If the equation `(1)/(x) + (1)/(x+a)=(1)/(lambda)+(1)/(lambda+a)` has real roots that are equal in magnitude and opposite in sign, then

Options

A. `lambda^(2) = 3a^(2)`
B. `lambda^(2) = 2a^(2)`
C. `lambda^(2) = a^(2)`
D. `a^(2) = 2 lambda^(2)`

Explanation

To solve the equation

\[ \frac{1}{x} + \frac{1}{x+a} = \frac{1}{\lambda} + \frac{1}{\lambda+a} \]

given that the roots are equal in magnitude and opposite in sign, we will follow these steps:

### Step 1: Understand the Roots
Since the roots are equal in magnitude and opposite in sign, if one root is \( \lambda \), the other must be \( -\lambda \).

### Step 2: Substitute \( x = -\lambda \)
Substituting \( x = -\lambda \) into the equation gives:

\[ \frac{1}{-\lambda} + \frac{1}{-\lambda + a} = \frac{1}{\lambda} + \frac{1}{\lambda + a} \]

### Step 3: Simplify the Left Side
The left side becomes:

\[ -\frac{1}{\lambda} + \frac{1}{a - \lambda} \]

### Step 4: Combine the Left Side
To combine the fractions on the left side, we find a common denominator:

\[ -\frac{(a - \lambda) + \lambda}{\lambda(a - \lambda)} = -\frac{a}{\lambda(a - \lambda)} \]

### Step 5: Simplify the Right Side
The right side can be combined similarly:

\[ \frac{1}{\lambda} + \frac{1}{\lambda + a} = \frac{(\lambda + a) + \lambda}{\lambda(\lambda + a)} = \frac{2\lambda + a}{\lambda(\lambda + a)} \]

### Step 6: Set the Two Sides Equal
Now we set the two sides equal:

\[ -\frac{a}{\lambda(a - \lambda)} = \frac{2\lambda + a}{\lambda(\lambda + a)} \]

### Step 7: Cross Multiply
Cross multiplying gives:

\[ -a(\lambda + a) = (2\lambda + a)(a - \lambda) \]

### Step 8: Expand Both Sides
Expanding both sides, we have:

Left Side:
\[ -a\lambda - a^2 \]

Right Side:
\[ 2a - 2\lambda^2 + a^2 - a\lambda \]

### Step 9: Combine Like Terms
Combining like terms leads to:

\[ -a\lambda - a^2 = 2a - 2\lambda^2 + a^2 - a\lambda \]

### Step 10: Rearranging the Equation
Rearranging gives:

\[ 0 = 2a - 2\lambda^2 + 2a \]

### Step 11: Simplify Further
This simplifies to:

\[ 2\lambda^2 = a^2 \]

### Final Relation
Thus, we find the relation:

\[ a^2 = 2\lambda^2 \]