Quadratic With Complex Roots

Question

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?

RankGuru · Accepted 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.

Quadratic With Complex Roots

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