If `a+b+c=0, a,b,c in Q` then roots of the equation `(b+c-a) x ^(2) + (c+a-c) =0` are:
Rational
Irrational
Imaginary
none of these .

Options

A. rational
B. irrational
C. imaginary
D. none of these

Answer

To solve the given quadratic equation and determine the nature of its roots, we will follow these steps:

### Step 1: Understand the given condition
We are given that \( a + b + c = 0 \) and \( a, b, c \in \mathbb{Q} \). This means that \( a, b, \) and \( c \) are rational numbers.

### Step 2: Rewrite the quadratic equation
The quadratic equation provided is:
\[ (b + c - a)x^2 + (c + a - c) = 0 \]
We can simplify the second term:
\[ c + a - c = a \]
Thus, the equation becomes:
\[ (b + c - a)x^2 + a = 0 \]

### Step 3: Substitute \( b + c \)
From the condition \( a + b + c = 0 \), we can express \( b + c \) as:
\[ b + c = -a \]
Now, substitute this into the equation:
\[ (-a - a)x^2 + a = 0 \]
This simplifies to:
\[ -2ax^2 + a = 0 \]

### Step 4: Factor out \( a \)
We can factor out \( a \) from the equation:
\[ a(-2x^2 + 1) = 0 \]
Since \( a \neq 0 \) (otherwise \( b + c \) would not be defined), we can divide both sides by \( a \):
\[ -2x^2 + 1 = 0 \]

### Step 5: Solve for \( x^2 \)
Rearranging gives:
\[ 2x^2 = 1 \]
\[ x^2 = \frac{1}{2} \]

### Step 6: Find the roots
Taking the square root of both sides, we find:
\[ x = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2} \]
Both roots \( \frac{\sqrt{2}}{2} \) and \( -\frac{\sqrt{2}}{2} \) are irrational numbers.

### Conclusion
The roots of the equation are irrational.

### Final Answer
The correct option is: **Irrational**