Question

if the roots of (b-c)x^2+(c-a)x+(a-b)=0 are equal, show that a, b and c are in AP

Answer 1

if roots are equal then D=b^2-4ac=0
hence
(c-a)^2-4 (a-b)(b-c)=0
(c-a)^2=4 (a-b)(b-c)

if we asume a, b and c is in AP
then a=x
b=x+d
c=x+2d
where x is first term and d is common difference .
now
LHS=(c-a)^2=(x+2d-x)^2=4d^2

RHS=4(a-b)(b-c)=4 (x-x-d)(x+d-x-2d)=4d^2
LHS=RHS
hence from above we look (c-a)^2=4(a-b)(b-c) possible only when a, b and c are in AP . hence a , b and c are in AP
when
(c-a)^2=4 (a-b)(b-c)