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The equation of smallest degree with real coefficients having 2 + 3i as one of the roots isa)x2 − 4x + 13 = 0b)x2 + 5x + 6 = 0c)x2 − 2x + 1 = 0d)x2 + 2x + 1 = 0Correct answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The equation of smallest degree with real coefficients having 2 + 3i as one of the roots isa)x2 − 4x + 13 = 0b)x2 + 5x + 6 = 0c)x2 − 2x + 1 = 0d)x2 + 2x + 1 = 0Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The equation of smallest degree with real coefficients having 2 + 3i as one of the roots isa)x2 − 4x + 13 = 0b)x2 + 5x + 6 = 0c)x2 − 2x + 1 = 0d)x2 + 2x + 1 = 0Correct answer is option 'A'. Can you explain this answer?.

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