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

(N/A) Let the two numbers be $a$ and $b$.
The geometric mean $(G.M.)$ is $\sqrt{ab}$.
According to the given condition:
$a+b = 6\sqrt{ab}$ --- $(1)$
Squaring both sides:
$(a+b)^2 = 36ab$
We know that $(a-b)^2 = (a+b)^2 - 4ab$.
Substituting the value from $(1)$:
$(a-b)^2 = 36ab - 4ab = 32ab$
$a-b = \sqrt{32}\sqrt{ab} = 4\sqrt{2}\sqrt{ab}$ --- $(2)$
Adding $(1)$ and $(2)$:
$2a = (6+4\sqrt{2})\sqrt{ab}$
$a = (3+2\sqrt{2})\sqrt{ab}$
Substituting $a$ into $(1)$:
$b = 6\sqrt{ab} - (3+2\sqrt{2})\sqrt{ab}$
$b = (3-2\sqrt{2})\sqrt{ab}$
Therefore,the ratio is:
$\frac{a}{b} = \frac{(3+2\sqrt{2})\sqrt{ab}}{(3-2\sqrt{2})\sqrt{ab}} = \frac{3+2\sqrt{2}}{3-2\sqrt{2}}$
Thus,the numbers are in the ratio $(3+2\sqrt{2}):(3-2\sqrt{2})$.