Question

Show that the roots of the quadratic equation given by(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0
are real and they can not be equal unless a=b=c

Answer 1

Expanding out the equation, we get

$x^2 - ax - bx+ab+x^2 -bx-cx+bc+x^2 -ax-cx+ac=0 \\ \\ 3x^2 -x(2a+2b+2c)+(ab+bc+ac)=0$

Considering the discriminant,

$(2a+2b+2c)^2 - 4(3)(ab+bc+ac) \\ \\ =4a^2 + 4b^2 + 4c^2 + 8(ab+bc+ac)-12(ab+bc+ac) \\ \\ =4(a^2 + b^2 + c^2 - ab - bc - ac) \\ \\ =2((a-b)^2 + (b-c)^2 + (c-a)^2)$

This is always non-negative, meaning the roots are real.

The roots are equal if and only if the discriminant is 0, which is when a=b=c.