Q. Both the roots of the given equation $(x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) = 0$ are always
A) Positive
B) Negative
C) Real
D) Imaginary
Explanation
(c)
Given, $3x^{2} - 2(a + b + c)x + (ab + bc + ca) = 0$
Now, $B^{2} - 4AC = 4\{(a + b + c)^{2} - 3(ab + bc + ca)\}$
$= 4\{ a^{2} + b^{2} + c^{2} - ab - bc - ac\}$
$= 2\left\{ (a - b)^{2} + (b - c)^{2} + (c - a)^{2} \right\} \geq 0$
Hence, both roots are always real
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.