Question

If the roots of the quadratic equation (a-b)x²+(b-c)x+(c-a)=0 are equal, prove that 2a=b+c.
I need quick answer

Answer 1

roots of the equation is given by -b +-√b²-4ac/2a
-(b-c)+√(b-c)²-4(c-a)(a-b)/2(a-b)=-(b-c)-√(b-c)²-4(c-a)(a-b)  both roots are equal
2√{(b-c)²-4(c-a)(a-b)}/2(a-b)=0
squaring both side 
(b-c)²-4(c-a)(a-b)=0
b²+c²-2bc=4(ac-bc-a²-ab)
b²+c²-2bc= 4ac-4bc-4a²-4ab
b²+c²+2bc=4ac-4ab-4a²
(b+c)²=4(ac-ab -a²)       ac=ab because both roots are equal
(b+c)²=-4a²
taking root on both side to make a²  +ve
sp √(b+c)²=√4a²
hence b+c=2a

Answer 2

plz refer to this attachment