If `alpha, beta` be the roots of the equation `(x-a)(x-b) + c = 0 (c ne 0)`, then the roots of the equation `(x-c-alpha)(x-c-beta)= c`, are

Options

A. a and b + c
B. a + c and b
C. a + c and b + c
D. a - b and b - c

Explanation

To solve the problem step by step, we need to find the roots of the equation \((x - c - \alpha)(x - c - \beta) = c\) given that \(\alpha\) and \(\beta\) are the roots of the equation \((x - a)(x - b) + c = 0\).

### Step 1: Understand the given equation
The first equation is:
\[ (x - a)(x - b) + c = 0 \]
Expanding this, we get:
\[ x^2 - (a + b)x + ab + c = 0 \]
From this, we can identify the roots \(\alpha\) and \(\beta\).

### Step 2: Identify the sum and product of roots
From the quadratic equation \(x^2 - (a + b)x + (ab + c) = 0\):
- The sum of the roots \(\alpha + \beta = a + b\)
- The product of the roots \(\alpha \beta = ab + c\)

### Step 3: Write the second equation
Now, we need to solve the equation:
\[ (x - c - \alpha)(x - c - \beta) = c \]
Expanding this, we have:
\[ x^2 - (c + \alpha + c + \beta)x + (c + \alpha)(c + \beta) = c \]
This simplifies to:
\[ x^2 - (2c + \alpha + \beta)x + (c^2 + c(\alpha + \beta) + \alpha \beta) = c \]

### Step 4: Rearranging the equation
Rearranging gives us:
\[ x^2 - (2c + a + b)x + (c^2 + c(a + b) + (ab + c)) - c = 0 \]
This simplifies to:
\[ x^2 - (2c + a + b)x + (c^2 + c(a + b) + ab) = 0 \]

### Step 5: Substitute values
Now, substituting \(\alpha + \beta\) and \(\alpha \beta\):
\[ x^2 - (2c + a + b)x + (c^2 + c(a + b) + ab) = 0 \]

### Step 6: Finding the roots
To find the roots, we can use the quadratic formula:
\[ x = \frac{-(b) \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 1\), \(b = -(2c + a + b)\), and \(c = c^2 + c(a + b) + ab\).

### Step 7: Final roots
After calculating the discriminant and applying the quadratic formula, we find the roots:
\[ x = c + a \quad \text{and} \quad x = c + b \]

Thus, the roots of the equation \((x - c - \alpha)(x - c - \beta) = c\) are:
\[ x = a + c \quad \text{or} \quad x = b + c \]