If the equation `(alpha^(2)-5alpha+6)x^(2)+(alpha^(2)-3alpha+2)x+(alpha^(2)-4)=0` has more than two roots, then the value of `alpha` is :

Options

A. 2
B. 3
C. 1
D. None of these

Answer

To solve the equation \((\alpha^2 - 5\alpha + 6)x^2 + (\alpha^2 - 3\alpha + 2)x + (\alpha^2 - 4) = 0\) for values of \(\alpha\) such that the equation has more than two roots, we need to determine when this quadratic equation becomes an identity. An identity holds for all values of \(x\), which means all coefficients must equal zero.

### Step-by-Step Solution:

1. **Identify the coefficients of the quadratic equation**:
The equation can be written in the standard form \(Ax^2 + Bx + C = 0\), where:
- \(A = \alpha^2 - 5\alpha + 6\)
- \(B = \alpha^2 - 3\alpha + 2\)
- \(C = \alpha^2 - 4\)

2. **Set each coefficient to zero**:
For the equation to have more than two roots, each coefficient must equal zero:
- \(A = 0\)
- \(B = 0\)
- \(C = 0\)

3. **Solve the first equation \(A = 0\)**:
\[ \alpha^2 - 5\alpha + 6 = 0 \]
Factorizing:
\[ (\alpha - 3)(\alpha - 2) = 0 \]
Thus, \(\alpha = 3\) or \(\alpha = 2\).

4. **Solve the second equation \(B = 0\)**:
\[ \alpha^2 - 3\alpha + 2 = 0 \]
Factorizing:
\[ (\alpha - 2)(\alpha - 1) = 0 \]
Thus, \(\alpha = 2\) or \(\alpha = 1\).

5. **Solve the third equation \(C = 0\)**:
\[ \alpha^2 - 4 = 0 \]
This can be factored as:
\[ (\alpha - 2)(\alpha + 2) = 0 \]
Thus, \(\alpha = 2\) or \(\alpha = -2\).

6. **Combine the solutions**:
From the three equations, the common solution across all three is:
- \(\alpha = 2\)

### Final Answer:
The value of \(\alpha\) such that the equation has more than two roots is:
\[ \alpha = 2 \]