Questão 19 - Prova Modelo A - AFA 2026

Como os coeficientes do polinômio são reais, se um número complexo é raiz, seu conjugado também será:

${\mathrm{x}}_1 = \sqrt{2}\mathrm{cis} \left(\frac{\mathrm{\pi }}{4}\right); {\mathrm{x}}_2 = \sqrt{2}\mathrm{cis} \left(-\frac{\mathrm{\pi }}{4}\right); {\mathrm{x}}_3$

Pelas relações de Girard, tem-se:

$$\begin{gathered} {\mathrm{x}}_1 · {\mathrm{x}}_2 · {\mathrm{x}}_3 = -\frac{-4}{1} = 4 = \sqrt{2}\mathrm{cis} \left(\frac{\mathrm{\pi }}{4}\right) \\ \sqrt{2}\mathrm{cis} \left(-\frac{\mathrm{\pi }}{4}\right) · {\mathrm{x}}_3 = 2 · {\mathrm{x}}_3 \\ {\mathrm{x}}_3 = 2 \\ {\mathrm{x}}_1 + {\mathrm{x}}_2 + {\mathrm{x}}_3 = -\mathrm{a} = \sqrt{2}\mathrm{cis} \left(\frac{\mathrm{\pi }}{4}\right) + \sqrt{2}\mathrm{cis} \left(-\frac{\mathrm{\pi }}{4}\right) + 2 = \\ = \sqrt{2} \left(\frac{\sqrt{2}}{2} + \frac{\mathrm{i}\sqrt{2}}{2}\right) + \sqrt{2}\left(\frac{\sqrt{2}}{2}-\frac{\mathrm{i}\sqrt{2}}{2}\right) + 2 = 4 \\ -\mathrm{a} = 4 \\ \mathrm{a} = -4 \end{gathered}$$

$$\begin{gathered} {\mathrm{x}}_1 · {\mathrm{x}}_2 + {\mathrm{x}}_1 · {\mathrm{x}}_3 + {\mathrm{x}}_2 · {\mathrm{x}}_3= \sqrt{2} \left(\frac{\sqrt{2}}{2}+\frac{\mathrm{i}\sqrt{2}}{2}\right) · \sqrt{2} \left(\frac{\sqrt{2}}{2}-\frac{\mathrm{i}\sqrt{2}}{2}\right) + \left(\sqrt{2 } \left(\frac{\sqrt{2}}{2} + \frac{\mathrm{i}\sqrt{2}}{2}\right) · 2 + 2 · \sqrt{2} \left(\frac{\sqrt{2}}{2} - \frac{\mathrm{i}\sqrt{2}}{2}\right)\right) = \\ = \sqrt{2} \left(\frac{\sqrt{2}}{2} + \frac{\mathrm{i}\sqrt{2}}{2}\right) · \sqrt{2} \left(\frac{\sqrt{2}}{2} - \frac{\mathrm{i}\sqrt{2}}{2}\right) + 2 · \left(\sqrt{2}\left(\frac{\sqrt{2}}{2} + \frac{\mathrm{i}\sqrt{2}}{2}\right) + \sqrt{2} \left(\frac{\sqrt{2}}{2} - \frac{\mathrm{i}\sqrt{2}}{2}\right)\right) = 2 + 2 · 2 = 6 = \mathrm{b} \end{gathered}$$

Como a = –4 e b = 6, tem-se:

$\mathrm{a} – \mathrm{b} = –4 – 6 = \boxed{–10}$