Question Number 175801 by Lekhraj last updated on 07/Sep/22
$${solve}\:{the}\:{follwing}\:{equation} \\ $$$${x}\sqrt{{x}\:}\:\:\:+\:\:{y}\sqrt{{y}\:}\:\:=\:\:\mathrm{3}\:\:\:{and}\:\:\:{x}\sqrt{{y}\:}\:\:+\:{y}\sqrt{{x}\:}\:\:=\:\:\mathrm{2} \\ $$$${someone}\:{solve}\:{the}\:{above}\:{equations}\:{in}\:{the}\:{following}\:{way}\: \\ $$$${x}^{\mathrm{3}} +\:{y}^{\mathrm{3}} +\:\mathrm{2}{xy}\sqrt{{xy}\:}\:\:=\:\mathrm{9}…..\left(\mathrm{1}\right)\:\:\:\:{and}\:\:\:{x}^{\mathrm{2}} {y}\:\:+\:\:{y}^{\mathrm{2}} {x}\:\:+\:\mathrm{2}{xy}\sqrt{{xy}\:}\:\:=\:\:\mathrm{4}……\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)\:−\:\left(\mathrm{2}\right)\:\:\:\Rightarrow\:\:\left({x}\:−\:{y}\right)\left({x}^{\mathrm{2}} \:−\:{y}^{\mathrm{2}} \:\right)\:=\:\mathrm{5} \\ $$$${hence}\:\:{x}\:=\:\mathrm{3}\:\:{and}\:\:{y}\:=\:\mathrm{2}\:\:{which}\:{is}\:{obiviusly}\:{does}\:{not}\:\:{satisfy}\:{the} \\ $$$$\:{original}\:{equations}. \\ $$$${where}\:{is}\:{the}\:{fallacy}\:\:{in}\:{the}\:{above}\:{solution}?\:\:\mathrm{Please}\:\mathrm{explain}. \\ $$
Commented by mr W last updated on 07/Sep/22
$$\left({x}\:−\:{y}\right)\left({x}^{\mathrm{2}} \:−\:{y}^{\mathrm{2}} \:\right)\:=\:\mathrm{5} \\ $$$${has}\:{infinitesolutions}!\:{not}\:{every} \\ $$$${solutionfrom}\:{it}\:{fulfills}\:{the}\:{original} \\ $$$${equations}.\:{x}=\mathrm{3}\:{and}\:{y}=\mathrm{2}\:{is}\:{only}\:{one} \\ $$$${possible}\:{solution}\:{of}\:\left({x}−{y}\right)\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \:\right)=\mathrm{5}. \\ $$
Commented by Frix last updated on 07/Sep/22
$$\mathrm{squaring}\:\mathrm{equations}\:\mathrm{leads}\:\mathrm{to}\:\mathrm{false}\:\mathrm{solutions} \\ $$
Commented by Lekhraj last updated on 08/Sep/22
$$\mathrm{Thank}\:\mathrm{you}. \\ $$
Commented by Lekhraj last updated on 08/Sep/22
$$\mathrm{What}\:\mathrm{about}\:\mathrm{cubing}\:\mathrm{equatins}\:? \\ $$
Commented by Frix last updated on 16/Sep/22
$$\mathrm{cubing}\:\mathrm{is}\:\mathrm{also}\:\mathrm{dangerous}\:\mathrm{for}\:\mathrm{complex}\:\mathrm{numbers} \\ $$
Answered by behi834171 last updated on 07/Sep/22
$$\boldsymbol{{x}}=\boldsymbol{{t}}^{\mathrm{2}} ,\boldsymbol{{y}}=\boldsymbol{{s}}^{\mathrm{2}} \\ $$$$\Rightarrow\begin{cases}{\boldsymbol{{t}}^{\mathrm{3}} +\boldsymbol{{s}}^{\mathrm{3}} =\mathrm{3}}\\{\boldsymbol{{st}}\left(\boldsymbol{{s}}+\boldsymbol{{t}}\right)=\mathrm{2}}\end{cases}\Rightarrow\begin{cases}{\left(\boldsymbol{{t}}+\boldsymbol{{s}}\right)\left(\boldsymbol{{t}}^{\mathrm{2}} +\boldsymbol{{s}}^{\mathrm{2}} −\boldsymbol{{ts}}\right)=\mathrm{3}}\\{\boldsymbol{{ts}}\left(\boldsymbol{{t}}+\boldsymbol{{s}}\right)=\mathrm{2}}\end{cases} \\ $$$$\underset{\boldsymbol{{ts}}=\boldsymbol{{q}}} {\overset{\boldsymbol{{t}}+\boldsymbol{{s}}=\boldsymbol{{p}}} {\Rightarrow}}\:\:\:\begin{cases}{\boldsymbol{{p}}\left(\boldsymbol{{p}}^{\mathrm{2}} −\mathrm{3}\boldsymbol{{q}}\right)=\mathrm{3}}\\{\boldsymbol{{pq}}=\mathrm{2}}\end{cases}\Rightarrow\boldsymbol{{p}}^{\mathrm{3}} −\mathrm{3}\boldsymbol{{pq}}=\mathrm{3}\Rightarrow \\ $$$$\Rightarrow\boldsymbol{{p}}^{\mathrm{3}} −\mathrm{6}=\mathrm{3}\Rightarrow\boldsymbol{{p}}^{\mathrm{3}} =\mathrm{9}\Rightarrow\boldsymbol{{p}}=\boldsymbol{{t}}+\boldsymbol{{s}}=\sqrt[{\mathrm{3}}]{\mathrm{9}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{q}}=\frac{\mathrm{2}}{\boldsymbol{{p}}}=\boldsymbol{{t}}.\boldsymbol{{s}}=\frac{\mathrm{2}}{\:\sqrt[{\mathrm{3}}]{\mathrm{9}}} \\ $$$$\Rightarrow\boldsymbol{{z}}^{\mathrm{2}} −\sqrt[{\mathrm{3}}]{\mathrm{9}}.\boldsymbol{{z}}+\frac{\mathrm{2}}{\:\sqrt[{\mathrm{3}}]{\mathrm{9}}}=\mathrm{0} \\ $$$$\Rightarrow\boldsymbol{{z}}_{\mathrm{1},\mathrm{2}} =\boldsymbol{{t}}\:\boldsymbol{{or}}\:\boldsymbol{{s}}\:=\frac{\sqrt[{\mathrm{3}}]{\mathrm{9}}\pm\sqrt{\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{3}}−\frac{\mathrm{8}}{\:\sqrt[{\mathrm{3}}]{\mathrm{9}}}}}{\mathrm{2}}\Rightarrow \\ $$$$\boldsymbol{{t}}\:\:\boldsymbol{{or}}\:\:\boldsymbol{{s}}=\frac{\mathrm{2}.\mathrm{1}\pm\mathrm{0}.\mathrm{7}}{\mathrm{2}}=\mathrm{1}.\mathrm{4}\:\:\:{or}\:\:\:\mathrm{0}.\mathrm{7} \\ $$$$\Rightarrow\boldsymbol{{x}}=\boldsymbol{{t}}^{\mathrm{2}} =\left(\mathrm{1}.\mathrm{4}\right)^{\mathrm{2}} \:\:\boldsymbol{{or}}\:\:\left(\mathrm{0}.\mathrm{7}\right)^{\mathrm{2}} \\ $$$$\boldsymbol{{y}}=\boldsymbol{{s}}^{\mathrm{2}} =\left(\mathrm{0}.\mathrm{7}\right)^{\mathrm{2}} \:\:\boldsymbol{{or}}\:\:\left(\mathrm{1}.\mathrm{4}\right)^{\mathrm{2}} \:\:\:\:\:.\blacksquare \\ $$
Answered by Frix last updated on 07/Sep/22
$${x}^{\mathrm{3}/\mathrm{2}} +{y}^{\mathrm{3}/\mathrm{2}} =\mathrm{3} \\ $$$${xy}^{\mathrm{1}/\mathrm{2}} +{x}^{\mathrm{1}/\mathrm{2}} {y}=\mathrm{2} \\ $$$${x}={u}^{\mathrm{2}} \wedge{y}={v}^{\mathrm{2}} \wedge{u}\geqslant\mathrm{0}\wedge{v}\geqslant\mathrm{0} \\ $$$${u}^{\mathrm{3}} +{v}^{\mathrm{3}} =\mathrm{3} \\ $$$${u}^{\mathrm{2}} {v}+{uv}^{\mathrm{2}} =\mathrm{2} \\ $$$$\left({u}^{\mathrm{3}} +{v}^{\mathrm{3}} \right)+\mathrm{3}\left({u}^{\mathrm{2}} {v}+{uv}^{\mathrm{2}} \right)=\mathrm{3}+\mathrm{3}×\mathrm{2} \\ $$$$\left({u}+{v}\right)^{\mathrm{3}} =\mathrm{9} \\ $$$${v}=\mathrm{9}^{\mathrm{1}/\mathrm{3}} −{u} \\ $$$${u}^{\mathrm{3}} +\left(\mathrm{9}^{\mathrm{1}/\mathrm{3}} −{u}\right)^{\mathrm{3}} =\mathrm{3} \\ $$$$\Rightarrow \\ $$$${u}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}/\mathrm{3}} {u}+\frac{\mathrm{2}}{\mathrm{3}^{\mathrm{2}/\mathrm{3}} }=\mathrm{0} \\ $$$${u}=\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{1}/\mathrm{3}} }\vee{u}=\frac{\mathrm{2}}{\mathrm{3}^{\mathrm{1}/\mathrm{3}} } \\ $$$$\Rightarrow \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}/\mathrm{3}} }\wedge{y}=\frac{\mathrm{4}}{\mathrm{3}^{\mathrm{2}/\mathrm{3}} }\:\mathrm{or}\:\mathrm{the}\:\mathrm{other}\:\mathrm{way}\:\mathrm{round} \\ $$
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
$$\boldsymbol{\mathrm{Thank}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{very}}\:\boldsymbol{\mathrm{much}}\:. \\ $$