Let A, G, and H are the A.M., G.M. and H.M. respectively of two unequal positive integers. Then, the equation `Ax^(2) - Gx - H = 0` has

Options

A. both roots as fractions
B. one root which is a negative fraction and other positive root
C. at least one root which is an integer
D. none of these

Explanation

To solve the problem, we need to analyze the quadratic equation given by \( Ax^2 - Gx - H = 0 \), where \( A, G, \) and \( H \) are the arithmetic mean (A.M.), geometric mean (G.M.), and harmonic mean (H.M.) of two unequal positive integers.

### Step-by-Step Solution:

1. **Understanding A, G, and H**:
- Let the two unequal positive integers be \( p \) and \( q \).
- The arithmetic mean \( A \) is given by:
\[ A = \frac{p + q}{2} \]
- The geometric mean \( G \) is given by:
\[ G = \sqrt{pq} \]
- The harmonic mean \( H \) is given by:
\[ H = \frac{2pq}{p + q} \]

2. **Establishing Relationships**:
- Since \( p \) and \( q \) are unequal positive integers, we know:
\[ A > G > H \]
- This implies that \( A \) is greater than \( G \) and \( G \) is greater than \( H \).

3. **Roots of the Quadratic Equation**:
- Let \( \alpha \) and \( \beta \) be the roots of the equation \( Ax^2 - Gx - H = 0 \).
- By Vieta's formulas:
- The sum of the roots \( \alpha + \beta = \frac{G}{A} \) (which is a positive fraction since \( G < A \)).
- The product of the roots \( \alpha \beta = -\frac{H}{A} \) (which is negative since \( H > 0 \)).

4. **Analyzing the Roots**:
- Since the sum \( \alpha + \beta \) is positive and the product \( \alpha \beta \) is negative, it follows that one root must be positive and the other must be negative.
- Therefore, we can conclude that:
- One root is a positive fraction.
- The other root is a negative fraction.

5. **Conclusion**:
- From the analysis, we can conclude that the equation \( Ax^2 - Gx - H = 0 \) has one root which is a negative fraction and the other root which is a positive fraction.
- Thus, the correct option is **(b) one root which is a negative fraction and the other positive root**.