Question Number 146936 by mathdanisur last updated on 16/Jul/21
$$\begin{cases}{{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} +{xy}=\mathrm{37}}\\{{y}^{\mathrm{2}} +\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{xy}=\mathrm{26}}\end{cases}\:\Rightarrow\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =? \\ $$
Answered by TheHoneyCat last updated on 16/Jul/21
$$\mathrm{let}\:\mathrm{S}=\begin{cases}{{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} +{xy}=\mathrm{37}}\\{{y}^{\mathrm{2}} +\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{xy}=\mathrm{26}}\end{cases} \\ $$$$\Leftrightarrow\begin{cases}{{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} +{xy}=\mathrm{37}}\\{−\mathrm{3}{y}^{\mathrm{2}} =\mathrm{26}−\mathrm{2}×\mathrm{37}}\end{cases} \\ $$$$\Leftrightarrow\begin{cases}{{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} +{xy}=\mathrm{37}}\\{{y}^{\mathrm{2}} =\mathrm{16}}\end{cases} \\ $$$$\Leftrightarrow\begin{cases}{{x}^{\mathrm{2}} +\mathrm{32}+{xy}=\mathrm{37}}\\{{y}^{\mathrm{2}} =\mathrm{16}}\end{cases} \\ $$$$\Leftrightarrow\begin{cases}{{x}^{\mathrm{2}} +{xy}−\mathrm{5}=\mathrm{0}}\\{{y}^{\mathrm{2}} =\mathrm{16}}\end{cases} \\ $$$$\mathrm{since}\:{y}^{\mathrm{2}} +\mathrm{4}×\mathrm{5}=\mathrm{36}>\mathrm{0}\:{x}\in\left\{\frac{−{y}\pm\sqrt{{y}^{\mathrm{2}} −\mathrm{4}×\mathrm{5}}}{\mathrm{2}}\right\} \\ $$$$\mathrm{so}: \\ $$$$\mathrm{S}\Leftrightarrow\begin{cases}{{x}\in\left\{−{y}/\mathrm{2}\pm\mathrm{3}\right\}}\\{{y}^{\mathrm{2}} =\mathrm{16}}\end{cases} \\ $$$$\mathrm{S}\Leftrightarrow\begin{cases}{{x}\in\left\{−\mathrm{2}\pm\mathrm{3},\mathrm{2}\pm\mathrm{3}\right\}}\\{{y}^{\mathrm{2}} =\mathrm{16}}\end{cases} \\ $$$$\mathrm{S}\Rightarrow\begin{cases}{{x}^{\mathrm{2}} \in\left\{\mathrm{2}^{\mathrm{2}} \pm\mathrm{2}×\mathrm{2}×\mathrm{3}+\mathrm{3}^{\mathrm{2}} \right\}}\\{{y}^{\mathrm{2}} =\mathrm{16}}\end{cases}{I}\:{just}\:{developped}\:\left({a}\pm{b}\right)^{\mathrm{2}} \\ $$$$\mathrm{thus}\:\mathrm{S}\Rightarrow\begin{cases}{{x}^{\mathrm{2}} \in\left\{\mathrm{13}\pm\mathrm{12}\right\}}\\{{y}^{\mathrm{2}} =\mathrm{16}}\end{cases} \\ $$$$\mathrm{so}\:\mathrm{S}\Rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \in\left\{\mathrm{17},\mathrm{41}\right\}_{\blacksquare} \\ $$$$ \\ $$$$ \\ $$$${By}\:{the}\:{way}\:\left({this}\:{has}\:{nothing}\:{to}\:{do}\:{with}\:{the}\:{question}\right) \\ $$$${those}\:{are}\:{two}\:{elipses}\:{so}\:{the}\:{number}\:{of}\:{solutions}\:{for}\:\left({x},{y}\right)\:{less}\:\left({or}\:{equal}\right)\:{to}\:\mathrm{4} \\ $$$${the}\:{two}\:{elispes}\:{beeing}\:{centered}\:{on}\:\left(\mathrm{0},\mathrm{0}\right)\:{the}\:{set}\:{of}\:{solutions}\:{is}\:{symetrical} \\ $$$$\mathrm{that}\:\mathrm{is}\:\mathrm{to}\:\mathrm{say}\:\mathrm{that}\:\mathrm{if}\:\left({x},{y}\right)\:\mathrm{is}\:\mathrm{solution}\:\left({x},{y}\right)\:\mathrm{is}\:\mathrm{too}. \\ $$$${you}\:{can}\:{then}\:{randomly}\:{try}\:{some}\:{values}\:{and}\:{notice}\:{that}\:: \\ $$$$\left({x},{y}\right)=\left(\mathrm{1},\mathrm{4}\right)\:{and}\:\left({x},{y}\right)=\left(\mathrm{5},−\mathrm{4}\right)\:{are}\:{solutions} \\ $$$${so}\:{you}\:{have}\:{directly}\:{what}\:{you}\:{wanted} \\ $$$$ \\ $$$${this}\:{method}\:{is}\:{not}\:{as}\:{great}\:{as}\:{what}\:{I}\:{first}\:{did}. \\ $$$${because}\:{is}\:{relies}\:{on}\:{you}\:{making}\:{a}\:{lucky}\:{guess} \\ $$$${however}\:{if}\:{you}\:{have}\:{access}\:{to}\:{geogebra}\:\left({or}\:{some}\:{equivalent}\right)\:{this}\:{makes}\:{the}\:{problem}\:{trivial} \\ $$$$ \\ $$$${plus},\:{you}\:{can}\:{mix}\:{the}\:{methods}. \\ $$$$\left({for}\:{instance},\:{once}\:{I}\:{have}\:{reached}\:{y}^{\mathrm{2}} =\mathrm{16}\:{with}\:{the}\:{first}\:{method},\:{you}\:{might}\:{want}\:{to}\:{check}\:{for}\:{obvious}\:{values}\:{of}\:{x}\right) \\ $$$$ \\ $$$$ \\ $$$${I}\:{hope}\:{you}\:{find}\:{that}\:{note}\:{interesting} \\ $$$$\left.:−\right) \\ $$
Commented by mathdanisur last updated on 16/Jul/21
$${thank}\:{you}\:{Ser} \\ $$
Answered by mr W last updated on 16/Jul/21
$${let}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\Rightarrow{x}={r}\:\mathrm{cos}\:\theta,\:{y}={r}\:\mathrm{sin}\:\theta \\ $$$${r}^{\mathrm{2}} +{r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta+{r}^{\mathrm{2}} \mathrm{cos}\:\theta\:\mathrm{sin}\:\theta=\mathrm{37}\:\:\:…\left({i}\right) \\ $$$${r}^{\mathrm{2}} +{r}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta+\mathrm{2}{r}^{\mathrm{2}} \mathrm{cos}\:\theta\:\mathrm{sin}\:\theta=\mathrm{26}\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)+\left({ii}\right): \\ $$$$\mathrm{3}{r}^{\mathrm{2}} +\mathrm{3}{r}^{\mathrm{2}} \mathrm{cos}\:\theta\:\mathrm{sin}\:\theta=\mathrm{63} \\ $$$${r}^{\mathrm{2}} +{r}^{\mathrm{2}} \:\frac{\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{2}}=\mathrm{21}\:\:\:…\left({iii}\right) \\ $$$$\Rightarrow\mathrm{sin}\:\mathrm{2}\theta=\frac{\mathrm{42}}{{r}^{\mathrm{2}} }−\mathrm{2} \\ $$$$\left({ii}\right)−\left({i}\right): \\ $$$${r}^{\mathrm{2}} \mathrm{cos}\:\mathrm{2}\theta+{r}^{\mathrm{2}} \frac{\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{2}}=−\mathrm{11}\:\:\:…\left({iv}\right) \\ $$$$\left({iv}\right)−\left({iii}\right): \\ $$$${r}^{\mathrm{2}} \mathrm{cos}\:\mathrm{2}\theta−{r}^{\mathrm{2}} =−\mathrm{32} \\ $$$$\Rightarrow\mathrm{cos}\:\mathrm{2}\theta=\mathrm{1}−\frac{\mathrm{32}}{{r}^{\mathrm{2}} } \\ $$$$\left(\frac{\mathrm{42}}{{r}^{\mathrm{2}} }−\mathrm{2}\right)^{\mathrm{2}} +\left(\mathrm{1}−\frac{\mathrm{32}}{{r}^{\mathrm{2}} }\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\left(\mathrm{42}−\mathrm{2}{r}^{\mathrm{2}} \right)^{\mathrm{2}} +\left({r}^{\mathrm{2}} −\mathrm{32}\right)^{\mathrm{2}} ={r}^{\mathrm{4}} \\ $$$${r}^{\mathrm{4}} −\mathrm{58}{r}^{\mathrm{2}} +\mathrm{697}=\mathrm{0} \\ $$$$\Rightarrow{r}^{\mathrm{2}} =\frac{\mathrm{58}\pm\sqrt{\mathrm{58}^{\mathrm{2}} −\mathrm{4}×\mathrm{697}}}{\mathrm{2}}=\frac{\mathrm{58}\pm\mathrm{24}}{\mathrm{2}}=\mathrm{41},\:\mathrm{17} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{17}\:{or}\:\mathrm{41} \\ $$
Commented by behi834171 last updated on 16/Jul/21
$${very}\:{nice}\:{dear}\:{master}! \\ $$
Commented by mathdanisur last updated on 16/Jul/21
$${thank}\:{you}\:{Ser}\:{cool} \\ $$
Commented by Tawa11 last updated on 23/Jul/21
$$\mathrm{great} \\ $$
Answered by behi834171 last updated on 17/Jul/21
$$\begin{cases}{\mathrm{3}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)+\mathrm{3}{xy}=\mathrm{63}}\\{{y}^{\mathrm{2}} −{x}^{\mathrm{2}} −{xy}=\mathrm{11}}\end{cases}\Rightarrow\mathrm{3}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)+\mathrm{3}\left({y}^{\mathrm{2}} −{x}^{\mathrm{2}} −\mathrm{11}\right)=\mathrm{63}\Rightarrow \\ $$$$\mathrm{6}{y}^{\mathrm{2}} =\mathrm{96}\Rightarrow{y}^{\mathrm{2}} =\mathrm{16}\Rightarrow{x}^{\mathrm{2}} \pm\mathrm{4}{x}=\mathrm{5}\Rightarrow \\ $$$$\left({x}\pm\mathrm{2}\right)^{\mathrm{2}} =\mathrm{9}\Rightarrow{x}\pm\mathrm{2}=\pm\mathrm{3}\Rightarrow{x}\in\left\{\mathrm{5},−\mathrm{5},+\mathrm{1},−\mathrm{1}\right\} \\ $$$$\Rightarrow{x}^{\mathrm{2}} =\mathrm{25}\:{or}\:\mathrm{1}\Rightarrow\left(\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} \right)=\mathrm{41}\:\boldsymbol{{or}}\:\mathrm{17}\:\blacksquare \\ $$
Commented by mathdanisur last updated on 16/Jul/21
$${thanks}\:{Ser}\:{cool} \\ $$