If the roots `(1)/(x+k+1)+(2)/(x+k+2)=1` are equal in magnitude but opposite in sign, then k =

Options

A. -1
B. 0
C. 1
D. 2

Answer

To solve the equation \(\frac{1}{x+k+1} + \frac{2}{x+k+2} = 1\) and find the value of \(k\) such that the roots are equal in magnitude but opposite in sign, we can follow these steps:

### Step 1: Combine the fractions
We start with the equation:
\[ \frac{1}{x+k+1} + \frac{2}{x+k+2} = 1 \]
To combine the fractions, we need a common denominator, which is \((x+k+1)(x+k+2)\). Thus, we can rewrite the equation as:
\[ \frac{(x+k+2) + 2(x+k+1)}{(x+k+1)(x+k+2)} = 1 \]

### Step 2: Simplify the numerator
Now, simplify the numerator:
\[ (x+k+2) + 2(x+k+1) = x+k+2 + 2x + 2k + 2 = 3x + 3k + 4 \]
So, we have:
\[ \frac{3x + 3k + 4}{(x+k+1)(x+k+2)} = 1 \]

### Step 3: Cross-multiply
Cross-multiplying gives us:
\[ 3x + 3k + 4 = (x+k+1)(x+k+2) \]

### Step 4: Expand the right-hand side
Expanding the right-hand side:
\[ (x+k+1)(x+k+2) = x^2 + (k+3)x + (k^2 + 3k + 2) \]

### Step 5: Set the equation to zero
Now, set the equation to zero:
\[ 3x + 3k + 4 = x^2 + (k+3)x + (k^2 + 3k + 2) \]
Rearranging gives:
\[ 0 = x^2 + (k+3 - 3)x + (k^2 + 3k + 2 - 4) \]
This simplifies to:
\[ 0 = x^2 + (k)x + (k^2 + 3k - 2) \]

### Step 6: Use the condition for roots
For the roots to be equal in magnitude but opposite in sign, the sum of the roots must be zero. The sum of the roots for the quadratic equation \(ax^2 + bx + c = 0\) is given by \(-\frac{b}{a}\):
\[ -\frac{k}{1} = 0 \implies k = 0 \]

### Conclusion
Thus, the value of \(k\) is:
\[ \boxed{0} \]