Question:

Step 1: Define the Arithmetic Mean (AM) and Geometric Mean (GM) for two positive numbers $a$ and $b$. $$ \text{AM} = \frac{a+b}{2} $$ $$ \text{GM} = \sqrt{ab} $$ Step 2: Apply the given condition that the AM is twice the GM. $$ \frac{a+b}{2} = 2\sqrt{ab} $$ $$ a+b = 4\sqrt{ab} $$ Step 3: Square both sides of the equation to eliminate the square root and simplify. $$ (a+b)^2 = (4\sqrt{ab})^2 $$ $$ a^2 + 2ab + b^2 = 16ab $$ $$ a^2 - 14ab + b^2 = 0 $$ Step 4: Divide the entire equation by $b^2$ (since $b \neq 0$) to express it in terms of the ratio $a/b$. Let $t = a/b$. $$ \frac{a^2}{b^2} - 14\frac{ab}{b^2} + \frac{b^2}{b^2} = 0 $$ $$ \left(\frac{a}{b}\right)^2 - 14\left(\frac{a}{b}\right) + 1 = 0 $$ $$ t^2 - 14t + 1 = 0 $$ Step 5: Solve the quadratic equation for $t$ using the quadratic formula $t = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$. $$ t = \frac{-(-14) \pm \sqrt{(-14)^2 - 4(1)(1)}}{2(1)} $$ $$ t = \frac{14 \pm \sqrt{196 - 4}}{2} $$ $$ t = \frac{14 \pm \sqrt{192}}{2} $$ Simplify $\sqrt{192}$: $\sqrt{192} = \sqrt{64 \times 3} = 8\sqrt{3}$. $$ t = \frac{14 \pm 8\sqrt{3}}{2} $$ $$ t = 7 \pm 4\sqrt{3} $$ Step 6: Determine the ratio $a:b$. The two possible values for $a/b$ are $7 + 4\sqrt{3}$ and $7 - 4\sqrt{3}$. Since $a \neq b$, $a/b \neq 1$. Both $7 + 4\sqrt{3}$ and $7 - 4\sqrt{3}$ satisfy this condition. Consider the ratio $(2 + \sqrt{3}) : (2 - \sqrt{3})$, which corresponds to $a/b = \frac{2 + \sqrt{3}}{2 - \sqrt{3}}$. Rationalize the denominator: $$ \frac{2 + \sqrt{3}}{2 - \sqrt{3}} = \frac{2 + \sqrt{3}}{2 - \sqrt{3}} \times \frac{2 + \sqrt{3}}{2 + \sqrt{3}} $$ $$ = \frac{(2 + \sqrt{3})^2}{2^2 - (\sqrt{3})^2} $$ $$ = \frac{4 + 4\sqrt{3} + 3}{4 - 3} $$ $$ = \frac{7 + 4\sqrt{3}}{1} $$ $$ = 7 + 4\sqrt{3} $$ Thus, one of the possible ratios for $a:b$ is $(2 + \sqrt{3}) : (2 - \sqrt{3})$.