \frac{v^2}{c^2}} \right)$$
$$x_B' =  -\gamma_v l $$
In $O'$, $S_B$ is at rest and $S_A$ has a velocity of $v'_A = -0.8c$.
Note that, due to the relativity of simultaneity, when the two spacecraft are separated by a distance $2l$ in $O$, there are not separated by a distance $\frac{2l}{\gamma_v}$ in $O'$ but, rather, a distance of $\frac{\gamma_v 2l}{1+ \frac{v^2}{c^2}}$.
This is why your reasoning fails.  Continuing, the velocity of $S_A$ in $O'$ is
$$v'_A = \frac{dx'_A}{dt'} = \frac{-2v}{1+ \frac{v^2}{c^2}} = -0.8c$$
which yields
$$v = 0.5c $$
