Question

The roots of the equation
$(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0$ are real and equal if

• A. $a>b>c$
• B. $a=b=c$
• C. $a<b<c$
• D. $a+b+c=0$

Answer 1

Answer: (B)

$a=b=c$

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

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

$D=b^2-4ac=0$ as it has equal roots.

$\left(2\right(a+b+c){)}^2-4(3\left)\right(ab+bc+ca)=0.$

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

$\Rightarrow \frac{1}{2}\left[\right(a-b{)}^2+(b-c{)}^2+(c-a{)}^2]=0.$

$\Rightarrow a-b=0;b-c=0;c-a=0.$

$\Rightarrow a=b=c$