If a,b,c are real, then both the roots of the equation (x-b)(x-c)+(x-c)(x-a)+(x-a)(x-b)=0 are always (A) positive (B) negative (C) real (D) imaginary.

Options

A. positive
B. negative
C. real
D. none of these

Answer

C

Explanation

The given equation can be written as `3x^(2) - 2 (a + b + c) + ab + bc + ca = 0`
Let D be its discriminant. Then, `D = 4 (a + b + c)^(2) - 12 (ab + bc + ca)`
`rArr" "D = 2 {(b-c)^(2) + (c-a)^(2) + (a-b)^(2)} gt 0`
Hence, the roots are real.