Problem: If the roots of the equation ${(x-a)(x-b)}$+${(x-b)(x-c)}$+${(x-c)(x-a)}$=$0$, (where a,b,c are real numbers) are equal , then
(A) $b^2-4ac=0$
(B) $a=b=c$
(C) a+b+c=0
(D) none of foregoing statements is correct
Answer: $(B)$
${(x-a)(x-b)}$+${(x-b)(x-c)}$+${(x-c)(x-a)}$=$0$
=> $x^2-{(a+b)}x$+$ab+x^2-{(b+c)}x$+$bc+x^2-{(c+a)}x+ca$=$0$
=> $3x^2-2{(a+b+c)}x$+$(ab+bc+ca)$=$0$
discriminant, of the equation is
=> $4{(a+b+c)^2}$-$4.3{(ab+bc+ca)}$=$0$
=> $a^2+b^2+c^2+2(ab+bc+ca)$-$3(ab+bc+ca)$=$0$
=> $a^2+b^2+c^2$-$(ab+bc+ca)$=$0$
=> $a=b=c$
So, option (B) is correct.
