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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 = ?

$$\begin{cases}{\mathrm{3x}^{\mathrm{2}} \:-\:\mathrm{2y}\:=\:-\:\frac{\mathrm{17}}{\mathrm{3}}}\\{\mathrm{y}^{\mathrm{2}} \:-\:\mathrm{6x}\:=\:\mathrm{7}}\end{cases}\:\:\:\Rightarrow\:\:\mathrm{xy}\:=\:? \\ $$

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)

$$\begin{cases}{\mathrm{3x}^{\mathrm{2}} \:-\:\mathrm{2y}\:=\:-\:\frac{\mathrm{17}}{\mathrm{3}}}\\{\mathrm{y}^{\mathrm{2}} \:-\:\mathrm{6x}\:=\:\mathrm{7}}\end{cases}\:\:\:\Rightarrow\:\:\mathrm{xy}\:=\:?\: \\ $$$$\begin{cases}{\mathrm{3x}^{\mathrm{2}} \:-\:\mathrm{2y}\:=\:-\:\frac{\mathrm{17}}{\mathrm{3}}\Rightarrow\mathrm{2y}=\mathrm{3x}^{\mathrm{2}} +\frac{\mathrm{17}}{\mathrm{3}}}\\{\mathrm{y}^{\mathrm{2}} \:-\:\mathrm{6x}\:=\:\mathrm{7}\Rightarrow\mathrm{y}^{\mathrm{2}} =\mathrm{6x}+\mathrm{7}}\end{cases}\:\:\:\: \\ $$$$\left(\frac{\mathrm{3}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{17}}{\mathrm{6}}\right)^{\mathrm{2}} =\mathrm{6x}+\mathrm{7} \\ $$$$\frac{\mathrm{9}}{\mathrm{4}}\mathrm{x}^{\mathrm{4}} +\mathrm{2}.\frac{\mathrm{3}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} .\frac{\mathrm{17}}{\mathrm{6}}+\frac{\mathrm{289}}{\mathrm{36}}−\mathrm{6x}−\mathrm{7}=\mathrm{0} \\ $$$$\frac{\mathrm{9}}{\mathrm{4}}\mathrm{x}^{\mathrm{4}} +\frac{\mathrm{51}}{\mathrm{6}}\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{289}}{\mathrm{36}}−\mathrm{6x}−\mathrm{7}=\mathrm{0} \\ $$$$\mathrm{81x}^{\mathrm{4}} +\mathrm{306x}^{\mathrm{2}} +\mathrm{289}−\mathrm{216x}−\mathrm{252}=\mathrm{0} \\ $$$$\mathrm{81x}^{\mathrm{4}} +\mathrm{306x}^{\mathrm{2}} −\mathrm{216x}+\mathrm{37}=\mathrm{0} \\ $$$$\left(\mathrm{3x}−\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{9x}^{\mathrm{2}} +\mathrm{6x}+\mathrm{37}\right)=\mathrm{0} \\ $$$$\mathrm{x}=\frac{\mathrm{1}}{\mathrm{3}}\:\:\:\mid\:\mathrm{x}=\frac{−\mathrm{6}\pm\sqrt{\mathrm{36}−\mathrm{4}\left(\mathrm{9}\right)\left(\mathrm{37}\right)}}{\mathrm{18}} \\ $$$$\mathrm{x}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{3}}\:\mid\:\mathrm{x}=\frac{−\mathrm{6}\pm\sqrt{\mathrm{36}\left(\mathrm{1}−\mathrm{37}\right)}}{\mathrm{18}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}_{\mathrm{2}} ,\mathrm{x}_{\mathrm{3}} =\frac{−\mathrm{1}\pm\mathrm{6}{i}}{\mathrm{3}} \\ $$$$\mathrm{y}=\frac{\mathrm{3}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{17}}{\mathrm{6}} \\ $$$$\mathrm{y}_{\mathrm{1}} =\frac{\mathrm{3}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} +\frac{\mathrm{17}}{\mathrm{6}}=\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{17}}{\mathrm{6}}=\mathrm{3} \\ $$$$\mathrm{xy}=\mathrm{x}_{\mathrm{1}} \mathrm{y}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{3}}×\mathrm{3}=\mathrm{1}\left(\mathrm{for}\:\mathrm{real}\:\mathrm{x}\:\&\mathrm{y}\right) \\ $$

Commented by mathdanisur last updated on 25/Aug/21

Thank You Ser

$$\mathrm{Thank}\:\mathrm{You}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$

Answered by john_santu last updated on 25/Aug/21

2y =3x^2 +((17)/3) ⇒(2y)^2 =(((9x^2 +17)/3))^2 4y^2 =(((9x^2 +17)/3))^2 24x+28=(((9x^2 +17)/3))^2

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

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