The equation 7x^7 + 14x^6 + 3x^4 + 12x^3 + 12x^2 + 5x + 7 = 0 has A. All complex roots B. At least one real root C. Four pairs of imaginary roots D. None of the above

GATE 1987 · Engineering Mathematics · Polynomials · medium

Answer: The equation has at least one real root.

1. Sign analysis at two points: p(0) = 7 > 0. For very large negative x the leading term 7x^7 dominates: 7*(-M)^7 = -7M^7 < 0, so p(x) < 0 for sufficiently large M. There is a sign change.
2. Conclude via IVT and conjugate-pair argument: By the Intermediate Value Theorem, since p is continuous and changes sign, it has at least one real root. Alternatively: complex roots of a real polynomial come in conjugate pairs, each pair costing 2 degrees. With degree 7 (odd), at least one root cannot be paired and must be real.