Both the roots of the equation (x − a)(x − b) + (x − b)(x − c) + (x − c)(x − a) = 0 are always:

Correct option is C

Given:

Equation: (x − a)(x − b) + (x − b)(x − c) + (x − c)(x − a) = 0

Solution:
Expand each term:

(x − a)(x − b) = x² − (a + b)x + ab
(x − b)(x − c) = x² − (b + c)x + bc
(x − c)(x − a) = x² − (c + a)x + ca

Adding all:
3x² − 2(a + b + c)x + (ab + bc + ca) = 0

This is a quadratic in x.

Discriminant,
D = [−2(a + b + c)]² − 4(3)(ab + bc + ca)
= 4[(a + b + c)² − 3(ab + bc + ca)]
= 4[a² + b² + c² − ab − bc − ca]
= 2[(a − b)² + (b − c)² + (c − a)²] ≥ 0

Hence D ≥ 0 for all real a, b, c.

So, both roots are always real.

Correct answer is (c) Real.