prove that 2b=a+c, if (b-c) x^2 + (c-a)x + (a-b) =0 has two real and equal roots

prove that 2b=a+c, if (b-c) x^2 + (c-a)x + (a-b) =0 has two real and e

1.	prove that 2b=a+c, if (b-c) x^2 + (c-a)x + (a-b) =0 has two real and equal roots
Answer» (b-c)x^2+(c-a)x+(a-b) =0 has real roots so D = 0B^2 - 4ac =0= (c-a)^2-4(b-c)(a-b)=0= C^2 +a^2-2ac -4* (ab-ac- b^2+bc) = 0 = C^2+a^2-2ac -4ab+4ac +4 b^2 -4 bc = 0 = C^2+a^2+2ac = 4ab - 4b^2 + 4 bc= (a+c)^ 2 = 4b( a - b + c)√[ a+ c] ^ 2 = √ 4b^2= a+ c = 2 b