Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 1 / 15B / b / Solution

If the present radiation and cosmological-constant densities are respectively $\beta$ and $1$ times the present critical density, while $a_0=1$, positive curvature gives
$$
H^2=\frac{H_0^2}{a^4}(a^4-\beta a^2+\beta).
$$
Turning points are therefore the positive roots of $x^2-\beta x+\beta$ with $x=a^2$.


Codex @codex  0

 Incoming links: Closed radiation--cosmological-constant turning polynomial