Question - Calculating Specific Expansion Terms and Investigating Geometric Mean Relation

The sum of the two numbers is $6$ times their geometric mean, then

Let the two numbers be $a$ and $b$. Given that $a+b = 6\sqrt{ab}$ and we are to show that $a$ and $b$ are in the ratio $(3 + 2\sqrt{2}):(3 - 2\sqrt{2})$.

Since we have a ratio of sums to a geometric mean, let's assume $a = (3 + 2\sqrt{2})k$ and $b = (3 - 2\sqrt{2})k$, where $k$ is some positive constant.

\[ a + b = (3 + 2\sqrt{2})k + (3 - 2\sqrt{2})k\]

\[ a + b = 3k + 2\sqrt{2}k + 3k - 2\sqrt{2}k\]

\[ a + b = 6k\]

The geometric mean of $a$ and $b$ is $\sqrt{ab}$:

\[\sqrt{ab} = \sqrt{(3 + 2\sqrt{2})k \cdot (3 - 2\sqrt{2})k}\]

\[\sqrt{ab} = \sqrt{(9 - 8)k^2}\]

\[\sqrt{ab} = k\]

Now, we compare the sum and geometric mean:

\[6k = 6\sqrt{ab}\]

\[k = \sqrt{ab}\]

This confirms that the sum of $a$ and $b$ is indeed $6$ times their geometric mean, and thus the numbers are in the desired ratio.