Question

If the roots of the equation
$\left(x-a\right)\left(x-b\right)+\left(x-b\right)\left(x-c\right)+\left(x-c\right)\left(x-a\right)=0$ 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

Answer: (D)

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,$D=0$
$\Rightarrow 4(a+b+c{)}^2-12(ab+bc+ca)=0$
$\Rightarrow (a+b+c{)}^2-3(ab+bc+ca)=0$
$\Rightarrow a^2+b^2+c^2+2ab+2bc$
$+2ca-3(ab+bc+ca)=0$
$\Rightarrow a^2+b^2+c^2=ab+bc+ca$