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 \( \frac{a}{x-a} + \frac{b}{x-b} = 1 \) and find the value of \( a + b \) given that the equation 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: Clear the fractions
Multiply through by \((x-a)(x-b)\) to eliminate the denominators:
\[ a(x-b) + b(x-a) = (x-a)(x-b) \]

### Step 3: Expand both sides
Expanding the left-hand side:
\[ ax - ab + bx - ba = ax + bx - ab - ab \]
This simplifies to:
\[ (a + b)x - ab \]

Expanding the right-hand side:
\[ x^2 - (a+b)x + ab \]

### Step 4: Set the equation to zero
Now, set the equation to zero:
\[ (a + b)x - ab = x^2 - (a + b)x + ab \]
Rearranging gives:
\[ x^2 - (2a + 2b)x + 2ab = 0 \]

### Step 5: Identify the coefficients
This is a quadratic equation of the form:
\[ Ax^2 + Bx + C = 0 \]
where \( A = 1 \), \( B = -(2a + 2b) \), and \( C = 2ab \).

### Step 6: Use the condition for roots
We know that the roots are equal in magnitude and opposite in sign, which implies:
\[ x_1 + x_2 = 0 \]
From Vieta's formulas, we know:
\[ x_1 + x_2 = -\frac{B}{A} = \frac{2a + 2b}{1} \]
Setting this equal to zero gives:
\[ 2a + 2b = 0 \]

### Step 7: Solve for \( a + b \)
Dividing both sides by 2:
\[ a + b = 0 \]

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