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Question Number 152034 by mathdanisur last updated on 25/Aug/21
${ ((3x^2 - 2y = - ((17)/3))),((y^2 - 6x = 7)) :} ⇒ xy = ?$
${\begin{cases} {3 x}^{2} - 2 y = - \frac{17}{3} \\ y^{2} - 6 x = 7 \end{cases} \Rightarrow xy = ?$
	Answered by Rasheed.Sindhi last updated on 25/Aug/21
	${ ((3x^2 - 2y = - ((17)/3))),((y^2 - 6x = 7)) :} ⇒ xy = ? { ((3x^2 - 2y = - ((17)/3)⇒2y=3x^2 +((17)/3))),((y^2 - 6x = 7⇒y^2 =6x+7)) :} ((3/2)x^2 +((17)/6))^2 =6x+7 (9/4)x^4 +2.(3/2)x^2 .((17)/6)+((289)/(36))−6x−7=0 (9/4)x^4 +((51)/6)x^2 +((289)/(36))−6x−7=0 81x^4 +306x^2 +289−216x−252=0 81x^4 +306x^2 −216x+37=0 (3x−1)^2 (9x^2 +6x+37)=0 x=(1/3) ∣ x=((−6±(√(36−4(9)(37))))/(18)) x_1 =(1/3) ∣ x=((−6±(√(36(1−37))))/(18)) x_2 ,x_3 =((−1±6i)/3) y=(3/2)x^2 +((17)/6) y_1 =(3/2)((1/3))^2 +((17)/6)=(1/6)+((17)/6)=3 xy=x_1 y_1 =(1/3)×3=1(for real x &y)$
	${\begin{cases} {3 x}^{2} - 2 y = - \frac{17}{3} \\ y^{2} - 6 x = 7 \end{cases} \Rightarrow xy = ?$ ${\begin{cases} {3 x}^{2} - 2 y = - \frac{17}{3} \Rightarrow 2 y = {3 x}^{2} + \frac{17}{3} \\ y^{2} - 6 x = 7 \Rightarrow y^{2} = 6 x + 7 \end{cases}$ ${(\frac{3}{2} x^{2} + \frac{17}{6})}^{2} = 6 x + 7$ $\frac{9}{4} x^{4} + 2 . \frac{3}{2} x^{2} . \frac{17}{6} + \frac{289}{36} - 6 x - 7 = 0$ $\frac{9}{4} x^{4} + \frac{51}{6} x^{2} + \frac{289}{36} - 6 x - 7 = 0$ ${81 x}^{4} + {306 x}^{2} + 289 - 216 x - 252 = 0$ ${81 x}^{4} + {306 x}^{2} - 216 x + 37 = 0$ ${(3 x - 1)}^{2} ({9 x}^{2} + 6 x + 37) = 0$ $x = \frac{1}{3} ∣ x = \frac{- 6 \pm \sqrt{36 - 4 (9) (37)}}{18}$ $x_{1} = \frac{1}{3} ∣ x = \frac{- 6 \pm \sqrt{36 (1 - 37)}}{18}$ $x_{2}, x_{3} = \frac{- 1 \pm 6 i}{3}$ $y = \frac{3}{2} x^{2} + \frac{17}{6}$ $y_{1} = \frac{3}{2} {(\frac{1}{3})}^{2} + \frac{17}{6} = \frac{1}{6} + \frac{17}{6} = 3$ $xy = x_{1} y_{1} = \frac{1}{3} \times 3 = 1 (for real x & y)$
	Commented by mathdanisur last updated on 25/Aug/21

	$Thank You S er$
	Answered by john_santu last updated on 25/Aug/21

	$2 y = {3 x}^{2} + \frac{17}{3} \Rightarrow {(2 y)}^{2} = {(\frac{{9 x}^{2} + 17}{3})}^{2}$ ${4 y}^{2} = {(\frac{{9 x}^{2} + 17}{3})}^{2}$ $24 x + 28 = {(\frac{{9 x}^{2} + 17}{3})}^{2}$
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