If the roots of the equation `(1)/(x+a)+(1)/(x+b)=(1)/(c)` are equal in magnitude but opposite in sign , then their prodcut is :

Answer

To solve the problem step by step, we start with the given equation:

\[ \frac{1}{x+a} + \frac{1}{x+b} = \frac{1}{c} \]

### Step 1: Combine the left-hand side
We can combine the fractions on the left-hand side:

\[ \frac{(x+b) + (x+a)}{(x+a)(x+b)} = \frac{1}{c} \]

This simplifies to:

\[ \frac{2x + (a + b)}{(x+a)(x+b)} = \frac{1}{c} \]

### Step 2: Cross-multiply
Cross-multiplying gives us:

\[ c(2x + a + b) = (x+a)(x+b) \]

### Step 3: Expand both sides
Now, we expand both sides:

Left-hand side:

\[ c(2x + a + b) = 2cx + c(a + b) \]

Right-hand side:

\[ (x+a)(x+b) = x^2 + (a+b)x + ab \]

Setting both sides equal gives:

\[ 2cx + c(a + b) = x^2 + (a + b)x + ab \]

### Step 4: Rearranging the equation
Rearranging the equation leads to:

\[ x^2 + (a + b - 2c)x + (ab - c(a + b)) = 0 \]

### Step 5: Roots are equal in magnitude but opposite in sign
Let the roots be \( \alpha \) and \( -\alpha \). The sum of the roots \( \alpha + (-\alpha) = 0 \), which means:

\[ \text{Sum of roots} = -\frac{b}{a} = 0 \]

This implies:

\[ a + b - 2c = 0 \quad \Rightarrow \quad 2c = a + b \quad \Rightarrow \quad c = \frac{a + b}{2} \]

### Step 6: Finding the product of the roots
The product of the roots \( \alpha \) and \( -\alpha \) is given by:

\[ \alpha \cdot (-\alpha) = -\alpha^2 \]

From the quadratic equation, the product of the roots is given by:

\[ \text{Product of roots} = \frac{c}{1} = c \]

Substituting \( c \):

\[ \text{Product of roots} = \frac{a + b}{2} \]

### Step 7: Final expression for the product
Since we know the product of the roots is also equal to \( -\alpha^2 \), we can equate:

\[ -\alpha^2 = \frac{a + b}{2} \]

Thus, the product of the roots is:

\[ \frac{ab - c(a + b)}{1} = \frac{ab - \frac{(a + b)^2}{2}}{1} \]

### Final Answer
The product of the roots is:

\[ \frac{-1}{2}(a^2 + b^2) \]