Question

The roots of the equation (x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b)=0 are always

Answer 1

Answer:Real

explanation-

(x−b)(x−c)+(x−a)(x−c)+(x−a)(x−b)=0

⇒3x2−2(a+b+c)x+(ab+bc+ca)=0

Now, discriminant =4(a+b+c)2−12(ab+bc+ca)

=4(a2+b2+c2−ab−bc−ca)

=2{(a−b)2+(b−c)2+(c−a)2}

which is always positive.

Hence, both roots are real.