The roots `alpha` and `beta` of the quadratic equation `x^(2) - 5x + 3(k-1) = 0` are such that `alpha- beta=1.` Find the value k

Answer

To solve the problem step by step, we need to find the value of \( k \) for the quadratic equation \( x^2 - 5x + 3(k-1) = 0 \) given that the roots \( \alpha \) and \( \beta \) satisfy \( \alpha - \beta = 1 \).

### Step 1: Identify the coefficients
The given quadratic equation is in the standard form \( ax^2 + bx + c = 0 \). Here, we have:
- \( a = 1 \)
- \( b = -5 \)
- \( c = 3(k-1) \)

### Step 2: Use the relationships of the roots
From Vieta's formulas, we know:
- The sum of the roots \( \alpha + \beta = -\frac{b}{a} = -\frac{-5}{1} = 5 \)
- The product of the roots \( \alpha \beta = \frac{c}{a} = \frac{3(k-1)}{1} = 3(k-1) \)

### Step 3: Set up equations based on the given condition
We are given that \( \alpha - \beta = 1 \). We can express \( \alpha \) and \( \beta \) in terms of their sum and difference:
- Let \( \alpha + \beta = 5 \) (from Step 2)
- Let \( \alpha - \beta = 1 \)

Now, we can solve these two equations simultaneously.

### Step 4: Solve for \( \alpha \) and \( \beta \)
Adding the two equations:
\[ (\alpha + \beta) + (\alpha - \beta) = 5 + 1 \]
\[ 2\alpha = 6 \implies \alpha = 3 \]

Now, substituting \( \alpha \) back to find \( \beta \):
\[ \alpha + \beta = 5 \implies 3 + \beta = 5 \implies \beta = 2 \]

### Step 5: Calculate the product of the roots
Now that we have \( \alpha = 3 \) and \( \beta = 2 \), we can find the product:
\[ \alpha \beta = 3 \cdot 2 = 6 \]

### Step 6: Set up the equation for \( k \)
From Step 2, we have:
\[ \alpha \beta = 3(k-1) \]
Setting this equal to the product we calculated:
\[ 6 = 3(k - 1) \]

### Step 7: Solve for \( k \)
Dividing both sides by 3:
\[ 2 = k - 1 \]
Adding 1 to both sides:
\[ k = 3 \]

### Final Answer
The value of \( k \) is \( 3 \).

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