If the equation `(a)/(x-a)+(b)/(x-b)=1` has two roots equal in magnitude and opposite in sign then the value of a + b is

Options

A. 0
B. 1
C. `-1`
D. none

Answer

To solve the equation \(\frac{a}{x-a} + \frac{b}{x-b} = 1\) under the condition that it has two roots equal in magnitude and opposite in sign, we can follow these steps:

### Step 1: Rewrite the Equation
Start by rewriting the equation:
\[ \frac{a}{x-a} + \frac{b}{x-b} = 1 \]

### Step 2: Combine the Fractions
To combine the fractions on the left-hand side, find a common denominator:
\[ \frac{a(x-b) + b(x-a)}{(x-a)(x-b)} = 1 \]
This simplifies to:
\[ \frac{ax - ab + bx - ba}{(x-a)(x-b)} = 1 \]
which can be rewritten as:
\[ \frac{(a+b)x - 2ab}{(x-a)(x-b)} = 1 \]

### Step 3: Cross-Multiply
Cross-multiply to eliminate the fraction:
\[ (a+b)x - 2ab = (x-a)(x-b) \]

### Step 4: Expand the Right Side
Expand the right-hand side:
\[ (a+b)x - 2ab = x^2 - (a+b)x + ab \]

### Step 5: Rearrange the Equation
Rearranging gives us:
\[ x^2 - (a+b)x + ab + 2ab = 0 \]
which simplifies to:
\[ x^2 - (a+b)x + 3ab = 0 \]

### Step 6: Analyze the Roots
For the roots to be equal in magnitude and opposite in sign, their sum must equal zero. The sum of the roots of the quadratic equation \(ax^2 + bx + c = 0\) is given by \(-\frac{b}{a}\). Here, \(b = -(a+b)\) and \(a = 1\).

Setting the sum of the roots to zero:
\[ -(a+b) = 0 \]
This implies:
\[ a + b = 0 \]

### Conclusion
Thus, the value of \(a + b\) is:
\[ \boxed{0} \]