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Question Number 61565 by naka3546 last updated on 04/Jun/19

x(y+1) + y(x−1) = 12 x(√y) + y(√x) = 21 x, y ∈ R x + y = ?

$${x}\left({y}+\mathrm{1}\right)\:+\:{y}\left({x}−\mathrm{1}\right)\:\:=\:\:\mathrm{12} \\ $$$${x}\sqrt{{y}}\:\:+\:\:{y}\sqrt{{x}}\:\:\:\:=\:\:\mathrm{21} \\ $$$${x},\:{y}\:\:\in\:\:\mathbb{R} \\ $$$${x}\:+\:{y}\:\:=\:\:? \\ $$

Commented by MJS last updated on 05/Jun/19

what′s the source of this question? it′s got no exact solution...

$$\mathrm{what}'\mathrm{s}\:\mathrm{the}\:\mathrm{source}\:\mathrm{of}\:\mathrm{this}\:\mathrm{question}? \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{got}\:\mathrm{no}\:\mathrm{exact}\:\mathrm{solution}... \\ $$

Commented by naka3546 last updated on 05/Jun/19

I don′t know, sir. ((12)/(xy)) = ((21)/(√(xy))) − 2 (√(xy)) x + y = ?

$${I}\:{don}'{t}\:{know},\:{sir}.\: \\ $$$$\frac{\mathrm{12}}{{xy}}\:\:=\:\:\frac{\mathrm{21}}{\sqrt{{xy}}}\:−\:\mathrm{2}\:\sqrt{{xy}}\:\:\: \\ $$$${x}\:+\:{y}\:\:=\:\:? \\ $$

Commented by MJS last updated on 05/Jun/19

this is only one equation in 2 variables but I don′t think you can get it from the first two... (√(xy))=t [t≥0] ((12)/t^2 )=((21)/t)−2t t^3 −((21)/2)t+6=0 t_0 =(√(14))sin ((1/3)arcsin ((6(√(14)))/(49))) ≈.591098 t_1 =(√(14))cos ((π/6)+(1/3)arcsin ((6(√(14)))/(49))) ≈2.90413 t_2 =−(√(14))sin ((π/3)+(1/3)arcsin ((6(√(14)))/(49))) ≈−3.49523 [not valid] ⇒ xy=.349397 ∨ xy=8.43398

$$\mathrm{this}\:\mathrm{is}\:\mathrm{only}\:\mathrm{one}\:\mathrm{equation}\:\mathrm{in}\:\mathrm{2}\:\mathrm{variables}\:\mathrm{but} \\ $$$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{you}\:\mathrm{can}\:\mathrm{get}\:\mathrm{it}\:\mathrm{from}\:\mathrm{the}\:\mathrm{first}\:\mathrm{two}... \\ $$$$\sqrt{{xy}}={t}\:\left[{t}\geqslant\mathrm{0}\right] \\ $$$$\frac{\mathrm{12}}{{t}^{\mathrm{2}} }=\frac{\mathrm{21}}{{t}}−\mathrm{2}{t} \\ $$$${t}^{\mathrm{3}} −\frac{\mathrm{21}}{\mathrm{2}}{t}+\mathrm{6}=\mathrm{0} \\ $$$${t}_{\mathrm{0}} =\sqrt{\mathrm{14}}\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\mathrm{6}\sqrt{\mathrm{14}}}{\mathrm{49}}\right)\:\approx.\mathrm{591098} \\ $$$${t}_{\mathrm{1}} =\sqrt{\mathrm{14}}\mathrm{cos}\:\left(\frac{\pi}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\mathrm{6}\sqrt{\mathrm{14}}}{\mathrm{49}}\right)\:\approx\mathrm{2}.\mathrm{90413} \\ $$$${t}_{\mathrm{2}} =−\sqrt{\mathrm{14}}\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\mathrm{6}\sqrt{\mathrm{14}}}{\mathrm{49}}\right)\:\approx−\mathrm{3}.\mathrm{49523}\:\left[\mathrm{not}\:\mathrm{valid}\right] \\ $$$$\Rightarrow\:{xy}=.\mathrm{349397}\:\vee\:{xy}=\mathrm{8}.\mathrm{43398} \\ $$

Answered by MJS last updated on 05/Jun/19

x>0∧y>0 (1) 2xy+x−y−12=0 ⇒ x=((y+12)/(2y+1)) let x=u^2 ∧y=v^2 ∧u>0∧v>0 (1) ⇒ u=((√(v^2 +12))/(√(2v^2 +1))) (2) ⇒ ((v^2 (√(v^2 +12)))/(√(2v^2 +1)))+((v(v^2 +12))/(2v^2 +1))=21 ((v(v^2 +12)+v^2 (√((v^2 +12)(2v^2 +1))))/(2v^2 +1))=21 v^2 (√((v^2 +12)(2v^2 +1)))=−v^3 +42v^2 −12v+21 v^4 (2v^4 +25v^2 +12)=(v^3 −42v^2 +12v−21)^2 ⇒ v^8 +12v^6 +42v^5 −888v^4 +525v^3 −954v^2 +252v−((441)/2)=0 this has got only one root >0 v=4.448447... ⇒ y=19.78864... ⇒ x=.7834092... x+y=20.57209...

$${x}>\mathrm{0}\wedge{y}>\mathrm{0} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{2}{xy}+{x}−{y}−\mathrm{12}=\mathrm{0}\:\Rightarrow\:{x}=\frac{{y}+\mathrm{12}}{\mathrm{2}{y}+\mathrm{1}} \\ $$$$\mathrm{let}\:{x}={u}^{\mathrm{2}} \wedge{y}={v}^{\mathrm{2}} \wedge{u}>\mathrm{0}\wedge{v}>\mathrm{0} \\ $$$$\left(\mathrm{1}\right)\:\Rightarrow\:{u}=\frac{\sqrt{{v}^{\mathrm{2}} +\mathrm{12}}}{\sqrt{\mathrm{2}{v}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\left(\mathrm{2}\right)\:\Rightarrow\:\frac{{v}^{\mathrm{2}} \sqrt{{v}^{\mathrm{2}} +\mathrm{12}}}{\sqrt{\mathrm{2}{v}^{\mathrm{2}} +\mathrm{1}}}+\frac{{v}\left({v}^{\mathrm{2}} +\mathrm{12}\right)}{\mathrm{2}{v}^{\mathrm{2}} +\mathrm{1}}=\mathrm{21} \\ $$$$\:\:\:\:\:\frac{{v}\left({v}^{\mathrm{2}} +\mathrm{12}\right)+{v}^{\mathrm{2}} \sqrt{\left({v}^{\mathrm{2}} +\mathrm{12}\right)\left(\mathrm{2}{v}^{\mathrm{2}} +\mathrm{1}\right)}}{\mathrm{2}{v}^{\mathrm{2}} +\mathrm{1}}=\mathrm{21} \\ $$$$\:\:\:\:\:{v}^{\mathrm{2}} \sqrt{\left({v}^{\mathrm{2}} +\mathrm{12}\right)\left(\mathrm{2}{v}^{\mathrm{2}} +\mathrm{1}\right)}=−{v}^{\mathrm{3}} +\mathrm{42}{v}^{\mathrm{2}} −\mathrm{12}{v}+\mathrm{21} \\ $$$$\:\:\:\:\:{v}^{\mathrm{4}} \left(\mathrm{2}{v}^{\mathrm{4}} +\mathrm{25}{v}^{\mathrm{2}} +\mathrm{12}\right)=\left({v}^{\mathrm{3}} −\mathrm{42}{v}^{\mathrm{2}} +\mathrm{12}{v}−\mathrm{21}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:{v}^{\mathrm{8}} +\mathrm{12}{v}^{\mathrm{6}} +\mathrm{42}{v}^{\mathrm{5}} −\mathrm{888}{v}^{\mathrm{4}} +\mathrm{525}{v}^{\mathrm{3}} −\mathrm{954}{v}^{\mathrm{2}} +\mathrm{252}{v}−\frac{\mathrm{441}}{\mathrm{2}}=\mathrm{0} \\ $$$$\mathrm{this}\:\mathrm{has}\:\mathrm{got}\:\mathrm{only}\:\mathrm{one}\:\mathrm{root}\:>\mathrm{0} \\ $$$${v}=\mathrm{4}.\mathrm{448447}... \\ $$$$\Rightarrow\:{y}=\mathrm{19}.\mathrm{78864}... \\ $$$$\Rightarrow\:{x}=.\mathrm{7834092}... \\ $$$${x}+{y}=\mathrm{20}.\mathrm{57209}... \\ $$

Answered by behi83417@gmail.com last updated on 05/Jun/19

x=p^2 ,y=q^2 ⇒ { ((p^2 −q^2 +2p^2 q^2 =12)),((pq(p+q)=21)) :} p+q=s,pq=t⇒ { ((s(√(s^2 −4t))+2t^2 =12)),((st=21)) :} s^2 (s^2 −4t)=144−48t^2 +4t^4 s^4 −84s=144−48×((441)/s^2 )+4×((194481)/s^4 ) s^8 −84s^5 −144s^4 +21168s^2 −777924=0 s=−4.631,5.334⇒t=−.4.535,3.937 ⇒ { ((p+q=−4.631)),((pq=−4.535)) :}⇒z^2 −4.631z−4.535=0 ⇒ { ((p=5.461⇒x=29.823)),((q=−0.83⇒y=0.689)) :}⇒x+y=30.512 ⇒ { ((p+q=5.334)),((pq=3.937⇒z^2 +5.334z+3.937=0)) :} ⇒ { ((p=−4.46⇒x=19.892)),((q=−0.884⇒y=0.778)) :}⇒x+y=20.67

$$\mathrm{x}=\mathrm{p}^{\mathrm{2}} ,\mathrm{y}=\mathrm{q}^{\mathrm{2}} \Rightarrow\begin{cases}{\mathrm{p}^{\mathrm{2}} −\mathrm{q}^{\mathrm{2}} +\mathrm{2p}^{\mathrm{2}} \mathrm{q}^{\mathrm{2}} =\mathrm{12}}\\{\mathrm{pq}\left(\mathrm{p}+\mathrm{q}\right)=\mathrm{21}}\end{cases} \\ $$$$\mathrm{p}+\mathrm{q}=\mathrm{s},\mathrm{pq}=\mathrm{t}\Rightarrow\begin{cases}{\mathrm{s}\sqrt{\mathrm{s}^{\mathrm{2}} −\mathrm{4t}}+\mathrm{2t}^{\mathrm{2}} =\mathrm{12}}\\{\mathrm{st}=\mathrm{21}}\end{cases} \\ $$$$\mathrm{s}^{\mathrm{2}} \left(\mathrm{s}^{\mathrm{2}} −\mathrm{4t}\right)=\mathrm{144}−\mathrm{48t}^{\mathrm{2}} +\mathrm{4t}^{\mathrm{4}} \\ $$$$\mathrm{s}^{\mathrm{4}} −\mathrm{84s}=\mathrm{144}−\mathrm{48}×\frac{\mathrm{441}}{\mathrm{s}^{\mathrm{2}} }+\mathrm{4}×\frac{\mathrm{194481}}{\mathrm{s}^{\mathrm{4}} } \\ $$$$\mathrm{s}^{\mathrm{8}} −\mathrm{84s}^{\mathrm{5}} −\mathrm{144s}^{\mathrm{4}} +\mathrm{21168s}^{\mathrm{2}} −\mathrm{777924}=\mathrm{0} \\ $$$$\mathrm{s}=−\mathrm{4}.\mathrm{631},\mathrm{5}.\mathrm{334}\Rightarrow\mathrm{t}=−.\mathrm{4}.\mathrm{535},\mathrm{3}.\mathrm{937} \\ $$$$\Rightarrow\begin{cases}{\mathrm{p}+\mathrm{q}=−\mathrm{4}.\mathrm{631}}\\{\mathrm{pq}=−\mathrm{4}.\mathrm{535}}\end{cases}\Rightarrow\mathrm{z}^{\mathrm{2}} −\mathrm{4}.\mathrm{631z}−\mathrm{4}.\mathrm{535}=\mathrm{0} \\ $$$$\Rightarrow\begin{cases}{\mathrm{p}=\mathrm{5}.\mathrm{461}\Rightarrow\mathrm{x}=\mathrm{29}.\mathrm{823}}\\{\mathrm{q}=−\mathrm{0}.\mathrm{83}\Rightarrow\mathrm{y}=\mathrm{0}.\mathrm{689}}\end{cases}\Rightarrow\mathrm{x}+\mathrm{y}=\mathrm{30}.\mathrm{512} \\ $$$$\Rightarrow\begin{cases}{\mathrm{p}+\mathrm{q}=\mathrm{5}.\mathrm{334}}\\{\mathrm{pq}=\mathrm{3}.\mathrm{937}\Rightarrow\mathrm{z}^{\mathrm{2}} +\mathrm{5}.\mathrm{334z}+\mathrm{3}.\mathrm{937}=\mathrm{0}}\end{cases} \\ $$$$\Rightarrow\begin{cases}{\mathrm{p}=−\mathrm{4}.\mathrm{46}\Rightarrow\mathrm{x}=\mathrm{19}.\mathrm{892}}\\{\mathrm{q}=−\mathrm{0}.\mathrm{884}\Rightarrow\mathrm{y}=\mathrm{0}.\mathrm{778}}\end{cases}\Rightarrow\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}=\mathrm{20}.\mathrm{67} \\ $$

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