The number k is such that `tan { arc tan (2) + arc tan (20 k) } = k`. The sum of all possible values of k is

Options

A. `- 19/40`
B. ` - 21/40`
C. 0
D. `1/5`

Answer

To solve the equation \( \tan(\tan^{-1}(2) + \tan^{-1}(20k)) = k \), we will follow these steps:

### Step 1: Rewrite the equation using inverse tangent properties
We know that \( \tan(\tan^{-1}(a) + \tan^{-1}(b)) = \frac{a + b}{1 - ab} \). Applying this to our equation:

\[ \tan(\tan^{-1}(2) + \tan^{-1}(20k)) = \frac{2 + 20k}{1 - (2)(20k)} \]

### Step 2: Set the equation equal to \( k \)
Now we set the expression equal to \( k \):

\[ \frac{2 + 20k}{1 - 40k} = k \]

### Step 3: Cross-multiply to eliminate the fraction
Cross-multiplying gives us:

\[ 2 + 20k = k(1 - 40k) \]

### Step 4: Expand the right side
Expanding the right side, we have:

\[ 2 + 20k = k - 40k^2 \]

### Step 5: Rearrange the equation
Rearranging all terms to one side results in:

\[ 40k^2 + 20k - k + 2 = 0 \]

This simplifies to:

\[ 40k^2 + 19k + 2 = 0 \]

### Step 6: Apply the quadratic formula
Using the quadratic formula \( k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 40, b = 19, c = 2 \):

\[ k = \frac{-19 \pm \sqrt{19^2 - 4 \cdot 40 \cdot 2}}{2 \cdot 40} \]

### Step 7: Calculate the discriminant
Calculating the discriminant:

\[ 19^2 - 4 \cdot 40 \cdot 2 = 361 - 320 = 41 \]

### Step 8: Substitute back into the formula
Now substituting back:

\[ k = \frac{-19 \pm \sqrt{41}}{80} \]

### Step 9: Find the sum of the roots
The sum of the roots of a quadratic equation \( ax^2 + bx + c = 0 \) is given by \( -\frac{b}{a} \). Thus, the sum of the roots here is:

\[ -\frac{19}{40} \]

### Final Answer
The sum of all possible values of \( k \) is:

\[ \boxed{-\frac{19}{40}} \]