If ` alpha, beta` are the roots of the equation ` ax^(2) -bx +c=0` then equation ` (a+cy)^(2) =b^(2) y` has the roots

Options

A. `1/alpha , 1/beta`
B. ` alpha^(2), beta^(2) `
C. `alpha/beta , beta/alpha`
D. `1/alpha^(2), 1/beta^(2) `

Answer

To solve the problem, we need to find the roots of the equation \( (a + cy)^2 = b^2 y \) given that \( \alpha \) and \( \beta \) are the roots of the equation \( ax^2 - bx + c = 0 \).

### Step-by-Step Solution:

1. **Start with the given equation**:
\[ (a + cy)^2 = b^2 y \]

2. **Expand the left-hand side**:
\[ a^2 + 2acy + c^2y^2 = b^2y \]

3. **Rearrange the equation**:
\[ c^2y^2 + (2ac - b^2)y + a^2 = 0 \]

4. **Identify coefficients**:
Here, we can identify the coefficients of the quadratic equation:
- \( A = c^2 \)
- \( B = 2ac - b^2 \)
- \( C = a^2 \)

5. **Use Vieta's formulas**:
According to Vieta's formulas, the sum and product of the roots \( m \) and \( n \) of the quadratic equation \( Ay^2 + By + C = 0 \) are given by:
\[ m + n = -\frac{B}{A} \quad \text{and} \quad mn = \frac{C}{A} \]

6. **Calculate the sum of the roots**:
\[ m + n = -\frac{2ac - b^2}{c^2} = \frac{b^2 - 2ac}{c^2} \]

7. **Calculate the product of the roots**:
\[ mn = \frac{a^2}{c^2} \]

8. **Relate the results to the roots \( \alpha \) and \( \beta \)**:
From the original quadratic equation \( ax^2 - bx + c = 0 \), we know:
- \( \alpha + \beta = \frac{b}{a} \)
- \( \alpha \beta = \frac{c}{a} \)

9. **Express \( m + n \) and \( mn \) in terms of \( \alpha \) and \( \beta \)**:
- Since \( m + n = \frac{b^2 - 2ac}{c^2} \) and \( mn = \frac{a^2}{c^2} \), we can express these in terms of \( \alpha \) and \( \beta \):
\[ m + n = \frac{b^2}{c^2} - \frac{2a}{c} \quad \text{and} \quad mn = \frac{a^2}{c^2} \]

10. **Final roots**:
The roots of the equation \( (a + cy)^2 = b^2 y \) are:
\[ m = \frac{1}{\beta^2}, \quad n = \frac{1}{\alpha^2} \]

### Summary:
The roots of the equation \( (a + cy)^2 = b^2 y \) are \( \frac{1}{\beta^2} \) and \( \frac{1}{\alpha^2} \).