Question

If $a,b,cϵR$ then the roots of the equation $(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0$ are equal if and only if

• A. $a=0,b=1,c=1$
• B. $a=b=c$
• C. $a=1,b=0,c=1$
• D. $a\ne 1,b=1,c=0$

Answer 1

The correct option is B $a=b=c$

The given question is,

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

$\therefore x^2-(a+b)x+ab+x^2-(b+c)x+bc+x^2-(a+c)x+ac=0$

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

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

Compare this equation with standard form of quadratic equation ${ax}^2+bx+c=0$, we get,

$a=3$, $b=-2(a+b+c)$ and $c=ab+bc+ac$

Now, we know that roots of quadratic equation are same if discriminant is zero.

$\therefore b^2-4ac=0$

$\therefore {[-2(a+b+c)]}^2-4\left(3\right)(ab+bc+ac)=0$

$\therefore 4(a^2+b^2+c^2+2ab+2bc+2ac)-12(ab+bc+ac)=0$

$\therefore 4a^2+4b^2+4c^2+8ab+8bc+8ac-12ab-12bc-12ac=0$

$\therefore 4a^2+4b^2+4c^2-4ab-4bc-4ac=0$

$\therefore 2(2a^2+2b^2+2c^2-2ab-2bc-2ac)=0$

$\therefore 2a^2+2b^2+2c^2-2ab-2bc-2ac=0$

$\therefore a^2+a^2+b^2+b^2+c^2+c^2-2ab-2bc-2ac=0$

Rearranging above equation, we get

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

$\therefore [{(a-b)}^2+{(b-c)}^2+{(a-c)}^2]=0$

This condition is possible if and only if,

${(a-b)}^2=0$, ${(b-c)}^2=0$ and ${(a-c)}^2=0$

i.e. $a-b=0$, $b-c=0$ and $a-c=0$

$\therefore a=b,b=c,a=c$

$\therefore a=b=c$

Thus, Answer is option (B)