Find the nature of the roots of the quadratic equation `(a-b)x^2+(b-c)x+(c-a)=0` without really knowing the exact values of the roots.

Options

A. (A) Rational
B. (B) Irrational
C. (C) Non-real
D. (D) Both (a) and (b)

Answer

To determine the nature of the roots of the quadratic equation \((a-b)x^2 + (b-c)x + (c-a) = 0\), we can use the discriminant method. The discriminant (\(D\)) of a quadratic equation \(Ax^2 + Bx + C = 0\) is given by the formula:

\[ D = B^2 - 4AC \]

Here, we identify:
- \(A = a - b\)
- \(B = b - c\)
- \(C = c - a\)

Now, we can calculate the discriminant \(D\):

1. **Calculate \(B^2\)**:
\[ B^2 = (b - c)^2 \]

2. **Calculate \(4AC\)**:
\[ 4AC = 4(a - b)(c - a) \]

3. **Substitute into the discriminant formula**:
\[ D = (b - c)^2 - 4(a - b)(c - a) \]

4. **Analyze the discriminant**:
- If \(D > 0\), the equation has two distinct real roots.
- If \(D = 0\), the equation has exactly one real root (a repeated root).
- If \(D < 0\), the equation has two complex (imaginary) roots.

5. **Evaluate the nature of the roots**:
- We can substitute specific values for \(a\), \(b\), and \(c\) to check the nature of the roots. However, we can also analyze the expression \(D\) without specific values.
- If we find that \(D\) is non-negative, we can conclude that the roots are real. If \(D\) is negative, the roots are complex.

6. **Conclusion**:
- Since the equation is structured such that substituting \(x = 1\) gives a sum of zero, it indicates that \(x = 1\) is a root. This suggests that the other root must also be rational because roots of polynomials with rational coefficients occur in pairs of rational or irrational numbers.

Thus, we conclude that both roots of the equation \((a-b)x^2 + (b-c)x + (c-a) = 0\) are rational.