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

If a,b and c are real numbers then the roots of the equation (x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0 are always

Both the roots of the equation (x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b)=0 are always a.positive b.real c.negative d.none of these

Both the roots of the equation (x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b)=0 are always (1980,1M) positive (b) negative (c) real (d) none of these

Show that the roots of the equation: (x-a) (x -b) +(x- b) (x-c)+ (x- c) (x -a) =0 are always real and these cannot be equal unless a=b=c

If the roots of the equation (x-b)(x-c)+(x-c)(x-a)+(x-a)(x-b)=0 are equal then