If `alphaa n dbeta` are roots of the equation `a x^2+b x+c=0,` then the roots of the equation `a(2x+1)^2-b(2x+1)(3-x)+c(3-x)^2=0` are `(2alpha+1)/(alpha-3),(2beta+1)/(beta-3)` b. `(3alpha+1)/(alpha-2),(3beta+1)/(beta-2)` c. `(2alpha-1)/(alpha-2),(2beta+1)/(beta-2)` d. none of these

Options

A. `(2alpha + 1)/(alpha - 3), (2beta + 1)/(beta - 3)`
B. `(3alpha + 1)/(alpha - 2), (2beta + 1)/(beta - 2)`
C. `(2alpha - 1)/(alpha - 2), (2beta + 1)/(beta - 2)`
D. none of these

Explanation

To solve the problem, we need to find the roots of the equation given that \(\alpha\) and \(\beta\) are the roots of the equation \(ax^2 + bx + c = 0\). The new equation we are dealing with is:

\[ a(2x + 1)^2 - b(2x + 1)(3 - x) + c(3 - x)^2 = 0 \]

### Step 1: Substitute the Roots
Since \(\alpha\) and \(\beta\) are roots of \(ax^2 + bx + c = 0\), we know that:

\[ a\alpha^2 + b\alpha + c = 0 \quad \text{and} \quad a\beta^2 + b\beta + c = 0 \]

### Step 2: Rewrite the New Equation
We can rewrite the new equation in terms of \(x\):

\[ a(2x + 1)^2 - b(2x + 1)(3 - x) + c(3 - x)^2 = 0 \]

### Step 3: Expand the New Equation
Expanding each term:

1. \(a(2x + 1)^2 = a(4x^2 + 4x + 1)\)
2. \(-b(2x + 1)(3 - x) = -b(6x - 2x^2 + 3 - x) = -b(-2x^2 + 5x + 3)\)
3. \(c(3 - x)^2 = c(9 - 6x + x^2)\)

Combining these gives:

\[ (4a - 2b + c)x^2 + (4a + 5b - 6c)x + (a + 3b + 9c) = 0 \]

### Step 4: Find the Roots
To find the roots of the new equation, we can use the relationships between the coefficients of the new equation and the original equation.

Using the relationships derived from the roots \(\alpha\) and \(\beta\):

1. For \(\alpha\):
\[ x = \frac{2\alpha + 1}{\alpha - 3} \]

2. For \(\beta\):
\[ x = \frac{2\beta + 1}{\beta - 3} \]

### Conclusion
Thus, the roots of the new equation are:

\[ \left(\frac{2\alpha + 1}{\alpha - 3}, \frac{2\beta + 1}{\beta - 3}\right) \]

This corresponds to option **a**.