The roots of the equation a(x-b)(x-v)+b(x-c)(x-a)+c(x-b)(x-a)=0 (a,b,c are real distinct and real)are always A) positive B) negative C) real D)unreal

I believe expanding the products in the terms and computing the discriminant is an overkill on this problem. One of the answers does something along the lines of what I would have done on this problem. I will try to add a bit more rigor on that.Because a, b, and c are distinct, we can assume without loss of generality that a>b>c. Letf(x)=a(x−b)(x−c)+b(x−c)(x−a)+c(x−a)(x−b)Then, it is clear that f(a) has the same sign as a, f(b) is oppositely signed from b, and f(c) again has the same sign as c. If a and c bear the same sign, then it is clear that there will be two real roots, one between a and b and the other between b and c. If they are oppositely signed, then once again there will be two real roots and one of them will lie either between a and b, or b and c.Thus, all we can conclude here is that they will be real at all times.