Show that the roots of the quadratic eqution: (b-c)x^2 + (c-a)x + (a-b)=0 are equal if c+a=2b

Show that the roots of the quadratic eqution: (b-c)x^2 + (c-a)x + (a-b

1.	Show that the roots of the quadratic eqution: (b-c)x^2 + (c-a)x + (a-b)=0 are equal if c+a=2b
Answer» b^2-4ac=0(c-a)^2-4(b-c)(a-b)=0c^2+a^2-2ac-4ab+4b^2+4ac-4bc=0a^2+4b^2+c^2-4ab-4bc+2ac=0a^2+(-2b)^2+c^2+2×a×(-2b)+2×(-2b)+2×(-2b)×c+2×a×c=0{a+(-2b)+c}^2=0a-2ab+c=02b=a+cHence,proved If roots are equalThen we have to prove that B^2-4AC=0=(c-a) ^2-4(b-c) (a-b) =c^2+a^2-2ac-4(ab-b^2-ac+bc)=a^2+c^2-2ac-4ab+4b^2+4ac-4bc=a^2+4b^2+c^2-4ab-4bc+2ac=(a-2b+c)^2. (eq1) But given that c+a=2bThen a-2b+c=0. (eq2)So, by (1) and (2)(0)^2=0So, B^2-4AC=0So., the equation has two equal roots Hi