CAIE FP1 2003 November — Question 6 9 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2003
SessionNovember
Marks9
PaperDownload PDF ↗
TopicRoots of polynomials
TypeSum of powers of roots
DifficultyChallenging +1.2 This is a Further Maths question requiring knowledge of symmetric functions of roots and modulus relationships. Part (a) uses standard formulas (sum of squares via Newton's identities: Σr² = (Σr)² - 2Σ(rs) = 0 - 2(1) = -2), then deduces complex roots from the negative result. Part (b) requires sign-checking f(-3) and f(-2) to locate α, then uses |complex roots|² = product/real root to bound the moduli. While systematic, it demands multiple techniques and careful reasoning beyond typical A-level, placing it moderately above average difficulty.
Spec4.02i Quadratic equations: with complex roots4.05a Roots and coefficients: symmetric functions

Find the sum of the squares of the roots of the equation $$x^3 + x + 12 = 0,$$ and deduce that only one of the roots is real. [4] The real root of the equation is denoted by \(\alpha\). Prove that \(-3 < \alpha < -2\), and hence prove that the modulus of each of the other roots lies between 2 and \(\sqrt{6}\). [5]

Find the sum of the squares of the roots of the equation
$$x^3 + x + 12 = 0,$$
and deduce that only one of the roots is real. [4]

The real root of the equation is denoted by $\alpha$. Prove that $-3 < \alpha < -2$, and hence prove that the modulus of each of the other roots lies between 2 and $\sqrt{6}$. [5]

\hfill \mbox{\textit{CAIE FP1 2003 Q6 [9]}}