Let X be a continuous r.v., with pdf $f_X(x) = kx(1-x), 0 < x <1$
I evaluated k = 6 and found the cdf $F_x(x) = 3x^2 - 2x^3$, but then I am asked to find the median. The equation $F_x(x) = 1/2$ brings me to $-4x^3 + 6x^2 -1 = 0$, for which I don't know any quick way to find its root (but I randomly plugged in $x = 1/2$ and got my answer). The question is, apart from the tedious cubic formula, is there another quick trick to solve general cubic equations, especially when I am not looking for integer solutions?
Cubic equations are indeed tedious.
I guess you are expected here to realize that the pdf is symmetric around $x=1/2$ : that is, $f_X(x-\frac12)=f_X(x+\frac12)$. Hence, the median is $1/2$, because $\int_0^{1/2} f_X(x) dx =\int_{1/2}^1 f_X(x) dx$