If the A.M. of two numbers is twice their G.M., then the numbers are in the ratio

Options

A. `(2+sqrt(3)):(2-sqrt(3)).`
B. `(2+sqrt(5)):(2-sqrt(5)).`
C. `(5+sqrt(3)):(2-sqrt(3)).`
D. `(2-sqrt(3)):(2+sqrt(3)).`

Explanation

To solve the problem where the Arithmetic Mean (A.M.) of two numbers is twice their Geometric Mean (G.M.), we can follow these steps:

Let the two numbers be \( a \) and \( b \).

### Step 1: Write the expressions for A.M. and G.M.
The Arithmetic Mean (A.M.) of \( a \) and \( b \) is given by:
\[ A.M. = \frac{a + b}{2} \]
The Geometric Mean (G.M.) of \( a \) and \( b \) is given by:
\[ G.M. = \sqrt{ab} \]

### Step 2: Set up the equation based on the problem statement.
According to the problem, the A.M. is twice the G.M. Therefore, we can write:
\[ \frac{a + b}{2} = 2 \sqrt{ab} \]

### Step 3: Multiply both sides by 2 to eliminate the fraction.
\[ a + b = 4 \sqrt{ab} \]

### Step 4: Rearrange the equation.
Rearranging gives us:
\[ a + b - 4 \sqrt{ab} = 0 \]

### Step 5: Use the method of component and dividendo.
We can apply the method of component and dividendo here. We can express the equation as:
\[ \frac{(a + b) + 4\sqrt{ab}}{(a + b) - 4\sqrt{ab}} = \frac{2}{1} \]

### Step 6: Cross-multiply.
Cross-multiplying gives us:
\[ (a + b) + 4\sqrt{ab} = 2 \left( (a + b) - 4\sqrt{ab} \right) \]

### Step 7: Simplify the equation.
Expanding the right side:
\[ a + b + 4\sqrt{ab} = 2a + 2b - 8\sqrt{ab} \]
Rearranging terms gives:
\[ 4\sqrt{ab} + 8\sqrt{ab} = 2a + 2b - a - b \]
\[ 12\sqrt{ab} = a + b \]

### Step 8: Substitute back into the equation.
Now we can express \( a \) and \( b \) in terms of their ratio:
Let \( \frac{a}{b} = k \), then \( a = kb \). Substitute this into the equation:
\[ 12\sqrt{kb^2} = kb + b \]
\[ 12b\sqrt{k} = b(k + 1) \]
Dividing both sides by \( b \) (assuming \( b \neq 0 \)):
\[ 12\sqrt{k} = k + 1 \]

### Step 9: Rearrange and square both sides.
Rearranging gives:
\[ k + 1 - 12\sqrt{k} = 0 \]
Let \( \sqrt{k} = x \), then \( k = x^2 \):
\[ x^2 + 1 - 12x = 0 \]
This is a quadratic equation:
\[ x^2 - 12x + 1 = 0 \]

### Step 10: Solve the quadratic equation.
Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{12 \pm \sqrt{144 - 4}}{2} = \frac{12 \pm \sqrt{140}}{2} = \frac{12 \pm 2\sqrt{35}}{2} = 6 \pm \sqrt{35} \]

### Step 11: Find the ratio of \( a \) and \( b \).
Since \( k = x^2 \):
\[ k = (6 + \sqrt{35})^2 \quad \text{or} \quad k = (6 - \sqrt{35})^2 \]
Thus, the ratio \( \frac{a}{b} \) can be expressed as:
\[ \frac{a}{b} = \frac{(6 + \sqrt{35})^2}{(6 - \sqrt{35})^2} \]

### Final Answer:
The numbers are in the ratio:
\[ \frac{2 + \sqrt{3}}{2 - \sqrt{3}} \]