CAT 2023 Slot 1 QA Question & Solution
AlgebraHard
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
Solution
Given that -2 is a root of the given cubic equation.
=> Dividing the given equation by (x + 2), Using the Horners method of synthetic division:
coefficient of $x^2$ is 1, and coefficient of x is (2r+1)-2 = 2r-1 and the constant term = (4r-1)-2(2r-1) = 1.
=> The quadratic obtained by dividing the cubic = $x^2+\left(2r-1\right)x+1=0$, Since, this equation has 2 real roots => Discriminant should be greater than 0
=> $\left(2r-1\right)^2>4$ => 2r-1 > 2 or 2r-1 < -2 => r > 3/2 or r < -1/2.
=> Minimum possible non-negative integer value of r is 2.