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Question Number 13647 by chux last updated on 22/May/17

x^2 +y^2 =5.....(1) 3x^2 +xy+y^2 =1.....(2) please help find x and y

$$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\mathrm{5}.....\left(\mathrm{1}\right) \\ $$$$\mathrm{3x}^{\mathrm{2}} +\mathrm{xy}+\mathrm{y}^{\mathrm{2}} =\mathrm{1}.....\left(\mathrm{2}\right) \\ $$$$ \\ $$$$ \\ $$$$\mathrm{please}\:\mathrm{help}\:\mathrm{find}\:\mathrm{x}\:\mathrm{and}\:\mathrm{y} \\ $$

Commented by ajfour last updated on 22/May/17

2x^2 +xy+(x^2 +y^2 )=1 2x^2 +xy+4=0 ⇒y=−(2x+(4/x)) ..(i) squaring, x^2 y^2 =(2x^2 +4)^2 x^2 (5−x^2 )=4(x^2 +2)^2 −(x^2 +2)^2 +9(x^2 +2)−14=4(x^2 +2)^2 5(x^2 +2)^2 −9(x^2 +2)+14=0 x^2 +2=((9±(√(81−280)))/(10)) x^2 =−((11)/(10))±(√(((1/(10)))^2 −2)) ....(ii) y=−(2x+(4/x)) .....(i) difficult for me beyond this point..

$$\mathrm{2}{x}^{\mathrm{2}} +{xy}+\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)=\mathrm{1} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +{xy}+\mathrm{4}=\mathrm{0}\:\:\Rightarrow{y}=−\left(\mathrm{2}{x}+\frac{\mathrm{4}}{{x}}\right)\:..\left({i}\right) \\ $$$${squaring}, \\ $$$${x}^{\mathrm{2}} {y}^{\mathrm{2}} =\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} \left(\mathrm{5}−{x}^{\mathrm{2}} \right)=\mathrm{4}\left({x}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{2}} \\ $$$$−\left({x}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{2}} +\mathrm{9}\left({x}^{\mathrm{2}} +\mathrm{2}\right)−\mathrm{14}=\mathrm{4}\left({x}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\mathrm{5}\left({x}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{2}} −\mathrm{9}\left({x}^{\mathrm{2}} +\mathrm{2}\right)+\mathrm{14}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} +\mathrm{2}=\frac{\mathrm{9}\pm\sqrt{\mathrm{81}−\mathrm{280}}}{\mathrm{10}} \\ $$$${x}^{\mathrm{2}} =−\frac{\mathrm{11}}{\mathrm{10}}\pm\sqrt{\left(\frac{\mathrm{1}}{\mathrm{10}}\right)^{\mathrm{2}} −\mathrm{2}}\:\:\:\:\:\:....\left({ii}\right) \\ $$$${y}=−\left(\mathrm{2}{x}+\frac{\mathrm{4}}{{x}}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:.....\left({i}\right) \\ $$$${difficult}\:{for}\:{me}\:{beyond}\:{this}\:{point}.. \\ $$

Commented by prakash jain last updated on 22/May/17

3x^2 +xy+y^2 =1 2x^2 +xy+5=1 2x^2 +xy=−4 ....(A) x^2 +y^2 =5⇒x=(√5)cos θ⇒y=(√5)sin θ substituting in (A) 10cos^2 θ+5cos θsin θ=−4 cos^2 θ=(cos 2θ+1)/2,2sin θcos θ=sin 2θ 5(cos 2θ+1)+(5/2)sin 2θ=−4 10cos 2θ+10+5sin 2θ=−8 10cos 2θ+5sin 2θ=−18 2cos 2θ+sin 2θ=((−18)/5) (2/(√5))cos 2θ+(1/(√5))sin 2θ=−((18)/(5(√5))) sin (2θ+sin^(−1) (2/(√5)))=−((18)/(5(√5)))>1 5(√5)≈11.18<18 no real solution.

$$\mathrm{3}{x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +{xy}+\mathrm{5}=\mathrm{1} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +{xy}=−\mathrm{4}\:\:\:....\left({A}\right) \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{5}\Rightarrow{x}=\sqrt{\mathrm{5}}\mathrm{cos}\:\theta\Rightarrow{y}=\sqrt{\mathrm{5}}\mathrm{sin}\:\theta \\ $$$${substituting}\:{in}\:\left({A}\right) \\ $$$$\mathrm{10cos}^{\mathrm{2}} \theta+\mathrm{5cos}\:\theta\mathrm{sin}\:\theta=−\mathrm{4} \\ $$$$\mathrm{cos}^{\mathrm{2}} \theta=\left(\mathrm{cos}\:\mathrm{2}\theta+\mathrm{1}\right)/\mathrm{2},\mathrm{2sin}\:\theta\mathrm{cos}\:\theta=\mathrm{sin}\:\mathrm{2}\theta \\ $$$$\mathrm{5}\left(\mathrm{cos}\:\mathrm{2}\theta+\mathrm{1}\right)+\frac{\mathrm{5}}{\mathrm{2}}\mathrm{sin}\:\mathrm{2}\theta=−\mathrm{4} \\ $$$$\mathrm{10cos}\:\mathrm{2}\theta+\mathrm{10}+\mathrm{5sin}\:\mathrm{2}\theta=−\mathrm{8} \\ $$$$\mathrm{10cos}\:\mathrm{2}\theta+\mathrm{5sin}\:\mathrm{2}\theta=−\mathrm{18} \\ $$$$\mathrm{2cos}\:\mathrm{2}\theta+\mathrm{sin}\:\mathrm{2}\theta=\frac{−\mathrm{18}}{\mathrm{5}} \\ $$$$\frac{\mathrm{2}}{\sqrt{\mathrm{5}}}\mathrm{cos}\:\mathrm{2}\theta+\frac{\mathrm{1}}{\sqrt{\mathrm{5}}}\mathrm{sin}\:\mathrm{2}\theta=−\frac{\mathrm{18}}{\mathrm{5}\sqrt{\mathrm{5}}} \\ $$$$\mathrm{sin}\:\left(\mathrm{2}\theta+\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{2}}{\sqrt{\mathrm{5}}}\right)=−\frac{\mathrm{18}}{\mathrm{5}\sqrt{\mathrm{5}}}>\mathrm{1} \\ $$$$\mathrm{5}\sqrt{\mathrm{5}}\approx\mathrm{11}.\mathrm{18}<\mathrm{18} \\ $$$${no}\:{real}\:{solution}. \\ $$

Commented by ajfour last updated on 22/May/17

line#9→line#10 −8−50≠−42

$${line}#\mathrm{9}\rightarrow{line}#\mathrm{10} \\ $$$$−\mathrm{8}−\mathrm{50}\neq−\mathrm{42} \\ $$

Commented by prakash jain last updated on 22/May/17

ok. will correct

$${ok}.\:{will}\:{correct} \\ $$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 22/May/17

(x/y)=t { ((t^2 +1=(5/y^2 ) (i))),((3t^2 +t+1=(1/y^2 ) (ii))) :} (i)−5(ii)⇒t^2 +1−15t^2 −5t−5=0 ⇒14t^2 +5t+4=0 t=((−5±(√(25−14×16)))/(28))=((−5±i(√(199)))/(28)) ⇒(x/y)=−0.18±0.5i (i^2 =−1) no real answers.

$$\frac{{x}}{{y}}={t} \\ $$$$\begin{cases}{{t}^{\mathrm{2}} +\mathrm{1}=\frac{\mathrm{5}}{{y}^{\mathrm{2}} }\:\:\:\:\:\:\:\:\left({i}\right)}\\{\mathrm{3}{t}^{\mathrm{2}} +{t}+\mathrm{1}=\frac{\mathrm{1}}{{y}^{\mathrm{2}} }\:\:\:\:\:\left({ii}\right)}\end{cases} \\ $$$$\left({i}\right)−\mathrm{5}\left({ii}\right)\Rightarrow{t}^{\mathrm{2}} +\mathrm{1}−\mathrm{15}{t}^{\mathrm{2}} −\mathrm{5}{t}−\mathrm{5}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{14}{t}^{\mathrm{2}} +\mathrm{5}{t}+\mathrm{4}=\mathrm{0} \\ $$$${t}=\frac{−\mathrm{5}\pm\sqrt{\mathrm{25}−\mathrm{14}×\mathrm{16}}}{\mathrm{28}}=\frac{−\mathrm{5}\pm{i}\sqrt{\mathrm{199}}}{\mathrm{28}} \\ $$$$\Rightarrow\frac{{x}}{{y}}=−\mathrm{0}.\mathrm{18}\pm\mathrm{0}.\mathrm{5}\boldsymbol{{i}}\:\:\:\:\left(\boldsymbol{{i}}^{\mathrm{2}} =−\mathrm{1}\right) \\ $$$$\boldsymbol{{no}}\:\boldsymbol{{real}}\:\boldsymbol{{answers}}. \\ $$

Commented by tawa tawa last updated on 22/May/17

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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