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

x3+(2r+1)x2+(4r1)x+2=0x^3 + (2r + 1)x^2 + (4r - 1)x + 2 = 0
has
2-2
as one of the roots. If the other two roots are real, then the minimum possible non-negative integer value of
rr
is:

Answer Format

Enter your answer as an integer value

cat 2023cubic equationsdiscriminantreal roots

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