Question

if 2a+3b+6c=0 then at least one root of the equation ax^2+bx+c=0 lies in the interval

Answer 1

     2 a + 3 b + 6 c = 0
           => b = - 2 (a+3c) / 3

a x² + b x + c = 0

   x  = [ -b +- √(b² - 4 a c) ] / 2a
      = [ 2 (a + 3c)/3  +- √[ 4/9 * (a+3c)² - 4 ac ] ] / 2 a
      = [  a + 3 c  +- √[ a² +9 c² + 6ac - 9ac ] ] / (3 a)
      = [  a + 3 c  +- √ [ a² + 9c² - 3ac ] ] / (3a)

EITHER,  (a-3c)² ≤ [a² +9c² - 3ac] ≤ (a+3c)²   OR,   (a+3c)² ≤ (a²+9c²-3ac) ≤ (a-3c)²

The boundaries are for one root are :
   [ a+3c + (a-3c) ] / (3a) = 2/3  and   [a+3c + (a+3c) ]/(3a) = 2(a+3c)/(3a) = -b/a

the boundaries or limits for the other root are: 
  [ a+3c - (a-3c) ] /(3a) = 2c/a   and  [a+3c - (a+3c) ]/(3a) = 2/3


So one root is between  2/3 and -b/a
the other is between  2/3  and  2c/a