The equations `x^2 - 4x + k = 0` and `x^2 + kx -4 = 0` where k is a real number have exactly one common root. What is the value of k.

Answer

To solve the problem, we need to find the value of \( k \) such that the equations \( x^2 - 4x + k = 0 \) and \( x^2 + kx - 4 = 0 \) have exactly one common root.

### Step-by-Step Solution:

1. **Let the common root be \( m \)**:
We assume that both equations share a common root, which we denote as \( m \).

2. **Substitute \( m \) into the first equation**:
Substitute \( m \) into the first equation:
\[ m^2 - 4m + k = 0 \quad \text{(1)} \]

3. **Substitute \( m \) into the second equation**:
Substitute \( m \) into the second equation:
\[ m^2 + km - 4 = 0 \quad \text{(2)} \]

4. **Subtract equation (2) from equation (1)**:
We subtract equation (2) from equation (1):
\[ (m^2 - 4m + k) - (m^2 + km - 4) = 0 \]
Simplifying this, we get:
\[ -4m + k - km + 4 = 0 \]
Rearranging gives:
\[ k - km - 4m + 4 = 0 \]

5. **Rearranging the equation**:
Rearranging the equation yields:
\[ k(1 - m) = 4m - 4 \]
Thus, we can express \( k \) as:
\[ k = \frac{4(m - 1)}{1 - m} \quad \text{(3)} \]

6. **Finding the condition for exactly one common root**:
For the equations to have exactly one common root, the discriminant of either equation must be zero. Let's use the first equation:
\[ D_1 = b^2 - 4ac = (-4)^2 - 4 \cdot 1 \cdot k = 16 - 4k \]
Setting the discriminant to zero for the first equation:
\[ 16 - 4k = 0 \]
Solving for \( k \):
\[ 4k = 16 \implies k = 4 \]

7. **Verifying the value of \( k \)**:
Now we need to check if \( k = 4 \) gives us exactly one common root. Substitute \( k = 4 \) back into both equations:
- First equation: \( x^2 - 4x + 4 = 0 \) simplifies to \( (x - 2)^2 = 0 \) which has a double root at \( x = 2 \).
- Second equation: \( x^2 + 4x - 4 = 0 \) has a discriminant \( 4^2 - 4 \cdot 1 \cdot (-4) = 16 + 16 = 32 \), which has two distinct roots.

Since both equations share the root \( x = 2 \) and the first equation has a double root, we confirm that there is exactly one common root.

### Final Answer:
The value of \( k \) is \( \boxed{4} \).