Prove that both the roots of the equation $(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0$ are real but they are equal only when $a=b=c$.

Given:

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

To do:

We have to prove that both the roots of the equation $(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0$ are real but they are equal only when $a=b=c$.

Solution:

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

$x^2-ax-bx+ab+x^2-bx-cx+bc+x^2-cx-ax+ac=0$

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

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

Comparing the given quadratic equation with the standard form of the quadratic equation $ax^2+bx+c=0$, we get,

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

The discriminant of the standard form of the quadratic equation $ax^2+bx+c=0$ is $D=b^2-4ac$.

$D=[-2(a+b+c)]^2-4(3)(ab+bc+ca)$

$D=4(a^2+b^2+c^2+2ab+2bc+2ca)-12(ab+bc+ca)$

$D=4(a^2+b^2+c^2+2ab+2bc+2ca-3ab-3bc-3ca)$

$D=4(a^2+b^2+c^2-ab-bc-ca)$

$D=2(2a^2+2b^2+2c^2-2ab-2bc-2ca)$

$D=2(a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+c^2)$

$D=2[(a-b)^2+(b-c)^2+(c-a)^2]$

$D>0$ or $D=0$ when $a=b=c$

Therefore, the roots of the equation $(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0$ are real but they are equal only when $a=b=c$.