both roots of the equation `(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0` are

Explanation

The given equation may be written as
`3x^(2)-2x(a+b+c)+(ab+bc+ca)=0.`
`:." "D=4(a+b+c)^(2)-12(ab+bc+ca)`
`=4[(a+b+c)^(2)-3(ab+bc+ca)]`
`=4(a^(2)+b^(2)+c^(2)-ab-bc-ca)`
`=2(2a^(2)+2b^(2)+2c^(2)-2ab-2bc-2ca)`
`=2[(a-b)^(2)+(b-c)^(2)+(c-a)^(2)]ge0`
`[because(a-b)^(2)ge0,(b-c)^(2)ge0" and "(c-a)^(2)ge0].`
This shows that both the roots of the given equation are real.
For equal roots, we must have D=0.
Now, `D=0implies(a-b)^(2)+(b-c)^(2)+(c-a)^(2)=0`
`implies" "(a-b)=0,(b-c)=0" and "(c-a)=0`
`implies" "a=b=c.`
Hence, the roots are equal only when a=b=c.