Question Number 175462 by ajfour last updated on 30/Aug/22
$${p}^{\mathrm{3}} +{q}^{\mathrm{3}} +{p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{0} \\ $$$${find}\:{p}+{q}\:{in}\:{terms}\:{of}\:{c}.\:\:{if}\:\:{c}^{\mathrm{2}} <\frac{\mathrm{4}}{\mathrm{9}}. \\ $$
Commented by mr W last updated on 01/Sep/22
$${i}\:{think}\:{there}\:{is}\:{no}\:{unique}\:{value}\:{p}+{q}, \\ $$$${but}\:{there}\:{are}\:\left({p}+{q}\right)_{{min}} \:{and}\:\left({p}+{q}\right)_{{max}} . \\ $$
Commented by ajfour last updated on 01/Sep/22
$${values}\:{may}\:{be}\:{multiple}\:{but}\:{are} \\ $$$${constants},\:{i}\:{think}. \\ $$
Commented by mr W last updated on 01/Sep/22
$${for}\:{p},{q}\in{R} \\ $$$$−\frac{\mathrm{2}}{\:\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{{c}^{\mathrm{2}} }\left(\sqrt{\mathrm{1}+\frac{\mathrm{8}}{\mathrm{27}{c}^{\mathrm{2}} }}+\mathrm{1}\right)}−\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{{c}^{\mathrm{2}} }\left(\sqrt{\mathrm{1}+\frac{\mathrm{8}}{\mathrm{27}{c}^{\mathrm{2}} }}−\mathrm{1}\right)}}\leqslant{p}+{q}<−\frac{\mathrm{2}}{\mathrm{3}} \\ $$
Answered by mr W last updated on 04/Sep/22
$${let}\:{s}={p}+{q} \\ $$$${s}^{\mathrm{3}} −\mathrm{3}{spq}+{s}^{\mathrm{2}} −\mathrm{2}{pq}+{c}^{\mathrm{2}} =\mathrm{0} \\ $$$${s}^{\mathrm{3}} −\mathrm{3}{sp}\left({s}−{p}\right)+{s}^{\mathrm{2}} −\mathrm{2}{p}\left({s}−{p}\right)+{c}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left(\mathrm{3}{s}+\mathrm{2}\right){p}^{\mathrm{2}} −{s}\left(\mathrm{3}{s}+\mathrm{2}\right){p}+{s}^{\mathrm{3}} +{s}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Delta={s}^{\mathrm{2}} \left(\mathrm{3}{s}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{3}{s}+\mathrm{2}\right)\left({s}^{\mathrm{3}} +{s}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\geqslant\mathrm{0} \\ $$$$\left(\mathrm{3}{s}+\mathrm{2}\right)\left({s}^{\mathrm{3}} +\mathrm{2}{s}^{\mathrm{2}} +\mathrm{4}{c}^{\mathrm{2}} \right)\leqslant\mathrm{0} \\ $$$$\Rightarrow\mathrm{3}{s}+\mathrm{2}<\mathrm{0}\:\Rightarrow{s}<−\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\Rightarrow{s}^{\mathrm{3}} +\mathrm{2}{s}^{\mathrm{2}} +\mathrm{4}{c}^{\mathrm{2}} \geqslant\mathrm{0}\:\Rightarrow{s}\geqslant−\frac{\mathrm{2}}{\:\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{{c}^{\mathrm{2}} }\left(\sqrt{\mathrm{1}+\frac{\mathrm{8}}{\mathrm{27}{c}^{\mathrm{2}} }}+\mathrm{1}\right)}−\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{{c}^{\mathrm{2}} }\left(\sqrt{\mathrm{1}+\frac{\mathrm{8}}{\mathrm{27}{c}^{\mathrm{2}} }}−\mathrm{1}\right)}} \\ $$$${i}.{e}.\:−\frac{\mathrm{2}}{\:\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{{c}^{\mathrm{2}} }\left(\sqrt{\mathrm{1}+\frac{\mathrm{8}}{\mathrm{27}{c}^{\mathrm{2}} }}+\mathrm{1}\right)}−\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{{c}^{\mathrm{2}} }\left(\sqrt{\mathrm{1}+\frac{\mathrm{8}}{\mathrm{27}{c}^{\mathrm{2}} }}−\mathrm{1}\right)}}\leqslant{p}+{q}<−\frac{\mathrm{2}}{\mathrm{3}} \\ $$
Commented by Tawa11 last updated on 15/Sep/22
$$\mathrm{Great}\:\mathrm{sir}. \\ $$