The equation x 3 + ( 2 r + 1 ) x 2 + ( 4 r − 1 ) x + 2 = 0 x^3 + (2r + 1)x^2 + (4r - 1)x + 2 = 0 x 3 + ( 2 r + 1 ) x 2 + ( 4 r − 1 ) x + 2 = 0 has − 2...
Question
The equation
has
as one of the roots. If the other two roots are real, then the minimum possible non-negative integer value of
is: