Question Number 79256 by ajfour last updated on 23/Jan/20
$$\mathrm{3}{s}^{\mathrm{2}} −\mathrm{2}{ps}−\mathrm{3}{cp}−\mathrm{1}=\mathrm{0}\:\:\:{and} \\ $$$$\mathrm{3}{s}−\mathrm{2}{p}−{sp}^{\mathrm{2}} −\mathrm{3}{cp}^{\mathrm{2}} =\mathrm{0} \\ $$$${find}\:{s}\:{and}\:{p}\:{both}\:{real}\:{in}\:{terms} \\ $$$${of}\:{c}\:\in\mathbb{R}. \\ $$
Answered by MJS last updated on 23/Jan/20
$$\left(\mathrm{1}\right)\:\:{c}=\frac{{s}^{\mathrm{2}} }{{p}}−\frac{\mathrm{2}{s}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}{p}} \\ $$$$\left(\mathrm{2}\right)\:\:{c}=\frac{{s}}{{p}^{\mathrm{2}} }−\frac{{s}}{\mathrm{3}}−\frac{\mathrm{2}}{\mathrm{3}{p}} \\ $$$$\frac{{s}^{\mathrm{2}} }{{p}}−\frac{\mathrm{2}{s}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}{p}}=\frac{{s}}{{p}^{\mathrm{2}} }−\frac{{s}}{\mathrm{3}}−\frac{\mathrm{2}}{\mathrm{3}{p}} \\ $$$$\mathrm{3}{ps}^{\mathrm{2}} −{p}^{\mathrm{2}} {s}−\mathrm{3}{s}+{p}=\mathrm{0} \\ $$$${s}=\frac{{p}}{\mathrm{3}}\vee{s}=\frac{\mathrm{1}}{{p}} \\ $$$${c}=−\frac{{p}^{\mathrm{2}} +\mathrm{3}}{\mathrm{9}{p}}\vee{c}=\frac{\mathrm{1}−{p}^{\mathrm{2}} }{{p}^{\mathrm{3}} } \\ $$$$\Rightarrow \\ $$$${p}^{\mathrm{2}} +\mathrm{9}{pc}+\mathrm{3}=\mathrm{0}\vee{p}^{\mathrm{3}} +\frac{\mathrm{1}}{{c}}{p}^{\mathrm{2}} −\frac{\mathrm{1}}{{c}}=\mathrm{0} \\ $$$$\mathrm{and}\:\mathrm{both}\:\mathrm{can}\:\mathrm{be}\:\mathrm{solved} \\ $$
Commented by mr W last updated on 24/Jan/20
$${more}\:{than}\:{fantastic}! \\ $$