Question

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

Answer 1

The correct answer is 2.

Answer 2

\( x^3 + (2r+1)x^2 + (4r − 1)x + 2 = 0 \)
Since -2 is one of the roots, the cubic equation can be factored as... 
\((x+2) (x^2 + (2x − 1)x + 1) = 0 \)
Since the other two roots are real, \((x^2 + (2x − 1)x + 1) =0\) has two real roots. 
That is the discriminant of \((x^2 + (2x − 1)x + 1) = 0\) is non-negative. 
\((2r - 1)^2 > 4 \)
\(r> 3/2\) or \(r\leq-\frac12 \)

Therefore the minimum possible non-negative integral value of r is 2.