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

Options

A. a, c
B. b, c
C. a, b
D. `a+c, b+c`

Answer

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

### Step-by-Step Solution:

1. **Start with the given equation**:
The original equation is \((x - a)(x - b) = c\).
This can be rewritten as:
\[ x^2 - (a + b)x + (ab - c) = 0 \]
Here, the roots \(\alpha\) and \(\beta\) are given by Vieta's formulas:
\(\alpha + \beta = a + b\) and \(\alpha \beta = ab - c\).

2. **Formulate the new equation**:
We need to analyze the equation \((x - \alpha)(x - \beta) + c = 0\).
Expanding this gives:
\[ (x - \alpha)(x - \beta) = x^2 - (\alpha + \beta)x + \alpha \beta \]
Substituting the values from the previous step:
\[ = x^2 - (a + b)x + (ab - c) \]
Therefore, the equation becomes:
\[ x^2 - (a + b)x + (ab - c) + c = 0 \]
Simplifying this gives:
\[ x^2 - (a + b)x + ab = 0 \]

3. **Identify the roots of the new equation**:
The roots of the equation \(x^2 - (a + b)x + ab = 0\) can be found using the quadratic formula:
\[ x = \frac{-(b) \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 1\), \(b = -(a + b)\), and \(c = ab\). Thus, we have:
\[ x = \frac{(a + b) \pm \sqrt{(a + b)^2 - 4 \cdot 1 \cdot ab}}{2 \cdot 1} \]
Simplifying the discriminant:
\[ (a + b)^2 - 4ab = a^2 + 2ab + b^2 - 4ab = a^2 - 2ab + b^2 = (a - b)^2 \]
Therefore, the roots become:
\[ x = \frac{(a + b) \pm |a - b|}{2} \]

4. **Final expression for the roots**:
The two possible roots are:
\[ x_1 = \frac{(a + b) + |a - b|}{2} \quad \text{and} \quad x_2 = \frac{(a + b) - |a - b|}{2} \]
Depending on the values of \(a\) and \(b\), this can be simplified further to yield the specific roots.

### Conclusion:
The roots of the equation \((x - \alpha)(x - \beta) + c = 0\) are given by:
\[ x = \frac{(a + b) + |a - b|}{2} \quad \text{and} \quad x = \frac{(a + b) - |a - b|}{2} \]