Prove that both the roots of the equation (x−a) (x−b)+(x−b) (x−c)+(x−c) (x−a)=0

Solution:

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

Here,

After simplifying the equation

$x^{2}-(a+b) x+a b+x^{2}-(b+c) x+b c+x^{2}-(c+a) x+c a=0$

$3 x^{2}-2(a+b+c) x+(a b+b c+c a)=0$

$a=3, b=2(a+b+c)$ and,$c=(a b+b c+c a)$

As we know that $D=b^{2}-4 a c$

Putting the value of $a=3, b=2(a+b+c)$ and, $c=(a b+b c+c a)$

$D=\{2(a+b+c)\}^{2}-4 \times 3 \times(a b+b c+c a)$

$=4\left(a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a\right)-12(a b+b c+c a)$

$=4\left(a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a-3 a b-3 b c-3 c a\right)$

$=4\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$

$D=4\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$

$=2\left[2 a^{2}+2 b^{2}+2 c^{2}-2 a b-2 a c-2 b c\right]$

$=2\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right]$

Since, $D>0$. So the solutions are real

Let $a=b=c$

Then

$D=4\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$

$=4\left(a^{2}+b^{2}+c^{2}-a a-b b-c c\right)$

$=4 \times 0$

Thus, the value of $D=0$

Therefore, the roots of the given equation are real and but they are equal only when,

Hence proved