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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

${(x-a)(x-b)}+{(x-b)(x-c)}+{(x-c)(x-a)}=0$

$=>x^2-{(a+b)}x+ab+x^2-{(b+c)}x+bc+x^2-{(c+a)}x+ca=0$

$=>3x^2-2{(a+b+c)}x+(ab+bc+ca)=0$

discriminant, of the equation is

$4{(a+b+c)^2}-4.3{(ab+bc+ca)}=o$

$=>a^2+b^2+c^2+2(ab+bc+ca)-3(ab+bc+ca)=0$

$=>a^2+b^2+c^2-(ab+bc+ca)=0$

$=>a=b=c$

So, option (B) is correct ….