Two capacitors of capacitances $C_1$ and $C_2$ have a total capacitance of 20 μF when connected in parallel and a total capacitance 4.8 μF when connected in series. Find $C_1$ and $C_2$.

Let $C_1$ and $C_2$ be the unknown capacitances.
${\mathrm{C}}_1+{\mathrm{C}}_2=18$   .........(i) when in parallel
$\frac{{\mathrm{C}}_1{\mathrm{C}}_2}{{\mathrm{C}}_1+{\mathrm{C}}_2}=4$   ..........(ii) when in series
Multiplying eqs. (i) and (ii),${\mathrm{C}}_1{\mathrm{C}}_2=72$
Now ${\mathrm{C}}_1-{\mathrm{C}}_2=\sqrt{{\left({\mathrm{C}}_1+{\mathrm{C}}_2\right)}^2+4{\mathrm{C}}_1{\mathrm{C}}_2}$
$=\sqrt{(18{)}^2-4\times 72}=\pm 6 .......\left(\mathrm{iii}\right)$
Solving eqs. (i) and (iii), we get
${\mathrm{C}}_1=12\mathrm{\mu F}\text{ or }6\mathrm{\mu F};{\mathrm{C}}_2=6\mathrm{\mu F}\text{ or }12\mathrm{\mu F}$