The sum of two numbers is $6$ times their geometric mean show that numbers are in the ratio $(3+2\sqrt{2}):(3-2\sqrt{2})$

Let the two numbers are $a$ and $b$.
Thus their $G.M=\sqrt{ab}$
According to the given condition, $a+b=6\sqrt{ab}....(1)$
$\Rightarrow {(a+b)}^2=36(ab)$
Also, $$\begin{gathered} {(a-b)}^2={(a+b)}^2-4ab=36ab-4ab=32ab \\ \Rightarrow a-b=\sqrt{32}\sqrt{ab}=4\sqrt{2}\sqrt{ab} \end{gathered}$$ ........(2)
Adding (1) and (2) , we obtain $2a=(6+4\sqrt{2})\sqrt{ab}$
$\Rightarrow a=(3+2\sqrt{2})\sqrt{ab}$
Substituting the value of $a$ in (1), we obtain $b=6\sqrt{ab}-(3+2\sqrt{2})\sqrt{ab}$
$$\begin{gathered} \Rightarrow b=(3-2\sqrt{2})\sqrt{ab} \\ \Rightarrow \frac{a}{b}=\frac{(3+2\sqrt{2)}\sqrt{ab}}{(3-2\sqrt{2})\sqrt{ab}}=\frac{3+2\sqrt{2}}{3-2\sqrt{2}} \end{gathered}$$
Thus , the required ratio is $(3+2\sqrt{2}):(3-2\sqrt{2})$