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Question Number 19729 by Joel577 last updated on 15/Aug/17

2x + 9y^2 = 4 2x^2 − 45y^2 + xy = 0 Find the value of xy

$$\mathrm{2}{x}\:+\:\mathrm{9}{y}^{\mathrm{2}} \:=\:\mathrm{4} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} \:−\:\mathrm{45}{y}^{\mathrm{2}} \:+\:{xy}\:=\:\mathrm{0} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{xy} \\ $$

Answered by mrW1 last updated on 15/Aug/17

2x^2 − 45y^2 + xy = 0 (x+5y)(2x−9y)=0 ⇒x+5y=0 or 2x−9y=0 { ((2x+9y^2 =4)),((x+5y=0)) :} −2×5y+9y^2 =4 9y^2 −10y−4=0 y=((10±(√(100+4×9×4)))/(2×9))=((5±(√(61)))/9) x=−5×((5±(√(61)))/9)=((−25∓5(√(61)))/9) { ((2x+9y^2 =4)),((2x−9y=0)) :} 9y+9y^2 =4 9y^2 +9y−4=0 y=((−9±(√(81+4×9×4)))/(2×9))=((−9±15)/(2×9))=((−3±5)/(2×3))=(1/3),−(4/3) x=(9/2)×(1/3)=(3/2), −(9/2)×(4/3)=−6 ⇒(x,y)=(−((25+5(√(61)))/9),((5+(√(61)))/9)) ⇒(x,y)=(−((25−5(√(61)))/9),((5−(√(61)))/9)) ⇒(x,y)=((3/2),(1/3)) ⇒(x,y)=(−6,−(4/3))

$$\mathrm{2}{x}^{\mathrm{2}} \:−\:\mathrm{45}{y}^{\mathrm{2}} \:+\:{xy}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{x}+\mathrm{5y}\right)\left(\mathrm{2x}−\mathrm{9y}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}+\mathrm{5y}=\mathrm{0}\:\mathrm{or}\:\mathrm{2x}−\mathrm{9y}=\mathrm{0} \\ $$$$ \\ $$$$\begin{cases}{\mathrm{2x}+\mathrm{9y}^{\mathrm{2}} =\mathrm{4}}\\{\mathrm{x}+\mathrm{5y}=\mathrm{0}}\end{cases} \\ $$$$−\mathrm{2}×\mathrm{5y}+\mathrm{9y}^{\mathrm{2}} =\mathrm{4} \\ $$$$\mathrm{9y}^{\mathrm{2}} −\mathrm{10y}−\mathrm{4}=\mathrm{0} \\ $$$$\mathrm{y}=\frac{\mathrm{10}\pm\sqrt{\mathrm{100}+\mathrm{4}×\mathrm{9}×\mathrm{4}}}{\mathrm{2}×\mathrm{9}}=\frac{\mathrm{5}\pm\sqrt{\mathrm{61}}}{\mathrm{9}} \\ $$$$\mathrm{x}=−\mathrm{5}×\frac{\mathrm{5}\pm\sqrt{\mathrm{61}}}{\mathrm{9}}=\frac{−\mathrm{25}\mp\mathrm{5}\sqrt{\mathrm{61}}}{\mathrm{9}} \\ $$$$ \\ $$$$\begin{cases}{\mathrm{2x}+\mathrm{9y}^{\mathrm{2}} =\mathrm{4}}\\{\mathrm{2x}−\mathrm{9y}=\mathrm{0}}\end{cases} \\ $$$$\mathrm{9y}+\mathrm{9y}^{\mathrm{2}} =\mathrm{4} \\ $$$$\mathrm{9y}^{\mathrm{2}} +\mathrm{9y}−\mathrm{4}=\mathrm{0} \\ $$$$\mathrm{y}=\frac{−\mathrm{9}\pm\sqrt{\mathrm{81}+\mathrm{4}×\mathrm{9}×\mathrm{4}}}{\mathrm{2}×\mathrm{9}}=\frac{−\mathrm{9}\pm\mathrm{15}}{\mathrm{2}×\mathrm{9}}=\frac{−\mathrm{3}\pm\mathrm{5}}{\mathrm{2}×\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{3}},−\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\mathrm{x}=\frac{\mathrm{9}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{3}}{\mathrm{2}},\:−\frac{\mathrm{9}}{\mathrm{2}}×\frac{\mathrm{4}}{\mathrm{3}}=−\mathrm{6} \\ $$$$ \\ $$$$\Rightarrow\left(\mathrm{x},\mathrm{y}\right)=\left(−\frac{\mathrm{25}+\mathrm{5}\sqrt{\mathrm{61}}}{\mathrm{9}},\frac{\mathrm{5}+\sqrt{\mathrm{61}}}{\mathrm{9}}\right) \\ $$$$\Rightarrow\left(\mathrm{x},\mathrm{y}\right)=\left(−\frac{\mathrm{25}−\mathrm{5}\sqrt{\mathrm{61}}}{\mathrm{9}},\frac{\mathrm{5}−\sqrt{\mathrm{61}}}{\mathrm{9}}\right) \\ $$$$\Rightarrow\left(\mathrm{x},\mathrm{y}\right)=\left(\frac{\mathrm{3}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$$\Rightarrow\left(\mathrm{x},\mathrm{y}\right)=\left(−\mathrm{6},−\frac{\mathrm{4}}{\mathrm{3}}\right) \\ $$

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