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

Question

Both the roots of the equation (x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b)=0 are always (A) positive (B) negative (C) real (D) None of these

Tardigrade Admin · Accepted Answer

(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=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.

Q. Both the roots of the equation (x−b)(x−c)+(x−a)(x−c)+(x−a)(x−b)=0 are always

positive

negative

real

None of these

(x−a)(x−b)+(x−b)(x−c)+(x−c)(x−a)=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.

Q. Both the roots of the equation $(x - b) (x - c) + (x - a) (x - c) + (x - a) (x - b) = 0$ are always