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. `-1`
B. `0`
C. `1`
D. None of these

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 with the given equation:
\[ \frac{a}{x-a} + \frac{b}{x-b} = 1 \]

### Step 2: Find a common denominator
The common denominator for the left-hand side is \((x-a)(x-b)\). Thus, we can rewrite the equation as:
\[ \frac{a(x-b) + b(x-a)}{(x-a)(x-b)} = 1 \]

### Step 3: Simplify the numerator
Expanding the numerator:
\[ a(x-b) + b(x-a) = ax - ab + bx - ab = (a+b)x - 2ab \]
So, we have:
\[ \frac{(a+b)x - 2ab}{(x-a)(x-b)} = 1 \]

### Step 4: Cross-multiply
Cross-multiplying gives:
\[ (a+b)x - 2ab = (x-a)(x-b) \]

### Step 5: Expand the right-hand side
Expanding the right-hand side:
\[ (x-a)(x-b) = x^2 - (a+b)x + ab \]
Thus, we equate:
\[ (a+b)x - 2ab = x^2 - (a+b)x + ab \]

### Step 6: Rearrange the equation
Rearranging gives us:
\[ x^2 - (a+b)x + ab + 2ab - (a+b)x = 0 \]
This simplifies to:
\[ x^2 - 2(a+b)x + 3ab = 0 \]

### Step 7: Analyze the roots
The roots of the quadratic equation \(Ax^2 + Bx + C = 0\) are given by the formula:
\[ x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \]
For our equation, \(A = 1\), \(B = -2(a+b)\), and \(C = 3ab\).

### Step 8: Condition for roots equal in magnitude and opposite in sign
If the roots are equal in magnitude and opposite in sign, then the sum of the roots must be zero. Therefore:
\[ -(-2(a+b)) = 0 \implies 2(a+b) = 0 \]

### Step 9: Solve for \(a + b\)
From \(2(a+b) = 0\), we find:
\[ a + b = 0 \]

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