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Question Number 111909 by john santu last updated on 05/Sep/20
solve { ((x(√x)+y(√y) = 5)),((x(√y) +y(√x) = 1)) :}
$${solve}\:\begin{cases}{{x}\sqrt{{x}}+{y}\sqrt{{y}}\:=\:\mathrm{5}}\\{{x}\sqrt{{y}}\:+{y}\sqrt{{x}}\:=\:\mathrm{1}}\end{cases} \\ $$
Answered by bemath last updated on 05/Sep/20
let (√x) = u ; (√y) = v { ((u^3 +v^3 = 5⇒(u+v)^3 −3uv(u+v)=5)),((u^2 v+uv^2 = 1⇒uv(u+v)=1)) :} ⇔ (u+v)^3 −3.1 = 5 ⇒u+v = (8)^(1/(3 )) ⇒u+v = 2 ∧ u.v = (1/2) ⇒(√x) +(√(y )) = 2 ; (√(xy)) = (1/2)→xy =(1/4) ⇒((√x)+(√y))^2 = 4 ⇒ x+y+2(√(xy)) = 4 ⇒ x+y = 3→y=3−x now x(3−x) = (1/4) ; −x^2 +3x = (1/4) x^2 −3x+(1/4) = 0 ⇒x = ((3 ± (√(9−1)))/2) ⇒x = ((3±2(√2))/2) →y=3−(((3±2(√2) )/2)) y= ((3 ∓ 2(√2))/2) .
$${let}\:\sqrt{{x}}\:=\:{u}\:;\:\sqrt{{y}}\:=\:{v} \\ $$$$\begin{cases}{{u}^{\mathrm{3}} +{v}^{\mathrm{3}} \:=\:\mathrm{5}\Rightarrow\left({u}+{v}\right)^{\mathrm{3}} −\mathrm{3}{uv}\left({u}+{v}\right)=\mathrm{5}}\\{{u}^{\mathrm{2}} {v}+{uv}^{\mathrm{2}} \:=\:\mathrm{1}\Rightarrow{uv}\left({u}+{v}\right)=\mathrm{1}}\end{cases} \\ $$$$\Leftrightarrow\:\left({u}+{v}\right)^{\mathrm{3}} −\mathrm{3}.\mathrm{1}\:=\:\mathrm{5}\:\Rightarrow{u}+{v}\:=\:\sqrt[{\mathrm{3}\:}]{\mathrm{8}} \\ $$$$\Rightarrow{u}+{v}\:=\:\mathrm{2}\:\wedge\:{u}.{v}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\sqrt{{x}}\:+\sqrt{{y}\:}\:=\:\mathrm{2}\:;\:\sqrt{{xy}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\rightarrow{xy}\:=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow\left(\sqrt{{x}}+\sqrt{{y}}\right)^{\mathrm{2}} \:=\:\mathrm{4} \\ $$$$\Rightarrow\:{x}+{y}+\mathrm{2}\sqrt{{xy}}\:=\:\mathrm{4} \\ $$$$\Rightarrow\:{x}+{y}\:=\:\mathrm{3}\rightarrow{y}=\mathrm{3}−{x} \\ $$$${now}\:{x}\left(\mathrm{3}−{x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{4}}\:;\:−{x}^{\mathrm{2}} +\mathrm{3}{x}\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${x}^{\mathrm{2}} −\mathrm{3}{x}+\frac{\mathrm{1}}{\mathrm{4}}\:=\:\mathrm{0}\:\Rightarrow{x}\:=\:\frac{\mathrm{3}\:\pm\:\sqrt{\mathrm{9}−\mathrm{1}}}{\mathrm{2}} \\ $$$$\Rightarrow{x}\:=\:\frac{\mathrm{3}\pm\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}}\:\rightarrow{y}=\mathrm{3}−\left(\frac{\mathrm{3}\pm\mathrm{2}\sqrt{\mathrm{2}}\:}{\mathrm{2}}\right) \\ $$$${y}=\:\frac{\mathrm{3}\:\mp\:\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}}\:. \\ $$$$ \\ $$
Answered by ajfour last updated on 05/Sep/20
p^3 +q^3 =5 ( let (√x)=p , (√y)=q ) p^2 q+pq^2 =1 ⇒(p+q){(p+q)^2 −3pq}=5 & pq(p+q)=1 let pq=m , p+q=s s(s^2 −(3/s))=5 ⇒ s^3 =8 , s=p+q= 2, 2ω, 2ω^2 m=(1/s) = (1/2), (ω^2 /2) , (ω/2) x+y=p^2 +q^2 =s^2 −2m xy=p^2 q^2 =m^2 x, y = (s^2 /2)−m±(√(((s^2 /2)−m)^2 −m^2 )) (s^2 /2)−m=2−(1/2)=(3/2) , = 2ω^2 −(ω^2 /2)=((3ω^2 )/2) , = ((3ω)/2) . x, y = (3/2)±(√((9/4)−(1/4))) =(3/2)±(√2) , ((3/2)±(√2))ω , ((3/2)±(√2))ω^2
$${p}^{\mathrm{3}} +{q}^{\mathrm{3}} =\mathrm{5}\:\:\:\:\:\:\:\left(\:{let}\:\:\:\sqrt{{x}}={p}\:,\:\:\sqrt{{y}}={q}\:\:\right) \\ $$$${p}^{\mathrm{2}} {q}+{pq}^{\mathrm{2}} =\mathrm{1} \\ $$$$\Rightarrow\left({p}+{q}\right)\left\{\left({p}+{q}\right)^{\mathrm{2}} −\mathrm{3}{pq}\right\}=\mathrm{5} \\ $$$$\&\:\:\:\:\:{pq}\left({p}+{q}\right)=\mathrm{1} \\ $$$${let}\:\:\:{pq}={m}\:\:,\:\:{p}+{q}={s} \\ $$$${s}\left({s}^{\mathrm{2}} −\frac{\mathrm{3}}{{s}}\right)=\mathrm{5} \\ $$$$\Rightarrow\:\:\:\:{s}^{\mathrm{3}} =\mathrm{8}\:\:\:,\:\:{s}={p}+{q}=\:\mathrm{2},\:\mathrm{2}\omega,\:\mathrm{2}\omega^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:{m}=\frac{\mathrm{1}}{{s}}\:=\:\frac{\mathrm{1}}{\mathrm{2}},\:\frac{\omega^{\mathrm{2}} }{\mathrm{2}}\:,\:\frac{\omega}{\mathrm{2}} \\ $$$${x}+{y}={p}^{\mathrm{2}} +{q}^{\mathrm{2}} ={s}^{\mathrm{2}} −\mathrm{2}{m} \\ $$$${xy}={p}^{\mathrm{2}} {q}^{\mathrm{2}} ={m}^{\mathrm{2}} \\ $$$${x},\:{y}\:=\:\frac{{s}^{\mathrm{2}} }{\mathrm{2}}−{m}\pm\sqrt{\left(\frac{{s}^{\mathrm{2}} }{\mathrm{2}}−{m}\right)^{\mathrm{2}} −{m}^{\mathrm{2}} } \\ $$$$\frac{{s}^{\mathrm{2}} }{\mathrm{2}}−{m}=\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}}\:\:, \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}\omega^{\mathrm{2}} −\frac{\omega^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{3}\omega^{\mathrm{2}} }{\mathrm{2}}\:, \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{3}\omega}{\mathrm{2}}\:. \\ $$$${x},\:{y}\:=\:\frac{\mathrm{3}}{\mathrm{2}}\pm\sqrt{\frac{\mathrm{9}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}}}\: \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mathrm{3}}{\mathrm{2}}\pm\sqrt{\mathrm{2}}\:,\:\left(\frac{\mathrm{3}}{\mathrm{2}}\pm\sqrt{\mathrm{2}}\right)\omega\:,\:\:\:\left(\frac{\mathrm{3}}{\mathrm{2}}\pm\sqrt{\mathrm{2}}\right)\omega^{\mathrm{2}} \\ $$$$ \\ $$

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