Question Number 207984 by Thomaseinstein last updated on 02/Jun/24
Answered by Frix last updated on 02/Jun/24
$${p},\:{q}\:>\mathrm{0} \\ $$$$\frac{\mathrm{1}}{{p}^{\mathrm{2}} }−\frac{\mathrm{1}}{{q}^{\mathrm{2}} }=\frac{\mathrm{4}}{\mathrm{3}}\:\Rightarrow\:{q}^{\mathrm{2}} =\frac{\mathrm{3}{p}^{\mathrm{2}} }{\:\mathrm{3}−\mathrm{4}{p}^{\mathrm{2}} } \\ $$$${p}+{q}=\mathrm{6}\:\Rightarrow\:{q}^{\mathrm{2}} =\left({p}−\mathrm{6}\right)^{\mathrm{2}} \\ $$$$\frac{\mathrm{3}{p}^{\mathrm{2}} }{\:\mathrm{3}−\mathrm{4}{p}^{\mathrm{2}} }=\left({p}−\mathrm{6}\right)^{\mathrm{2}} \\ $$$${p}^{\mathrm{4}} −\mathrm{12}{p}^{\mathrm{3}} +\mathrm{36}{p}^{\mathrm{2}} +\mathrm{9}{p}−\mathrm{27}=\mathrm{0} \\ $$$${p}\approx.\mathrm{854016106031} \\ $$$$\Rightarrow \\ $$$${q}\approx\mathrm{5}.\mathrm{14598389396} \\ $$$$\Rightarrow \\ $$$${m}\approx.\mathrm{729343509360} \\ $$$${n}\approx\mathrm{26}.\mathrm{4811502370} \\ $$