Consider the equation `(a+c-b)x^2+2cx+(b+c-a)=0` where a,b,c are distinct real number. suppose that both the roots of the equation are rational then

Options

A. `f(x) = 0` has both roots as irrational numbers
B. `f(x) = 0` has both roots as rational numbers
C. `f(x) = 0` has imaginary roots
D. `f(x) = 0` has atleast one integer roots

Explanation

To solve the equation \((a+c-b)x^2 + 2cx + (b+c-a) = 0\) under the condition that both roots are rational, we can follow these steps:

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

### Step 2: Use the condition for rational roots
For a quadratic equation to have rational roots, the discriminant must be a perfect square. The discriminant \(\Delta\) is given by:
\[ \Delta = B^2 - 4AC \]
Substituting the values of \(A\), \(B\), and \(C\):
\[ \Delta = (2c)^2 - 4(a+c-b)(b+c-a) \]
\[ \Delta = 4c^2 - 4[(a+c-b)(b+c-a)] \]

### Step 3: Simplify the expression for the discriminant
Now, we simplify the expression:
\[ \Delta = 4c^2 - 4[(a+c-b)(b+c-a)] \]
Expanding the product:
\[ (a+c-b)(b+c-a) = ab + ac - a^2 - b^2 + bc - ab - ac + b^2 + c^2 - ac \]
This simplifies to:
\[ = c^2 - a^2 - b^2 + ab \]
Thus, we have:
\[ \Delta = 4c^2 - 4(c^2 - a^2 - b^2 + ab) \]
\[ = 4c^2 - 4c^2 + 4a^2 + 4b^2 - 4ab \]
\[ = 4(a^2 + b^2 - ab) \]

### Step 4: Set the discriminant as a perfect square
For the roots to be rational, we need:
\[ 4(a^2 + b^2 - ab) = k^2 \quad \text{for some integer } k \]
This implies:
\[ a^2 + b^2 - ab = \frac{k^2}{4} \]
This means that \(a^2 + b^2 - ab\) must be a perfect square.

### Step 5: Check for integer roots
Next, we can check if \(x = -1\) is a root of the equation:
Substituting \(x = -1\) into the equation:
\[ (a+c-b)(-1)^2 + 2c(-1) + (b+c-a) = 0 \]
This simplifies to:
\[ a + c - b - 2c + b + c - a = 0 \]
\[ 0 = 0 \]
This shows that \(x = -1\) is indeed a root.

### Conclusion
Since we have established that both roots are rational and that \(x = -1\) is an integer root, we conclude:
- The equation has both roots as rational numbers (Option B).
- The equation has at least one integer root (Option D).