If the roots of the equation `1/(x+p) + 1/(x+q) = 1/r` are equal in magnitude but opposite in sign and its product is ` alpha `

Options

A. p + q =r
B. p +q =2r
C. `alpha^(2) = ( p^(2) +q^(2))/2`
D. `alpha= ( -p^(2) +q^(2))/2`

Answer

To solve the equation \( \frac{1}{x+p} + \frac{1}{x+q} = \frac{1}{r} \) given that the roots are equal in magnitude but opposite in sign and their product is \( \alpha \), we can follow these steps:

### Step 1: Rewrite the Equation
Start with the given equation:
\[ \frac{1}{x+p} + \frac{1}{x+q} = \frac{1}{r} \]

### Step 2: Find a Common Denominator
The common denominator for the left side is \((x+p)(x+q)\):
\[ \frac{(x+q) + (x+p)}{(x+p)(x+q)} = \frac{1}{r} \]
This simplifies to:
\[ \frac{2x + p + q}{(x+p)(x+q)} = \frac{1}{r} \]

### Step 3: Cross Multiply
Cross-multiplying gives:
\[ r(2x + p + q) = (x+p)(x+q) \]

### Step 4: Expand the Right Side
Expanding the right side:
\[ r(2x + p + q) = x^2 + (p+q)x + pq \]

### Step 5: Rearrange the Equation
Rearranging gives:
\[ x^2 + (p + q - 2r)x + (pq - r(p + q)) = 0 \]

### Step 6: Identify the Roots
Let the roots of this quadratic equation be \( \alpha \) and \( -\alpha \) (since they are equal in magnitude but opposite in sign).

### Step 7: Use Vieta's Formulas
According to Vieta's formulas:
- The sum of the roots \( \alpha + (-\alpha) = 0 \) gives:
\[ p + q - 2r = 0 \implies p + q = 2r \]

### Step 8: Calculate the Product of the Roots
The product of the roots is given by:
\[ \alpha \cdot (-\alpha) = -\alpha^2 = pq - r(p + q) \]
Substituting \( p + q = 2r \):
\[ -\alpha^2 = pq - r(2r) \implies -\alpha^2 = pq - 2r^2 \]

### Step 9: Solve for \( \alpha^2 \)
This gives:
\[ \alpha^2 = 2r^2 - pq \]

### Step 10: Final Expression for \( \alpha \)
Thus, we have:
\[ \alpha = \sqrt{2r^2 - pq} \]

### Summary
The roots of the equation are equal in magnitude but opposite in sign, and their product is given by:
\[ \alpha = \sqrt{2r^2 - pq} \]