= Solution
Put $x=a^2$. Apart from the positive factor $H_0^2/a^4$,
$$
H^2\quad\hbox{has the sign of}\quad
F(x)=x^2-\beta x+\beta.
$$
For $\beta>4$, its discriminant $\beta(\beta-4)$ is positive and its two roots are
$$
x_\pm=\frac{\beta\pm\sqrt{\beta^2-4\beta}}2.
$$
Both are positive. Moreover $F(0)>0$ and $F(2)=4-\beta<0$, so
$$
0<x_-<2<x_+.
$$
A universe expanding from $a=0$ follows the allowed region $0\leq a^2\leq x_-$ and reaches $a^2=x_-$ with $H=0$. There,
$$
\dot H
=-\beta\frac{H_0^2}{a^4}(2-a^2)<0,
$$
so $H$ crosses from positive to negative: expansion reverses into contraction.
As $a\to0$ on the contracting branch,
$$
H\sim-\frac{H_0\sqrt\beta}{a^2},
\qquad
\dot a=aH\sim-\frac{H_0\sqrt\beta}{a}.
$$
\b[Hence $a\,da\sim-H_0\sqrt\beta\,dt$, so zero scale factor is reached in finite time. The universe therefore undergoes the <supercritical closed radiation--cosmological-constant recollapse> to a <Big Crunch>.]
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