The roots of the equation `px^(2)-2(p+1)x+3p=0` are `alpha and beta`. If `alpha-beta=2`, calculate the value of `alpha,beta` and p.

Answer

To solve the equation \( px^2 - 2(p+1)x + 3p = 0 \) given that the roots \( \alpha \) and \( \beta \) satisfy \( \alpha - \beta = 2 \), we will follow these steps:

### Step 1: Identify the coefficients
The coefficients of the quadratic equation are:
- \( a = p \)
- \( b = -2(p + 1) \)
- \( c = 3p \)

### Step 2: Use the relationships of roots
From Vieta's formulas, we know:
1. The sum of the roots \( \alpha + \beta = -\frac{b}{a} = \frac{2(p + 1)}{p} \)
2. The product of the roots \( \alpha \beta = \frac{c}{a} = \frac{3p}{p} = 3 \)

### Step 3: Express \( \alpha \) and \( \beta \)
Given \( \alpha - \beta = 2 \), we can express \( \alpha \) and \( \beta \) in terms of their sum and difference:
- Let \( \alpha + \beta = S \) and \( \alpha - \beta = D \).
- Then, we have:
\[ \alpha = \frac{S + D}{2} = \frac{\frac{2(p + 1)}{p} + 2}{2} = \frac{2(p + 1) + 2p}{2p} = \frac{4p + 2}{2p} = \frac{2p + 1}{p} \]
\[ \beta = \frac{S - D}{2} = \frac{\frac{2(p + 1)}{p} - 2}{2} = \frac{2(p + 1) - 2p}{2p} = \frac{2}{2p} = \frac{1}{p} \]

### Step 4: Calculate the product of the roots
Now, we can set up the equation using the product of the roots:
\[ \alpha \beta = 3 \]
Substituting the values of \( \alpha \) and \( \beta \):
\[ \left(\frac{2p + 1}{p}\right) \left(\frac{1}{p}\right) = 3 \]
This simplifies to:
\[ \frac{2p + 1}{p^2} = 3 \]

### Step 5: Solve for \( p \)
Cross-multiplying gives:
\[ 2p + 1 = 3p^2 \]
Rearranging this gives:
\[ 3p^2 - 2p - 1 = 0 \]

### Step 6: Factor or use the quadratic formula
We can factor this quadratic equation:
\[ 3p^2 - 3p + p - 1 = 0 \]
Factoring gives:
\[ (3p + 1)(p - 1) = 0 \]
Thus, \( p = -\frac{1}{3} \) or \( p = 1 \).

### Step 7: Find \( \alpha \) and \( \beta \) for each value of \( p \)
1. For \( p = 1 \):
\[ \alpha = \frac{2(1) + 1}{1} = 3, \quad \beta = \frac{1}{1} = 1 \]
So, \( \alpha = 3 \), \( \beta = 1 \).

2. For \( p = -\frac{1}{3} \):
\[ \alpha = \frac{2(-\frac{1}{3}) + 1}{-\frac{1}{3}} = \frac{-\frac{2}{3} + 1}{-\frac{1}{3}} = \frac{\frac{1}{3}}{-\frac{1}{3}} = -1 \]
\[ \beta = \frac{1}{-\frac{1}{3}} = -3 \]
So, \( \alpha = -1 \), \( \beta = -3 \).

### Conclusion
The valid solution that satisfies \( \alpha - \beta = 2 \) is:
- \( p = 1 \)
- \( \alpha = 3 \)
- \( \beta = 1 \)