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Solution

Let the numbers be $a$ and $b$ with G.M. $\sqrt{ab}$. Given: $a + b = 6\sqrt{ab}$. Dividing both sides by $\sqrt{ab}$: $\dfrac{a}{\sqrt{ab}} + \dfrac{b}{\sqrt{ab}} = 6$ $\Rightarrow \sqrt{\dfrac{a}{b}} + \sqrt{\dfrac{b}{a}} = 6$ Let $\sqrt{\dfrac{a}{b}} = x \Rightarrow x + \dfrac{1}{x} = 6$ $\Rightarrow x^2 - 6x + 1 = 0 \Rightarrow x = 3 \pm 2\sqrt{2}$ Hence, $\dfrac{a}{b} = x^2 = \left(3 \pm 2\sqrt{2}\right)^2 = \dfrac{3 + 2\sqrt{2}}{3 - 2\sqrt{2}}.$