Mathematics - Quadratic Equation Question with Solution | TestHub

From Vieta's formulas, $\alpha +\beta =-P$ and $\alpha \beta =Q$. We are given ${\alpha }^2+{\beta }^2=7$ and ${\alpha }^3+{\beta }^3=18$. We know ${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta =(-P{)}^2-2Q=P^2-2Q$. So, $P^2-2Q=7$ (Eq 1). Also, ${\alpha }^3+{\beta }^3=(\alpha +\beta )(({\alpha }^2+{\beta }^2)-\alpha \beta )=(-P)(7-Q)$. So, $-P(7-Q)=18$ (Eq 2). From (Eq 1), $Q=\frac{P^2-7}{2}$. Substitute $Q$ into (Eq 2): $-P(7-\frac{P^2-7}{2})=18⟹-P(\frac{14-P^2+7}{2})=18⟹-P(21-P^2)=36⟹P^3-21P-36=0$. Since $P$ is an integer, we test integer factors of 36. For $P=-3$, $(-3{)}^3-21(-3)-36=-27+63-36=0$. So $P=-3$ is a solution. Substitute $P=-3$ into $Q=\frac{P^2-7}{2}$: $Q=\frac{(-3{)}^2-7}{2}=\frac{9-7}{2}=1$. Both $P=-3$ and $Q=1$ are integers. Thus, $P+Q=-3+1=-2$.