For `a, b, c in Q` and `b + c ne a`, the roots of `ax^(2)-(a+b+c)x+(b+c)=0`

Options

A. Rational and unequal
B. rational and equal
C. complex numbers
D. Canot be determined

Answer

To solve the quadratic equation \( ax^2 - (a + b + c)x + (b + c) = 0 \) where \( a, b, c \in \mathbb{Q} \) and \( b + c \neq a \), we will follow these steps:

### Step 1: Identify the coefficients
The given quadratic equation can be expressed in the standard form \( Ax^2 + Bx + C = 0 \) where:
- \( A = a \)
- \( B = -(a + b + c) \)
- \( C = b + c \)

### Step 2: Calculate the discriminant
The discriminant \( D \) of a quadratic equation \( Ax^2 + Bx + C = 0 \) is given by the formula:
\[ D = B^2 - 4AC \]
Substituting the values of \( A \), \( B \), and \( C \):
\[ D = [-(a + b + c)]^2 - 4(a)(b + c) \]
This simplifies to:
\[ D = (a + b + c)^2 - 4a(b + c) \]

### Step 3: Expand the discriminant
Now, we will expand \( D \):
\[ D = (a^2 + 2ab + 2ac + b^2 + c^2 + 2bc) - (4ab + 4ac) \]
Combining like terms:
\[ D = a^2 + b^2 + c^2 + 2bc - 2ab - 2ac \]

### Step 4: Factor the discriminant
Notice that the expression can be rewritten as:
\[ D = (a - b - c)^2 \]

### Step 5: Analyze the discriminant
Given that \( b + c \neq a \), it follows that:
\[ a - b - c \neq 0 \]
This implies that \( D \) is positive since it is a square of a non-zero number.

### Step 6: Conclusion about the roots
Since the discriminant \( D \) is positive, the roots of the quadratic equation are real and unequal.

### Final Answer
The roots of the equation \( ax^2 - (a + b + c)x + (b + c) = 0 \) are rational and unequal.

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