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Question Number 175801 by Lekhraj last updated on 07/Sep/22

solve the follwing equation x(√(x )) + y(√(y )) = 3 and x(√(y )) + y(√(x )) = 2 someone solve the above equations in the following way x^3 + y^3 + 2xy(√(xy )) = 9.....(1) and x^2 y + y^2 x + 2xy(√(xy )) = 4......(2) (1) − (2) ⇒ (x − y)(x^2 − y^2 ) = 5 hence x = 3 and y = 2 which is obiviusly does not satisfy the original equations. where is the fallacy in the above solution? Please explain.

$${solve}\:{the}\:{follwing}\:{equation} \\ $$$${x}\sqrt{{x}\:}\:\:\:+\:\:{y}\sqrt{{y}\:}\:\:=\:\:\mathrm{3}\:\:\:{and}\:\:\:{x}\sqrt{{y}\:}\:\:+\:{y}\sqrt{{x}\:}\:\:=\:\:\mathrm{2} \\ $$$${someone}\:{solve}\:{the}\:{above}\:{equations}\:{in}\:{the}\:{following}\:{way}\: \\ $$$${x}^{\mathrm{3}} +\:{y}^{\mathrm{3}} +\:\mathrm{2}{xy}\sqrt{{xy}\:}\:\:=\:\mathrm{9}.....\left(\mathrm{1}\right)\:\:\:\:{and}\:\:\:{x}^{\mathrm{2}} {y}\:\:+\:\:{y}^{\mathrm{2}} {x}\:\:+\:\mathrm{2}{xy}\sqrt{{xy}\:}\:\:=\:\:\mathrm{4}......\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)\:−\:\left(\mathrm{2}\right)\:\:\:\Rightarrow\:\:\left({x}\:−\:{y}\right)\left({x}^{\mathrm{2}} \:−\:{y}^{\mathrm{2}} \:\right)\:=\:\mathrm{5} \\ $$$${hence}\:\:{x}\:=\:\mathrm{3}\:\:{and}\:\:{y}\:=\:\mathrm{2}\:\:{which}\:{is}\:{obiviusly}\:{does}\:{not}\:\:{satisfy}\:{the} \\ $$$$\:{original}\:{equations}. \\ $$$${where}\:{is}\:{the}\:{fallacy}\:\:{in}\:{the}\:{above}\:{solution}?\:\:\mathrm{Please}\:\mathrm{explain}. \\ $$

Commented by mr W last updated on 07/Sep/22

(x − y)(x^2 − y^2 ) = 5 has infinitesolutions! not every solutionfrom it fulfills the original equations. x=3 and y=2 is only one possible solution of (x−y)(x^2 −y^2 )=5.

$$\left({x}\:−\:{y}\right)\left({x}^{\mathrm{2}} \:−\:{y}^{\mathrm{2}} \:\right)\:=\:\mathrm{5} \\ $$$${has}\:{infinitesolutions}!\:{not}\:{every} \\ $$$${solutionfrom}\:{it}\:{fulfills}\:{the}\:{original} \\ $$$${equations}.\:{x}=\mathrm{3}\:{and}\:{y}=\mathrm{2}\:{is}\:{only}\:{one} \\ $$$${possible}\:{solution}\:{of}\:\left({x}−{y}\right)\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \:\right)=\mathrm{5}. \\ $$

Commented by Frix last updated on 07/Sep/22

squaring equations leads to false solutions

$$\mathrm{squaring}\:\mathrm{equations}\:\mathrm{leads}\:\mathrm{to}\:\mathrm{false}\:\mathrm{solutions} \\ $$

Commented by Lekhraj last updated on 08/Sep/22

Thank you.

$$\mathrm{Thank}\:\mathrm{you}. \\ $$

Commented by Lekhraj last updated on 08/Sep/22

What about cubing equatins ?

$$\mathrm{What}\:\mathrm{about}\:\mathrm{cubing}\:\mathrm{equatins}\:? \\ $$

Commented by Frix last updated on 16/Sep/22

cubing is also dangerous for complex numbers

$$\mathrm{cubing}\:\mathrm{is}\:\mathrm{also}\:\mathrm{dangerous}\:\mathrm{for}\:\mathrm{complex}\:\mathrm{numbers} \\ $$

Answered by behi834171 last updated on 07/Sep/22

x=t^2 ,y=s^2 ⇒ { ((t^3 +s^3 =3)),((st(s+t)=2)) :}⇒ { (((t+s)(t^2 +s^2 −ts)=3)),((ts(t+s)=2)) :} ⇒_(ts=q) ^(t+s=p) { ((p(p^2 −3q)=3)),((pq=2)) :}⇒p^3 −3pq=3⇒ ⇒p^3 −6=3⇒p^3 =9⇒p=t+s=(9)^(1/3) q=(2/p)=t.s=(2/( (9)^(1/3) )) ⇒z^2 −(9)^(1/3) .z+(2/( (9)^(1/3) ))=0 ⇒z_(1,2) =t or s =(((9)^(1/3) ±(√(3(3)^(1/3) −(8/( (9)^(1/3) )))))/2)⇒ t or s=((2.1±0.7)/2)=1.4 or 0.7 ⇒x=t^2 =(1.4)^2 or (0.7)^2 y=s^2 =(0.7)^2 or (1.4)^2 .■

$$\boldsymbol{{x}}=\boldsymbol{{t}}^{\mathrm{2}} ,\boldsymbol{{y}}=\boldsymbol{{s}}^{\mathrm{2}} \\ $$$$\Rightarrow\begin{cases}{\boldsymbol{{t}}^{\mathrm{3}} +\boldsymbol{{s}}^{\mathrm{3}} =\mathrm{3}}\\{\boldsymbol{{st}}\left(\boldsymbol{{s}}+\boldsymbol{{t}}\right)=\mathrm{2}}\end{cases}\Rightarrow\begin{cases}{\left(\boldsymbol{{t}}+\boldsymbol{{s}}\right)\left(\boldsymbol{{t}}^{\mathrm{2}} +\boldsymbol{{s}}^{\mathrm{2}} −\boldsymbol{{ts}}\right)=\mathrm{3}}\\{\boldsymbol{{ts}}\left(\boldsymbol{{t}}+\boldsymbol{{s}}\right)=\mathrm{2}}\end{cases} \\ $$$$\underset{\boldsymbol{{ts}}=\boldsymbol{{q}}} {\overset{\boldsymbol{{t}}+\boldsymbol{{s}}=\boldsymbol{{p}}} {\Rightarrow}}\:\:\:\begin{cases}{\boldsymbol{{p}}\left(\boldsymbol{{p}}^{\mathrm{2}} −\mathrm{3}\boldsymbol{{q}}\right)=\mathrm{3}}\\{\boldsymbol{{pq}}=\mathrm{2}}\end{cases}\Rightarrow\boldsymbol{{p}}^{\mathrm{3}} −\mathrm{3}\boldsymbol{{pq}}=\mathrm{3}\Rightarrow \\ $$$$\Rightarrow\boldsymbol{{p}}^{\mathrm{3}} −\mathrm{6}=\mathrm{3}\Rightarrow\boldsymbol{{p}}^{\mathrm{3}} =\mathrm{9}\Rightarrow\boldsymbol{{p}}=\boldsymbol{{t}}+\boldsymbol{{s}}=\sqrt[{\mathrm{3}}]{\mathrm{9}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{q}}=\frac{\mathrm{2}}{\boldsymbol{{p}}}=\boldsymbol{{t}}.\boldsymbol{{s}}=\frac{\mathrm{2}}{\:\sqrt[{\mathrm{3}}]{\mathrm{9}}} \\ $$$$\Rightarrow\boldsymbol{{z}}^{\mathrm{2}} −\sqrt[{\mathrm{3}}]{\mathrm{9}}.\boldsymbol{{z}}+\frac{\mathrm{2}}{\:\sqrt[{\mathrm{3}}]{\mathrm{9}}}=\mathrm{0} \\ $$$$\Rightarrow\boldsymbol{{z}}_{\mathrm{1},\mathrm{2}} =\boldsymbol{{t}}\:\boldsymbol{{or}}\:\boldsymbol{{s}}\:=\frac{\sqrt[{\mathrm{3}}]{\mathrm{9}}\pm\sqrt{\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{3}}−\frac{\mathrm{8}}{\:\sqrt[{\mathrm{3}}]{\mathrm{9}}}}}{\mathrm{2}}\Rightarrow \\ $$$$\boldsymbol{{t}}\:\:\boldsymbol{{or}}\:\:\boldsymbol{{s}}=\frac{\mathrm{2}.\mathrm{1}\pm\mathrm{0}.\mathrm{7}}{\mathrm{2}}=\mathrm{1}.\mathrm{4}\:\:\:{or}\:\:\:\mathrm{0}.\mathrm{7} \\ $$$$\Rightarrow\boldsymbol{{x}}=\boldsymbol{{t}}^{\mathrm{2}} =\left(\mathrm{1}.\mathrm{4}\right)^{\mathrm{2}} \:\:\boldsymbol{{or}}\:\:\left(\mathrm{0}.\mathrm{7}\right)^{\mathrm{2}} \\ $$$$\boldsymbol{{y}}=\boldsymbol{{s}}^{\mathrm{2}} =\left(\mathrm{0}.\mathrm{7}\right)^{\mathrm{2}} \:\:\boldsymbol{{or}}\:\:\left(\mathrm{1}.\mathrm{4}\right)^{\mathrm{2}} \:\:\:\:\:.\blacksquare \\ $$

Answered by Frix last updated on 07/Sep/22

x^(3/2) +y^(3/2) =3 xy^(1/2) +x^(1/2) y=2 x=u^2 ∧y=v^2 ∧u≥0∧v≥0 u^3 +v^3 =3 u^2 v+uv^2 =2 (u^3 +v^3 )+3(u^2 v+uv^2 )=3+3×2 (u+v)^3 =9 v=9^(1/3) −u u^3 +(9^(1/3) −u)^3 =3 ⇒ u^2 −3^(2/3) u+(2/3^(2/3) )=0 u=(1/3^(1/3) )∨u=(2/3^(1/3) ) ⇒ x=(1/3^(2/3) )∧y=(4/3^(2/3) ) or the other way round

$${x}^{\mathrm{3}/\mathrm{2}} +{y}^{\mathrm{3}/\mathrm{2}} =\mathrm{3} \\ $$$${xy}^{\mathrm{1}/\mathrm{2}} +{x}^{\mathrm{1}/\mathrm{2}} {y}=\mathrm{2} \\ $$$${x}={u}^{\mathrm{2}} \wedge{y}={v}^{\mathrm{2}} \wedge{u}\geqslant\mathrm{0}\wedge{v}\geqslant\mathrm{0} \\ $$$${u}^{\mathrm{3}} +{v}^{\mathrm{3}} =\mathrm{3} \\ $$$${u}^{\mathrm{2}} {v}+{uv}^{\mathrm{2}} =\mathrm{2} \\ $$$$\left({u}^{\mathrm{3}} +{v}^{\mathrm{3}} \right)+\mathrm{3}\left({u}^{\mathrm{2}} {v}+{uv}^{\mathrm{2}} \right)=\mathrm{3}+\mathrm{3}×\mathrm{2} \\ $$$$\left({u}+{v}\right)^{\mathrm{3}} =\mathrm{9} \\ $$$${v}=\mathrm{9}^{\mathrm{1}/\mathrm{3}} −{u} \\ $$$${u}^{\mathrm{3}} +\left(\mathrm{9}^{\mathrm{1}/\mathrm{3}} −{u}\right)^{\mathrm{3}} =\mathrm{3} \\ $$$$\Rightarrow \\ $$$${u}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}/\mathrm{3}} {u}+\frac{\mathrm{2}}{\mathrm{3}^{\mathrm{2}/\mathrm{3}} }=\mathrm{0} \\ $$$${u}=\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{1}/\mathrm{3}} }\vee{u}=\frac{\mathrm{2}}{\mathrm{3}^{\mathrm{1}/\mathrm{3}} } \\ $$$$\Rightarrow \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}/\mathrm{3}} }\wedge{y}=\frac{\mathrm{4}}{\mathrm{3}^{\mathrm{2}/\mathrm{3}} }\:\mathrm{or}\:\mathrm{the}\:\mathrm{other}\:\mathrm{way}\:\mathrm{round} \\ $$

Commented by Lekhraj last updated on 08/Sep/22

Commented by Lekhraj last updated on 08/Sep/22

Commented by Lekhraj last updated on 08/Sep/22

Thank you very much .

$$\boldsymbol{\mathrm{Thank}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{very}}\:\boldsymbol{\mathrm{much}}\:. \\ $$

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