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


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

$2 x + 9 y^{2} = 4$ $2 x^{2} - 45 y^{2} + x y = 0$ $Find the value of x y$
	Answered by mrW1 last updated on 15/Aug/17
	$2x^2 − 45y^2 + xy = 0 (x+5y)(2x−9y)=0 ⇒x+5y=0 or 2x−9y=0 { ((2x+9y^2 =4)),((x+5y=0)) :} −2×5y+9y^2 =4 9y^2 −10y−4=0 y=((10±(√(100+4×9×4)))/(2×9))=((5±(√(61)))/9) x=−5×((5±(√(61)))/9)=((−25∓5(√(61)))/9) { ((2x+9y^2 =4)),((2x−9y=0)) :} 9y+9y^2 =4 9y^2 +9y−4=0 y=((−9±(√(81+4×9×4)))/(2×9))=((−9±15)/(2×9))=((−3±5)/(2×3))=(1/3),−(4/3) x=(9/2)×(1/3)=(3/2), −(9/2)×(4/3)=−6 ⇒(x,y)=(−((25+5(√(61)))/9),((5+(√(61)))/9)) ⇒(x,y)=(−((25−5(√(61)))/9),((5−(√(61)))/9)) ⇒(x,y)=((3/2),(1/3)) ⇒(x,y)=(−6,−(4/3))$
	$2 x^{2} - 45 y^{2} + x y = 0$ $(x + 5 y) (2 x - 9 y) = 0$ $\Rightarrow x + 5 y = 0 or 2 x - 9 y = 0$ ${\begin{cases} 2 x + {9 y}^{2} = 4 \\ x + 5 y = 0 \end{cases}$ $- 2 \times 5 y + {9 y}^{2} = 4$ ${9 y}^{2} - 10 y - 4 = 0$ $y = \frac{10 \pm \sqrt{100 + 4 \times 9 \times 4}}{2 \times 9} = \frac{5 \pm \sqrt{61}}{9}$ $x = - 5 \times \frac{5 \pm \sqrt{61}}{9} = \frac{- 25 \mp 5 \sqrt{61}}{9}$ ${\begin{cases} 2 x + {9 y}^{2} = 4 \\ 2 x - 9 y = 0 \end{cases}$ $9 y + {9 y}^{2} = 4$ ${9 y}^{2} + 9 y - 4 = 0$ $y = \frac{- 9 \pm \sqrt{81 + 4 \times 9 \times 4}}{2 \times 9} = \frac{- 9 \pm 15}{2 \times 9} = \frac{- 3 \pm 5}{2 \times 3} = \frac{1}{3}, - \frac{4}{3}$ $x = \frac{9}{2} \times \frac{1}{3} = \frac{3}{2}, - \frac{9}{2} \times \frac{4}{3} = - 6$ $\Rightarrow (x, y) = (- \frac{25 + 5 \sqrt{61}}{9}, \frac{5 + \sqrt{61}}{9})$ $\Rightarrow (x, y) = (- \frac{25 - 5 \sqrt{61}}{9}, \frac{5 - \sqrt{61}}{9})$ $\Rightarrow (x, y) = (\frac{3}{2}, \frac{1}{3})$ $\Rightarrow (x, y) = (- 6, - \frac{4}{3})$
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