Question Number 208241 by Frix last updated on 08/Jun/24
$$\mathrm{Solve}\:\mathrm{for}\:{p},\:{q},\:{r} \\ $$$${p}+{q}+{r}=\alpha \\ $$$${p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} =\beta \\ $$$${pq}={r} \\ $$
Answered by mr W last updated on 08/Jun/24
$${p}^{\mathrm{2}} +{q}^{\mathrm{2}} +\left({pq}\right)^{\mathrm{2}} =\beta \\ $$$$\left({p}+{q}\right)^{\mathrm{2}} +\left({pq}\right)^{\mathrm{2}} −\mathrm{2}\left({pq}\right)=\beta \\ $$$$\left(\alpha−{pq}\right)^{\mathrm{2}} +\left({pq}\right)^{\mathrm{2}} −\mathrm{2}\left({pq}\right)=\beta \\ $$$$\mathrm{2}\left({pq}\right)^{\mathrm{2}} −\mathrm{2}\left(\alpha+\mathrm{1}\right)\left({pq}\right)+\alpha^{\mathrm{2}} −\beta=\mathrm{0} \\ $$$$\Rightarrow{pq}=\frac{\alpha+\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{2}\alpha−\alpha^{\mathrm{2}} +\mathrm{2}\beta}}{\mathrm{2}}={r} \\ $$$$\Rightarrow{p}+{q}=\alpha−{r} \\ $$$${p},\:{q}\:{are}\:{roots}\:{of} \\ $$$${x}^{\mathrm{2}} −\left(\alpha−{r}\right){x}+{r}=\mathrm{0} \\ $$$$\Rightarrow{p},\:{q}=\frac{\alpha−{r}\pm\sqrt{\left(\alpha−{r}\right)^{\mathrm{2}} −\mathrm{4}{r}}}{\mathrm{2}} \\ $$
Commented by Frix last updated on 08/Jun/24
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