Why ab-bc,bc-ca,ca-ab=0?????
Answers
Answered by
1
for example - the discriminant is zero .
this gives
(c-a)² - 4 (b-c)(a-b) = 0
EXPAND
(C²-2AC+A²) +4B² -4(C+A)B + 4AC = 0
4B² - 4²C+A)B + (C²+2AC+A²) = 0
(2B)² - 2(A+C)(2B) + (C+A)² = 0
A PERFECT SQUARE
(AB-(C+A))² = )
WITH ROOT
2B = C+A ,i.e.b = (c+a)/2
this gives
(c-a)² - 4 (b-c)(a-b) = 0
EXPAND
(C²-2AC+A²) +4B² -4(C+A)B + 4AC = 0
4B² - 4²C+A)B + (C²+2AC+A²) = 0
(2B)² - 2(A+C)(2B) + (C+A)² = 0
A PERFECT SQUARE
(AB-(C+A))² = )
WITH ROOT
2B = C+A ,i.e.b = (c+a)/2
Similar questions