If every pair of equations `x^2+ax+bc=0, x^2+bx+ca=0` and `x^2+cx+ab=0` has a common root then their sum is

Options

A. the sum of the three common roots is `- (1//2)(a + b + c)`
B. the sum of the three common roots is `2(a + b + c)`
C. one of the values of the product of the three common
roots is abc
D. the product of the three common roots is ` a^(2) b^(2) c^(2)`

Explanation

To solve the problem, we need to analyze the three quadratic equations given and find the sum of their roots under the condition that each pair of equations has a common root.

### Step 1: Define the equations and common roots
Let the three equations be:
1. \( x^2 + ax + bc = 0 \) (let's denote its common root as \( \alpha \))
2. \( x^2 + bx + ca = 0 \) (let's denote its common root as \( \beta \))
3. \( x^2 + cx + ab = 0 \) (let's denote its common root as \( \gamma \))

### Step 2: Establish relationships using Vieta's formulas
From Vieta's formulas, we know:
- For the first equation: \( \alpha + \beta = -a \) (sum of roots)
- For the second equation: \( \beta + \gamma = -b \) (sum of roots)
- For the third equation: \( \gamma + \alpha = -c \) (sum of roots)

### Step 3: Add the equations
Adding all three equations:
\[ (\alpha + \beta) + (\beta + \gamma) + (\gamma + \alpha) = -a - b - c \]
This simplifies to:
\[ 2(\alpha + \beta + \gamma) = -a - b - c \]

### Step 4: Solve for the sum of the roots
Dividing both sides by 2, we find:
\[ \alpha + \beta + \gamma = -\frac{(a + b + c)}{2} \]

### Conclusion
Thus, the sum of the roots of the three equations is:
\[ \alpha + \beta + \gamma = -\frac{(a + b + c)}{2} \]

### Final Answer
The sum of the roots is \( -\frac{(a + b + c)}{2} \).

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