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Quadratic With Complex Roots

Mediummathematics

If one root of the real-coefficient quadratic x^2 + px + q = 0 is the complex number 2 + 3i, then the ordered pair (p, q) of real coefficients equals which value?

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About This Question

Subject
mathematics
Chapter
complex numbers and quadratic equations
Topic
quadratic with complex roots
Difficulty
Medium
Year
2025
Tags
advanced-calculus-drillconjugate-root-theoremreal-coefficientsvietas-formulascomplex-roots

Solution

Correct Answer:

Complex roots of a polynomial with real coefficients occur in conjugate pairs, a theorem that immediately determines the second root in JEE Advanced problems. If 2 + 3i is a root, then its conjugate 2 - 3i is also a root. The sum of the roots is (2 + 3i) + (2 - 3i) = 4, and the product is (2 + 3i)(2 - 3i) = 4 - (3i)^2 = 4 + 9 = 13. For x^2 + px + q = 0, the sum of roots is -p and the product is q. Hence -p = 4 giving p = -4, and q = 13. Option (4, 13) has the wrong sign of p. Option (-4, -13) wrongly takes the product as negative. Option (4, -5) miscomputes both. Hence (p, q) = (-4, 13). Plausibility check: the quadratic x^2 - 4x + 13 has discriminant 16 - 52 = -36 < 0, confirming non-real conjugate roots, and its roots (4 ± 6i)/2 = 2 ± 3i match the given root exactly. The conjugate-root theorem applies to any polynomial with real coefficients, so non-real roots always arrive in mirror-image pairs across the real axis. This pairing lets a student reconstruct missing roots and the polynomial's coefficients from a single complex root, and checking the discriminant afterwards provides a quick confirmation that the roots are indeed a genuine conjugate pair rather than two reals.

This medium difficulty mathematics question is from the chapter complex numbers and quadratic equations, covering the topic of quadratic with complex roots. It appeared in the 2025 exam.

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