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

Le the two numbers be $a$ and $b$

$G.M.$ $=\sqrt{a b}$

According to the given condition,

$a+b=6 \sqrt{a b}$        ..........$(1)$

$\Rightarrow(a+b)^{2}=36(a b)$

Also,

$(a-b)^{2}=(a+b)^{2}-4 a b=36 a b-4 a b=32 a b$

$\Rightarrow a-b=\sqrt{32} \sqrt{a b}$

$=4 \sqrt{2} \sqrt{a b}$         .........$(2)$

Adding $(1)$ and $(2),$ we obtain

$2 a=(6+4 \sqrt{2}) \sqrt{a b}$

$a=(3+2 \sqrt{2}) \sqrt{a b}$

Substituting the value of $a$ in $(1),$ we obtain

$b=6 \sqrt{a b}-(3+2 \sqrt{2}) \sqrt{a b}$

$\Rightarrow b=(3-2 \sqrt{2}) \sqrt{a b}$

$\frac{a}{b}=\frac{(3+2 \sqrt{2}) \sqrt{a b}}{(3-2 \sqrt{2}) \sqrt{a b}}=\frac{3+2 \sqrt{2}}{3-2 \sqrt{2}}$

Thus, the required ratio is $(3+2 \sqrt{2}):(3-2 \sqrt{2})$