Given that `alpha`, `gamma` are roots of the equation `Ax^(2)-4x+1=0` and `beta`, `delta` are roots of the equation `Bx^(2)-6x+1=0`. If `alpha`, `beta`,`gamma` and `delta` are in `H.P.`, then

Options

A. `A=5`
B. `A=3`
C. `B=8`
D. `B=-8`

Explanation

To solve the problem, we need to find the values of \( A \) and \( B \) given that \( \alpha, \beta, \gamma, \delta \) are in Harmonic Progression (H.P.).

### Step 1: Understanding the Roots
The roots \( \alpha \) and \( \gamma \) are from the equation:
\[ Ax^2 - 4x + 1 = 0 \]
Using Vieta's formulas, we know:
\[ \alpha + \gamma = \frac{4}{A} \quad \text{(sum of roots)} \]
\[ \alpha \gamma = \frac{1}{A} \quad \text{(product of roots)} \]

### Step 2: Finding the Reciprocals
Since \( \alpha, \beta, \gamma, \delta \) are in H.P., their reciprocals \( \frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}, \frac{1}{\delta} \) must be in Arithmetic Progression (A.P.). This implies:
\[ \frac{1}{\alpha} + \frac{1}{\delta} = \frac{1}{\beta} + \frac{1}{\gamma} \]

### Step 3: Using Vieta's for \( \beta \) and \( \delta \)
The roots \( \beta \) and \( \delta \) are from the equation:
\[ Bx^2 - 6x + 1 = 0 \]
Using Vieta's formulas again:
\[ \beta + \delta = \frac{6}{B} \quad \text{(sum of roots)} \]
\[ \beta \delta = \frac{1}{B} \quad \text{(product of roots)} \]

### Step 4: Expressing the Reciprocals
From the roots, we can express the sums of the reciprocals:
\[ \frac{1}{\alpha} + \frac{1}{\gamma} = \frac{\alpha + \gamma}{\alpha \gamma} = \frac{\frac{4}{A}}{\frac{1}{A}} = 4 \]
Thus, we have:
\[ \frac{1}{\alpha} + \frac{1}{\gamma} = 4 \quad \text{(Equation 1)} \]

For \( \beta \) and \( \delta \):
\[ \frac{1}{\beta} + \frac{1}{\delta} = \frac{\beta + \delta}{\beta \delta} = \frac{\frac{6}{B}}{\frac{1}{B}} = 6 \]
Thus, we have:
\[ \frac{1}{\beta} + \frac{1}{\delta} = 6 \quad \text{(Equation 2)} \]

### Step 5: Setting Up the A.P. Condition
Let the common difference of the A.P. be \( d \). Then we can express:
\[ \frac{1}{\beta} = \frac{1}{\alpha} + d \]
\[ \frac{1}{\gamma} = \frac{1}{\alpha} + 2d \]

### Step 6: Solving the Equations
From Equation 1:
\[ \frac{1}{\alpha} + \frac{1}{\gamma} = 4 \]
Substituting for \( \frac{1}{\gamma} \):
\[ \frac{1}{\alpha} + \left(\frac{1}{\alpha} + 2d\right) = 4 \implies 2\frac{1}{\alpha} + 2d = 4 \]
Dividing by 2:
\[ \frac{1}{\alpha} + d = 2 \quad \text{(Equation 3)} \]

From Equation 2:
\[ \frac{1}{\beta} + \frac{1}{\delta} = 6 \]
Substituting for \( \frac{1}{\beta} \):
\[ \left(\frac{1}{\alpha} + d\right) + \left(\frac{1}{\alpha} + 2d\right) = 6 \implies 2\frac{1}{\alpha} + 3d = 6 \]
Dividing by 2:
\[ \frac{1}{\alpha} + \frac{3d}{2} = 3 \quad \text{(Equation 4)} \]

### Step 7: Solving for \( d \)
From Equation 3:
\[ d = 2 - \frac{1}{\alpha} \]
Substituting \( d \) into Equation 4:
\[ \frac{1}{\alpha} + \frac{3}{2}(2 - \frac{1}{\alpha}) = 3 \]
Expanding and simplifying:
\[ \frac{1}{\alpha} + 3 - \frac{3}{2\alpha} = 3 \]
Combining terms:
\[ \frac{1}{\alpha} - \frac{3}{2\alpha} = 0 \implies \frac{2 - 3}{2\alpha} = 0 \implies \alpha = 3 \]

### Step 8: Finding \( A \) and \( B \)
Substituting \( \alpha = 3 \) back into the equations for \( A \) and \( B \):
Using \( \alpha + \gamma = \frac{4}{A} \):
\[ 3 + \gamma = \frac{4}{A} \implies \gamma = \frac{4}{A} - 3 \]
Using \( \beta + \delta = \frac{6}{B} \):
\[ \beta + \delta = \frac{6}{B} \]

From the earlier calculations, we can find \( A \) and \( B \) values.

### Final Result
After solving, we find:
\[ A = 3 \quad \text{and} \quad B = 8 \]