Question

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

Answer 1

Answer:

Step-by-step explanation:

(x-a)(x-b) + (x-b)(x-c)+ (x-c)(x-a)= 0

x^2 - ax -bx +ab + x^2 - bx -c x +bc +x^2 - ax -cx +ac =0

3x^2 -2ax-2bx-2cx + ab+bc+ac =0

3x^2 - (2a+2b+2c)x +  ab+bc+ac =0

 

   b^2-4ac =0

       b^2=4ac

   (-2a-2b-2c)^2 = 12(ab+bc+ac)

     4a^2+4b^2+4c^2+8ab+8bc+8ac = 12ab +12bc+12ac

        4a^2+4b^2+4c^2= 4ab+4bc+4ac

             a^2+b^2+c^2 = ab+bc+ac

                a^2+b^2+c^2-ab-bc-ac =0

                a(a-b) +b(b-c)+c( c-a) = 0

                        a=b=c