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
$(x-a)(x-b)+(x-b)(x-c) + (x-c)(x-a)=0$
$\Rightarrow x^{2}-(a+b) x+a b+x^{2}-(b+c) x+b c + x^{2}-(c+a) x+c a=0$
$\Rightarrow 3 x^{2}-2(a+b+c) x+(a b+b c+c a)=0$
Since, roots of the above equation are equal. Then, discriminant,$D=0$
$\Rightarrow 4(a+b+c)^{2}-12(a b+b c+c a)=0$
$\Rightarrow (a+b+c)^{2}-3(a b+b c+c a)=0$
$\Rightarrow a^{2}+b^{2}+c^{2}+2 a b+2 b c$
$+2 c a-3(a b+b c+c a)=0$
$\Rightarrow a^{2}+b^{2}+c^{2}=a b+b c+c a$