Question Number 11356 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 21/Mar/17
Answered by mrW1 last updated on 21/Mar/17
$${p}=\sqrt{{x}} \\ $$$${q}=\sqrt{{y}} \\ $$$${p}^{\mathrm{2}} +{q}={a} \\ $$$${q}^{\mathrm{2}} +{p}={b} \\ $$$${q}={a}−{p}^{\mathrm{2}} \\ $$$$\left({a}−{p}^{\mathrm{2}} \right)^{\mathrm{2}} +{p}={b} \\ $$$${p}^{\mathrm{4}} −\mathrm{2}{ap}^{\mathrm{2}} +{p}+{a}^{\mathrm{2}} −{b}=\mathrm{0} \\ $$$${similarly} \\ $$$${q}^{\mathrm{4}} −\mathrm{2}{bq}^{\mathrm{2}} +{q}+{b}^{\mathrm{2}} −{a}=\mathrm{0} \\ $$$$ \\ $$$${working}... \\ $$