Question Number 86541 by peter frank last updated on 29/Mar/20
Commented by jagoll last updated on 29/Mar/20
$$\mathrm{Q}^{\mathrm{2}} \:=\:\mathrm{PR} \\ $$$$\mathrm{4Q}\:=\:\mathrm{P}+\mathrm{R}\:\Rightarrow\:\mathrm{4ar}\:=\:\mathrm{a}+\mathrm{ar}^{\mathrm{2}} \\ $$$$\mathrm{a}\left(\mathrm{r}^{\mathrm{2}} −\mathrm{4r}+\mathrm{1}\right)\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\left(\mathrm{r}−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{r}\:=\:\mathrm{2}\:\pm\:\sqrt{\mathrm{3}} \\ $$