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

\( x^3 + (2r+1)x^2 + (4r − 1)x + 2 = 0 \) 

Since \( -2 \) is one of the roots, the cubic equation can be factored as:
\((x+2)(x^2 + (2r - 1)x + 1) = 0\)

For the other two roots to be real, the quadratic \( x^2 + (2r - 1)x + 1 = 0 \) must have real roots.

So, the discriminant should be non-negative:
\( (2r - 1)^2 - 4 \geq 0 \)
\( (2r - 1)^2 \geq 4 \)

Taking square root on both sides:
\( |2r - 1| \geq 2 \)

So either \( 2r - 1 \geq 2 \Rightarrow r \geq \frac{3}{2} \) or \( 2r - 1 \leq -2 \Rightarrow r \leq -\frac{1}{2} \)

Therefore, the minimum possible non-negative integral value of \( r \) is 2.