Question Number 118386 by mohammad17 last updated on 17/Oct/20
$${if}:{x}+\sqrt{{y}}={a}\:,\:\sqrt{{x}}+{y}={b}\:{find}\:{x},{y} \\ $$
Answered by benjo_mathlover last updated on 17/Oct/20
$${let}\:\sqrt{{y}}\:=\:{t}\:\wedge\:\sqrt{{x}}\:=\:{q} \\ $$$$\Rightarrow\begin{cases}{{q}^{\mathrm{2}} +{t}={a}\:…\left(×{b}\right)}\\{{q}+{t}^{\mathrm{2}} ={b}\:…\left(×{a}\right)}\end{cases}\: \\ $$$$\Rightarrow\begin{cases}{{bq}^{\mathrm{2}} +{bt}={ab}}\\{{aq}+{at}^{\mathrm{2}} ={ab}}\end{cases} \\ $$$$\Rightarrow{bq}^{\mathrm{2}} −{aq}+{bt}−{at}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{bq}^{\mathrm{2}} −{aq}\:−\left({at}^{\mathrm{2}} −{bt}\right)=\mathrm{0} \\ $$$$\Rightarrow{b}\left({q}^{\mathrm{2}} −\frac{{a}}{{b}}{q}\right)−{a}\left({t}^{\mathrm{2}} −\frac{{b}}{{a}}{t}\right)=\mathrm{0} \\ $$$$\Rightarrow{b}\left\{\left({q}−\frac{{a}}{\mathrm{2}{b}}\right)^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}{b}^{\mathrm{2}} }\right\}−{a}\left\{\left({t}−\frac{{b}}{\mathrm{2}{a}}\right)^{\mathrm{2}} −\frac{{b}^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} }\right\}\:=\:\mathrm{0} \\ $$$$ \\ $$