a+b+c=0, then roots of ax^2+bx+c=0 is

1.imagenary 2.real 3.coincidental 4.zero

• roots are real.
  
  x = [-b +/- sqrt(b^2 - 4ac)] / 2a ------------- [1]
  Because
  a+b+c = 0
  b = -(a+c)
  b^2 = (a+c)^2 = a^2+c^2 +2ac
  
  b^2 -4ac = a^2+c^2-2ac = (a-c)^2 --------- [2]
  
  solving [1] and [2], we get,
  x = [a+c +/- (a-c)] / 2a
  
  so
  x = 1,
  and
  x = c/a ....... both are real roots
• 13 years agoHelpfull: Yes(2) No(0)