If the equation `x^(2) - 3x + b = 0` and `x^(3) - 4x^(2) + qx = 0`, where `b ne 0, q ne 0` have one common root and the second equation has two equal roots, then find the value of `(q + b)`.

Explanation

To solve the problem, we need to find the values of \( b \) and \( q \) from the given equations and then compute \( q + b \).

### Step 1: Identify the equations and their roots
The first equation is:
\[ x^2 - 3x + b = 0 \]
Let the roots of this equation be \( \alpha \) and \( \beta \).

The second equation is:
\[ x^3 - 4x^2 + qx = 0 \]
Factoring out \( x \), we get:
\[ x(x^2 - 4x + q) = 0 \]
This shows that one root is \( 0 \) and the other two roots are the roots of the quadratic \( x^2 - 4x + q = 0 \).

### Step 2: Determine the condition for equal roots
Since the second equation has two equal roots, the discriminant of the quadratic must be zero:
\[ D = b^2 - 4ac = (-4)^2 - 4(1)(q) = 16 - 4q = 0 \]
Setting the discriminant to zero gives:
\[ 16 - 4q = 0 \]
Solving for \( q \):
\[ 4q = 16 \implies q = 4 \]

### Step 3: Use the common root to find \( b \)
Since both equations have one common root, we can assume that the common root is \( \alpha \). From the first equation, we can express \( b \) in terms of \( \alpha \):
\[ \alpha^2 - 3\alpha + b = 0 \implies b = 3\alpha - \alpha^2 \]

### Step 4: Find the roots of the second equation
The roots of the second equation are \( 0, \alpha, \alpha \). Since we already found \( q = 4 \), we can substitute this into the quadratic:
\[ x^2 - 4x + 4 = 0 \]
Factoring gives:
\[ (x - 2)^2 = 0 \]
Thus, the repeated root is \( \alpha = 2 \).

### Step 5: Substitute \( \alpha \) back to find \( b \)
Now substituting \( \alpha = 2 \) into the expression for \( b \):
\[ b = 3(2) - (2)^2 = 6 - 4 = 2 \]

### Step 6: Calculate \( q + b \)
Now that we have \( q = 4 \) and \( b = 2 \):
\[ q + b = 4 + 2 = 6 \]

### Final Answer
The value of \( q + b \) is:
\[ \boxed{6} \]