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Question Number 86541 by peter frank last updated on 29/Mar/20

Commented by jagoll last updated on 29/Mar/20

$${\mathrm{Q}}^2=\mathrm{PR}$$$$4\mathrm{Q}=\mathrm{P}+\mathrm{R}\Rightarrow 4\mathrm{ar}=\mathrm{a}+{\mathrm{ar}}^2$$$$\mathrm{a}({\mathrm{r}}^2-4\mathrm{r}+1)=0$$$$\Rightarrow {(\mathrm{r}-2)}^2-3=0$$$$\mathrm{r}=2\pm \sqrt{3}$$

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