Question

If the roots of the equation $(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0$ are equal, then $a^2+b^2+c^2$ is equal to

• A. $a+b+c$
• B. $2a+b+c$
• C. $3abc$
• D. $ab+bc+ca$
• E. $abc$

Answer 1

The correct option is D $ab+bc+ca$

Given equation is
$(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0$

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

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

Since, roots of the above equation are equal.

Then, discriminant, $D=0$

$\Rightarrow 4{(a+b+c)}^2-12(ab+bc+ca)=0$

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

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

$\Rightarrow a^2+b^2+c^2=ab+bc+ca$