Let p, q and r be real numbers `(p ne q,r ne 0),` such that the roots of the equation `1/(x+p) + 1/(x+q) =1/r` are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to.

Options

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

Answer

To solve the problem, we need to analyze the given equation and derive the required sum of squares of the roots step by step.

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

### Step 2: Clear the fractions
To eliminate the fractions, we can multiply through by \((x+p)(x+q)r\):
\[ r(x+q) + r(x+p) = (x+p)(x+q) \]

### Step 3: Simplify the equation
Expanding both sides:
\[ rx + rq + rx + rp = x^2 + (p+q)x + pq \]
This simplifies to:
\[ 2rx + r(p+q) = x^2 + (p+q)x + pq \]

### Step 4: Rearranging the equation
Rearranging gives us:
\[ x^2 + (p+q - 2r)x + (pq - r(p+q)) = 0 \]

### Step 5: Roots of the quadratic equation
Let the roots of the quadratic equation be \(\alpha\) and \(\beta\). Since the roots are equal in magnitude but opposite in sign, we have:
\[ \alpha + \beta = 0 \]
This implies:
\[ \beta = -\alpha \]

### Step 6: Use Vieta's formulas
From Vieta's formulas, we know:
\[ \alpha + \beta = -(p + q - 2r) = 0 \implies p + q - 2r = 0 \implies p + q = 2r \]
And:
\[ \alpha \beta = pq - r(p + q) \]

### Step 7: Substitute \(p + q\)
Substituting \(p + q = 2r\) into the equation for \(\alpha \beta\):
\[ \alpha \beta = pq - r(2r) = pq - 2r^2 \]

### Step 8: Find the sum of squares of the roots
The sum of squares of the roots is given by:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta \]
Since \(\alpha + \beta = 0\):
\[ \alpha^2 + \beta^2 = 0^2 - 2\alpha \beta = -2\alpha \beta \]
Substituting \(\alpha \beta = pq - 2r^2\):
\[ \alpha^2 + \beta^2 = -2(pq - 2r^2) = -2pq + 4r^2 \]

### Final Result
Thus, the sum of squares of the roots is:
\[ \alpha^2 + \beta^2 = 4r^2 - 2pq \]