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 $a=b=c$.
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
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