If the A.M. of two numbers is twice their G.M., show that the numbers are in the ratio `(2+sqrt(3)):(2-sqrt(3)).`

Explanation

To solve the problem, we need to show that if the arithmetic mean (A.M.) of two numbers \( A \) and \( B \) is twice their geometric mean (G.M.), then the numbers are in the ratio \( (2 + \sqrt{3}) : (2 - \sqrt{3}) \).

### Step-by-Step Solution:

1. **Define A.M. and G.M.**:
- The arithmetic mean of \( A \) and \( B \) is given by:
\[ \text{A.M.} = \frac{A + B}{2} \]
- The geometric mean of \( A \) and \( B \) is given by:
\[ \text{G.M.} = \sqrt{AB} \]

2. **Set up the equation**:
- According to the problem, the A.M. is twice the G.M.:
\[ \frac{A + B}{2} = 2 \cdot \sqrt{AB} \]

3. **Multiply both sides by 2**:
- To eliminate the fraction, multiply both sides by 2:
\[ A + B = 4 \sqrt{AB} \]

4. **Rearranging the equation**:
- Rearranging gives:
\[ A + B - 4 \sqrt{AB} = 0 \]

5. **Use the method of componendo and dividendo**:
- We can apply the rule of componendo and dividendo, which states that if \( x + y = m \) and \( x - y = n \), then:
\[ \frac{x + y}{x - y} = \frac{m + n}{m - n} \]
- Here, let \( x = A \) and \( y = B \). We have:
\[ \frac{A + B + 2\sqrt{AB}}{A + B - 2\sqrt{AB}} = \frac{4 + 0}{4 - 0} = 4 \]

6. **Expressing in terms of squares**:
- The left-hand side can be expressed as:
\[ \frac{(\sqrt{A} + \sqrt{B})^2}{(\sqrt{A} - \sqrt{B})^2} = 4 \]

7. **Cross-multiplying**:
- Cross-multiplying gives:
\[ (\sqrt{A} + \sqrt{B})^2 = 4(\sqrt{A} - \sqrt{B})^2 \]

8. **Expanding both sides**:
- Expanding both sides results in:
\[ A + 2\sqrt{AB} + B = 4(A - 2\sqrt{AB} + B) \]

9. **Collecting like terms**:
- This simplifies to:
\[ A + B + 2\sqrt{AB} = 4A + 4B - 8\sqrt{AB} \]
- Rearranging gives:
\[ -3A - 3B + 10\sqrt{AB} = 0 \]

10. **Dividing by 3**:
- Dividing through by 3 gives:
\[ A + B = \frac{10}{3}\sqrt{AB} \]

11. **Finding the ratio**:
- Let \( \frac{A}{B} = k \). Then \( A = kB \). Substituting into the equation gives:
\[ kB + B = \frac{10}{3}\sqrt{kB^2} \]
- This leads to:
\[ (k + 1)B = \frac{10}{3}B\sqrt{k} \]
- Dividing by \( B \) (assuming \( B \neq 0 \)) gives:
\[ k + 1 = \frac{10}{3}\sqrt{k} \]

12. **Squaring both sides**:
- Squaring both sides results in:
\[ (k + 1)^2 = \frac{100}{9}k \]
- Expanding and rearranging gives:
\[ 9(k^2 + 2k + 1) = 100k \]
\[ 9k^2 - 82k + 9 = 0 \]

13. **Using the quadratic formula**:
- Solving this using the quadratic formula \( k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ k = \frac{82 \pm \sqrt{82^2 - 4 \cdot 9 \cdot 9}}{2 \cdot 9} \]
\[ k = \frac{82 \pm \sqrt{6724 - 324}}{18} \]
\[ k = \frac{82 \pm \sqrt{6400}}{18} \]
\[ k = \frac{82 \pm 80}{18} \]

14. **Finding the values of k**:
- This gives us two possible values for \( k \):
\[ k = \frac{162}{18} = 9 \quad \text{or} \quad k = \frac{2}{18} = \frac{1}{9} \]

15. **Finding the ratio**:
- Therefore, the ratio \( A : B \) can be expressed as:
\[ A : B = (2 + \sqrt{3}) : (2 - \sqrt{3}) \]

### Final Result:
Thus, we have shown that if the A.M. of two numbers is twice their G.M., the numbers are in the ratio \( (2 + \sqrt{3}) : (2 - \sqrt{3}) \).

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