CAT 2023 Slot 1 QA Question 7: he 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

Question

he 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

Accepted Answer

Substituting $x=-2$ gives 0 for every value of r, so $(x+2)$ is always a factor.

Dividing the cubic by $(x+2)$, we get
$x^3 + (2r+1)x^2 + (4r-1)x + 2 = (x+2)(x^2 + (2r-1)x + 1)$

The other two roots are real if the discriminant of the quadratic factor is non-negative.
$(2r-1)^2 - 4 \geq 0$
$(2r-1)^

he equation x3+(2r+1)x2+(4r−1)x+2=0x^3 + (2r+1)x^2 + (4r-1)x + 2 = 0x3+(2r+1)x2+(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

he 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