Question Number 192440 by cortano12 last updated on 18/May/23
$$\:\:\:\:\mathrm{Given}\:\begin{cases}{\mathrm{A}=\frac{\left(\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left(\mathrm{pq}\right)^{\mathrm{2}} +\left(\mathrm{pr}\right)^{\mathrm{2}} +\left(\mathrm{qr}\right)^{\mathrm{2}} }}\\{\mathrm{B}=\frac{\mathrm{q}^{\mathrm{2}} −\mathrm{pr}}{\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} }\:}\end{cases}\:\:\:\:\:\: \\ $$$$\:\mathrm{If}\:\mathrm{p}+\mathrm{q}+\mathrm{r}=\mathrm{0}\:\mathrm{then}\:\mathrm{A}^{\mathrm{2}} −\mathrm{4B}=? \\ $$$$ \\ $$
Answered by Frix last updated on 18/May/23
$${r}=−\left({p}+{q}\right)\:\Rightarrow\:{A}=\mathrm{4}\wedge{B}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${A}^{\mathrm{2}} −\mathrm{4}{B}=\mathrm{14} \\ $$