If the sum of any two roots of the equation $x^3+p x^2+q x+r=0$ is zero, then
- $r=p q$
- $p q^2=r$
- $r^2=p q$
- $p q r=1$
Solution

[From Eqs. (i) and (ii)] $ \gamma=-p $ Substituting this value in Eq. (iv) $ \alpha \beta(-p)=-r \Rightarrow \alpha \beta=\frac{r}{p} $ Substituting this value in Eq. (iii) $ \begin{aligned} \frac{r}{p}+r(\alpha+\beta) & =+q \\ \frac{r}{p} & =+q \\ r & =p q \end{aligned} \quad[\because \alpha+\beta=0] $
Asked in: AP EAMCET 2018 (23 Apr Shift 1)