If the equations `x^(2)+abx+c=0 and x^(2)+acx+b=0` have a common root then their other roots satisfy the equation `x^(2)-a(b+c)x+a^(2)bc=0`

Answer

To solve the problem, we need to analyze the two quadratic equations given and find the conditions under which they have a common root.

Let the common root be denoted as \( r \).

### Step 1: Set up the equations with the common root
For the first equation \( x^2 + abx + c = 0 \):
\[ r^2 + abr + c = 0 \quad \text{(1)} \]

For the second equation \( x^2 + acx + b = 0 \):
\[ r^2 + acr + b = 0 \quad \text{(2)} \]

### Step 2: Subtract the two equations
Subtract equation (2) from equation (1):
\[ (r^2 + abr + c) - (r^2 + acr + b) = 0 \]
This simplifies to:
\[ (ab - ac)r + (c - b) = 0 \]
Rearranging gives:
\[ (ab - ac)r = b - c \quad \text{(3)} \]

### Step 3: Analyze the implications
From equation (3), we can express \( r \):
\[ r = \frac{b - c}{ab - ac} \quad \text{(assuming } ab \neq ac\text{)} \]

### Step 4: Find the other roots of the equations
Let the other root of the first equation be \( s_1 \) and the other root of the second equation be \( s_2 \).

Using Vieta's formulas:
- For the first equation:
\[ r + s_1 = -ab \quad \Rightarrow \quad s_1 = -ab - r \quad \text{(4)} \]
- For the second equation:
\[ r + s_2 = -ac \quad \Rightarrow \quad s_2 = -ac - r \quad \text{(5)} \]

### Step 5: Substitute \( r \) into \( s_1 \) and \( s_2 \)
Substituting equation (3) into equations (4) and (5):
\[ s_1 = -ab - \frac{b - c}{ab - ac} \quad \text{(6)} \]
\[ s_2 = -ac - \frac{b - c}{ab - ac} \quad \text{(7)} \]

### Step 6: Form the new quadratic equation
The new quadratic equation whose roots are \( s_1 \) and \( s_2 \) can be formed using Vieta's formulas:
\[ x^2 - (s_1 + s_2)x + s_1s_2 = 0 \]

### Step 7: Calculate \( s_1 + s_2 \) and \( s_1s_2 \)
Calculating \( s_1 + s_2 \):
\[ s_1 + s_2 = (-ab - r) + (-ac - r) = -ab - ac - 2r \]
Substituting \( r \):
\[ s_1 + s_2 = -ab - ac - 2\left(\frac{b - c}{ab - ac}\right) \]

Calculating \( s_1s_2 \):
Using the product of roots:
\[ s_1s_2 = (-ab - r)(-ac - r) \]

### Step 8: Establish the final quadratic equation
After calculating \( s_1 + s_2 \) and \( s_1s_2 \), we can write the final quadratic equation as:
\[ x^2 - (s_1 + s_2)x + s_1s_2 = 0 \]

### Conclusion
The other roots satisfy the equation:
\[ x^2 - a(b+c)x + a^2bc = 0 \]