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

Options

A. `-2( a ^(2) + b ^(2))`
B. ` (a6(2) + b^(2))`
C. `-(( a^(2) + b^(2))/(2 ))`
D. `-ab`

Answer

To solve the problem, we need to analyze the given equation and find the product of the roots when they are equal in magnitude but opposite in sign.

### Step-by-step Solution:

1. **Start with the given equation**:
\[ \frac{1}{x+a} + \frac{1}{x+b} = \frac{1}{c} \]

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

3. **Simplify the numerator**:
Combine the terms in the numerator:
\[ \frac{2x + (a+b)}{(x+a)(x+b)} = \frac{1}{c} \]

4. **Cross-multiply to eliminate the fractions**:
This gives us:
\[ c(2x + (a+b)) = (x+a)(x+b) \]

5. **Expand the right side**:
Expanding \((x+a)(x+b)\):
\[ x^2 + (a+b)x + ab \]
So we have:
\[ c(2x + (a+b)) = x^2 + (a+b)x + ab \]

6. **Rearranging the equation**:
Rearranging gives us:
\[ x^2 + (a+b - 2c)x + ab - c(a+b) = 0 \]

7. **Identify the coefficients**:
The quadratic equation is in the form \(Ax^2 + Bx + C = 0\):
- \(A = 1\)
- \(B = a + b - 2c\)
- \(C = ab - c(a+b)\)

8. **Condition for roots**:
The roots are equal in magnitude but opposite in sign, which implies that the sum of the roots is zero:
\[ \alpha + \beta = 0 \]
For a quadratic equation, the sum of the roots is given by \(-\frac{B}{A}\):
\[ -\frac{B}{1} = 0 \implies B = 0 \]
Therefore, we have:
\[ a + b - 2c = 0 \implies c = \frac{a+b}{2} \]

9. **Finding the product of the roots**:
The product of the roots \(\alpha \beta\) is given by \(\frac{C}{A}\):
\[ \alpha \beta = \frac{ab - c(a+b)}{1} \]
Substituting \(c = \frac{a+b}{2}\):
\[ \alpha \beta = ab - \frac{a+b}{2}(a+b) \]
Simplifying:
\[ \alpha \beta = ab - \frac{(a+b)^2}{2} \]

10. **Final expression for the product**:
Thus, we can express the product of the roots as:
\[ \alpha \beta = ab - \frac{a^2 + 2ab + b^2}{2} = ab - \frac{a^2 + b^2 + 2ab}{2} = -\frac{a^2 + b^2}{2} \]

### Conclusion:
The product of the roots is:
\[ \alpha \beta = -\frac{a^2 + b^2}{2} \]