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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

CAT · 2023 · Quant Slot 1
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

Options

2
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Answer: 2

Explanation:
Given that −2 is a root, divide the polynomial x^3 + (2r + 1)x^2 + (4r − 1)x + 2 by (x + 2). The factorization becomes: (x + 2)(x^2 + (2r − 1)x + 1) = 0 So the other two roots come from the quadratic: x^2 + (2r − 1)x + 1 = 0 For these roots to be real, the discriminant ≥ 0: Discriminant = (2r − 1)^2 − 411 = 4r^2 − 4r − 3 Condition: 4r^2 − 4r − 3 ≥ 0 Solve 4r^2 − 4r − 3 = 0: Roots are r = −1/2 and r = 3/2 Since the parabola opens upward, 4r^2 − 4r − 3 ≥ 0 for r ≤ −1/2 or r ≥ 3/2 We need the minimum non-negative integer r satisfying this. The smallest non-negative integer ≥ 3/2 is r = 2. Final Answer: 2

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