If the roots of the equation `(a)/(x+a+k)+(b)/(x+b+k)=2` are equal in magnitude but opposite in sign, then the value of k is

Options

A. `-(a+b)/(4)`
B. `(a+b)/(4)`
C. `(a+b)/(2)`
D. 0

Answer

To solve the equation \(\frac{a}{x+a+k} + \frac{b}{x+b+k} = 2\) under the condition that the roots are equal in magnitude but opposite in sign, we can follow these steps:

### Step 1: Rearranging the Equation
Start by simplifying the equation:
\[ \frac{a}{x+a+k} + \frac{b}{x+b+k} = 2 \]

### Step 2: Finding a Common Denominator
We can combine the fractions on the left side:
\[ \frac{a(x+b+k) + b(x+a+k)}{(x+a+k)(x+b+k)} = 2 \]

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

### Step 4: Expanding Both Sides
Expand both sides of the equation:
- Left side:
\[ ax + ab + ak + bx + ba + bk = (a+b)x + (ab + ak + ba + bk) \]
- Right side:
\[ 2(x^2 + (a+b+k)x + (ak + bk + ab)) \]

### Step 5: Setting the Equation to Zero
Combine all terms to one side:
\[ (a+b)x + (ab + ak + ba + bk) - 2(x^2 + (a+b+k)x + (ak + bk + ab)) = 0 \]

### Step 6: Collecting Like Terms
Rearranging gives:
\[ -2x^2 + (a+b - 2(a+b+k))x + (ab + ak + ba + bk - 2(ak + bk + ab)) = 0 \]
This simplifies to:
\[ -2x^2 + (a+b - 2(a+b+k))x + (ab + ak + ba + bk - 2ak - 2bk - 2ab) = 0 \]

### Step 7: Finding the Coefficients
Let:
\[ A = -2, \quad B = a + b - 2(a + b + k), \quad C = ab + ak + ba + bk - 2ak - 2bk - 2ab \]

### Step 8: Condition for Roots
Since the roots are equal in magnitude but opposite in sign, the sum of the roots must be zero:
\[ \frac{-B}{A} = 0 \implies B = 0 \]

### Step 9: Solving for k
Set \(B = 0\):
\[ a + b - 2(a + b + k) = 0 \]
This simplifies to:
\[ a + b = 2a + 2b + 2k \implies 0 = a + b + 2k \implies k = -\frac{a + b}{2} \]

### Final Answer
Thus, the value of \(k\) is:
\[ k = -\frac{a + b}{2} \]