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

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