Menu Close

2x-2-3y-2-6xy-12-x-2-y-2-4-

Tinku Tara 0 Comments
Pin




Question Number 211548 by MrGaster last updated on 12/Sep/24
{ ((2x^2 +3y^2 −6xy=12)),((x^2 −y^2 =4)) :}
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\begin{cases}{\mathrm{2}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{3}\boldsymbol{\mathrm{y}}^{\mathrm{2}} −\mathrm{6}\boldsymbol{{x}\mathrm{y}}=\mathrm{12}}\\{\boldsymbol{{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{y}}^{\mathrm{2}} =\mathrm{4}}\end{cases} \\ $$$$ \\ $$
Answered by Frix last updated on 12/Sep/24
(2) ⇒ y^2 =x^2 −4 (1) ⇒ y=((5x^2 −24)/(6x)) Inserting in (1) or (2) gives x^4 +((96x^2 )/(11))−((576)/(11))=0 x^2 =−((48)/(11))±((24(√(15)))/(11)) ...
$$\left(\mathrm{2}\right)\:\Rightarrow\:{y}^{\mathrm{2}} ={x}^{\mathrm{2}} −\mathrm{4} \\ $$$$\left(\mathrm{1}\right)\:\Rightarrow\:{y}=\frac{\mathrm{5}{x}^{\mathrm{2}} −\mathrm{24}}{\mathrm{6}{x}} \\ $$$$\mathrm{Inserting}\:\mathrm{in}\:\left(\mathrm{1}\right)\:\mathrm{or}\:\left(\mathrm{2}\right)\:\mathrm{gives} \\ $$$${x}^{\mathrm{4}} +\frac{\mathrm{96}{x}^{\mathrm{2}} }{\mathrm{11}}−\frac{\mathrm{576}}{\mathrm{11}}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} =−\frac{\mathrm{48}}{\mathrm{11}}\pm\frac{\mathrm{24}\sqrt{\mathrm{15}}}{\mathrm{11}} \\ $$$$… \\ $$
Answered by Rasheed.Sindhi last updated on 13/Sep/24
{ ((2x^2 +3y^2 −6xy=12...(i))),((x^2 −y^2 =4....(ii))) :} 2x^2 +3y^2 −6xy=3×4 2x^2 +3y^2 −6xy=3×(x^2 −y^2 ) x^2 +6xy−6y^2 =0............A x=((−6y±(√(36y^2 +24y^2 )) )/2) x=((−6y±2y(√(15)) )/2)=(−3±(√(15)) )y x^2 =(24∓6(√(15)) )y^2 (ii)⇒(24∓6(√(15)) )y^2 −y^2 =4 (23∓6(√(15)) )y^2 =4 y^2 =(4/(23∓6(√(15)) ))∙((23±6(√(15)) )/(23±6(√(15)) )) y^2 =((92±24(√(15)) )/(529−36(15))) =−((92)/(11))∓((24(√(15)) )/(11)) A⇒6y^2 −6xy−x^2 =0 y=((6x±(√(36x^2 +24x^2 )) )/(12)) y=((6x±2x(√(15)) )/(12))=(((3±(√(15)))/6))x y^2 =((24±6(√(15)) )/(36))x^2 (ii)⇒x^2 −((24±6(√(15)) )/(36))x^2 =4 (1−((24±6(√(15)) )/(36)))x^2 =4 (((12∓6(√(15)) )/(36)))x^2 =4 x^2 =((144)/(12∓6(√(15)) ))∙((12±6(√(15)) )/(12±6(√(15)) )) =((1728±864(√(15)))/(144−36(15)))=((−1728∓864(√(15)))/(396)) =−((48)/(11))∓((24(√(15)))/(11)) (x^2 ,y^2 )=(−((48)/(11))±((24(√(15)))/(11)) , −((92)/(11))±((24(√(15)) )/(11))) (x,y)={((√(−((48)/(11))+((24(√(15)))/(11)))) ,(√(−((92)/(11))+((24(√(15)) )/(11)))) ), (−(√(−((48)/(11))+((24(√(15)))/(11)))) ,−(√(−((92)/(11))+((24(√(15)) )/(11)))) ), ((√(−((48)/(11))−((24(√(15)))/(11)))) ,(√(−((92)/(11))−((24(√(15)) )/(11)))) ), (−(√(−((48)/(11))−((24(√(15)))/(11)))) ,−(√(−((92)/(11))−((24(√(15)) )/(11)))) )}
$$\:\:\:\:\begin{cases}{\mathrm{2}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{3}\boldsymbol{\mathrm{y}}^{\mathrm{2}} −\mathrm{6}\boldsymbol{{x}\mathrm{y}}=\mathrm{12}…\left({i}\right)}\\{\boldsymbol{{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{y}}^{\mathrm{2}} =\mathrm{4}….\left({ii}\right)}\end{cases} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{y}^{\mathrm{2}} −\mathrm{6}{xy}=\mathrm{3}×\mathrm{4} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{y}^{\mathrm{2}} −\mathrm{6}{xy}=\mathrm{3}×\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right) \\ $$$${x}^{\mathrm{2}} +\mathrm{6}{xy}−\mathrm{6}{y}^{\mathrm{2}} =\mathrm{0}…………\boldsymbol{\mathrm{A}} \\ $$$${x}=\frac{−\mathrm{6}{y}\pm\sqrt{\mathrm{36}{y}^{\mathrm{2}} +\mathrm{24}{y}^{\mathrm{2}} }\:}{\mathrm{2}} \\ $$$${x}=\frac{−\mathrm{6}{y}\pm\mathrm{2}{y}\sqrt{\mathrm{15}}\:\:}{\mathrm{2}}=\left(−\mathrm{3}\pm\sqrt{\mathrm{15}}\:\right){y} \\ $$$${x}^{\mathrm{2}} =\left(\mathrm{24}\mp\mathrm{6}\sqrt{\mathrm{15}}\:\right){y}^{\mathrm{2}} \\ $$$$\left({ii}\right)\Rightarrow\left(\mathrm{24}\mp\mathrm{6}\sqrt{\mathrm{15}}\:\right){y}^{\mathrm{2}} −{y}^{\mathrm{2}} =\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{23}\mp\mathrm{6}\sqrt{\mathrm{15}}\:\right){y}^{\mathrm{2}} =\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{y}^{\mathrm{2}} =\frac{\mathrm{4}}{\mathrm{23}\mp\mathrm{6}\sqrt{\mathrm{15}}\:}\centerdot\frac{\mathrm{23}\pm\mathrm{6}\sqrt{\mathrm{15}}\:}{\mathrm{23}\pm\mathrm{6}\sqrt{\mathrm{15}}\:} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{y}^{\mathrm{2}} =\frac{\mathrm{92}\pm\mathrm{24}\sqrt{\mathrm{15}}\:}{\mathrm{529}−\mathrm{36}\left(\mathrm{15}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\frac{\mathrm{92}}{\mathrm{11}}\mp\frac{\mathrm{24}\sqrt{\mathrm{15}}\:}{\mathrm{11}} \\ $$$$\boldsymbol{\mathrm{A}}\Rightarrow\mathrm{6}{y}^{\mathrm{2}} −\mathrm{6}{xy}−{x}^{\mathrm{2}} =\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:{y}=\frac{\mathrm{6}{x}\pm\sqrt{\mathrm{36}{x}^{\mathrm{2}} +\mathrm{24}{x}^{\mathrm{2}} }\:}{\mathrm{12}} \\ $$$$\:\:\:\:\:\:\:\:{y}=\frac{\mathrm{6}{x}\pm\mathrm{2}{x}\sqrt{\mathrm{15}}\:}{\mathrm{12}}=\left(\frac{\mathrm{3}\pm\sqrt{\mathrm{15}}}{\mathrm{6}}\right){x} \\ $$$$\:\:\:\:\:\:{y}^{\mathrm{2}} =\frac{\mathrm{24}\pm\mathrm{6}\sqrt{\mathrm{15}}\:}{\mathrm{36}}{x}^{\mathrm{2}} \\ $$$$\left({ii}\right)\Rightarrow{x}^{\mathrm{2}} −\frac{\mathrm{24}\pm\mathrm{6}\sqrt{\mathrm{15}}\:}{\mathrm{36}}{x}^{\mathrm{2}} =\mathrm{4} \\ $$$$\left(\mathrm{1}−\frac{\mathrm{24}\pm\mathrm{6}\sqrt{\mathrm{15}}\:}{\mathrm{36}}\right){x}^{\mathrm{2}} =\mathrm{4} \\ $$$$\left(\frac{\mathrm{12}\mp\mathrm{6}\sqrt{\mathrm{15}}\:}{\mathrm{36}}\right){x}^{\mathrm{2}} =\mathrm{4} \\ $$$${x}^{\mathrm{2}} =\frac{\mathrm{144}}{\mathrm{12}\mp\mathrm{6}\sqrt{\mathrm{15}}\:}\centerdot\frac{\mathrm{12}\pm\mathrm{6}\sqrt{\mathrm{15}}\:}{\mathrm{12}\pm\mathrm{6}\sqrt{\mathrm{15}}\:} \\ $$$$\:\:\:=\frac{\mathrm{1728}\pm\mathrm{864}\sqrt{\mathrm{15}}}{\mathrm{144}−\mathrm{36}\left(\mathrm{15}\right)}=\frac{−\mathrm{1728}\mp\mathrm{864}\sqrt{\mathrm{15}}}{\mathrm{396}} \\ $$$$\:\:\:\:=−\frac{\mathrm{48}}{\mathrm{11}}\mp\frac{\mathrm{24}\sqrt{\mathrm{15}}}{\mathrm{11}} \\ $$$$\left({x}^{\mathrm{2}} ,{y}^{\mathrm{2}} \right)=\left(−\frac{\mathrm{48}}{\mathrm{11}}\pm\frac{\mathrm{24}\sqrt{\mathrm{15}}}{\mathrm{11}}\:,\:−\frac{\mathrm{92}}{\mathrm{11}}\pm\frac{\mathrm{24}\sqrt{\mathrm{15}}\:}{\mathrm{11}}\right) \\ $$$$\left({x},{y}\right)=\left\{\left(\sqrt{−\frac{\mathrm{48}}{\mathrm{11}}+\frac{\mathrm{24}\sqrt{\mathrm{15}}}{\mathrm{11}}}\:,\sqrt{−\frac{\mathrm{92}}{\mathrm{11}}+\frac{\mathrm{24}\sqrt{\mathrm{15}}\:}{\mathrm{11}}}\:\right),\right. \\ $$$$\left(−\sqrt{−\frac{\mathrm{48}}{\mathrm{11}}+\frac{\mathrm{24}\sqrt{\mathrm{15}}}{\mathrm{11}}}\:,−\sqrt{−\frac{\mathrm{92}}{\mathrm{11}}+\frac{\mathrm{24}\sqrt{\mathrm{15}}\:}{\mathrm{11}}}\:\right), \\ $$$$\left(\sqrt{−\frac{\mathrm{48}}{\mathrm{11}}−\frac{\mathrm{24}\sqrt{\mathrm{15}}}{\mathrm{11}}}\:,\sqrt{−\frac{\mathrm{92}}{\mathrm{11}}−\frac{\mathrm{24}\sqrt{\mathrm{15}}\:}{\mathrm{11}}}\:\right), \\ $$$$\left.\left(−\sqrt{−\frac{\mathrm{48}}{\mathrm{11}}−\frac{\mathrm{24}\sqrt{\mathrm{15}}}{\mathrm{11}}}\:,−\sqrt{−\frac{\mathrm{92}}{\mathrm{11}}−\frac{\mathrm{24}\sqrt{\mathrm{15}}\:}{\mathrm{11}}}\:\right)\right\} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *