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Question Number 163746 by HongKing last updated on 10/Jan/22

x^2 + 5y^2 - 4xy + 6y + 9 = 0 find xy=?

$$\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{5y}^{\mathrm{2}} \:-\:\mathrm{4xy}\:+\:\mathrm{6y}\:+\:\mathrm{9}\:=\:\mathrm{0} \\ $$$$\mathrm{find}\:\:\mathrm{xy}=? \\ $$

Answered by nurtani last updated on 10/Jan/22

x^2 +4y^2 −4xy+y^2 +6y+9=0x2 x^2 +(2y)^2 −4xy+(y+3)^2 −9+9=0 (x−2y)^2 +4xy−4xy+(y+3)^2 =0 (x−2y)^2 +(y+3)^2 =0 ⇒x−2y=0 ∧ y+3=0 ⇒ x=2y ∧ y=−3 ∴ xy=2y(y) =2y^2 =2(−3)^2 =2(9)=18

$${x}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} −\mathrm{4}{xy}+{y}^{\mathrm{2}} +\mathrm{6}{y}+\mathrm{9}=\mathrm{0}{x}\mathrm{2} \\ $$$${x}^{\mathrm{2}} +\left(\mathrm{2}{y}\right)^{\mathrm{2}} −\mathrm{4}{xy}+\left({y}+\mathrm{3}\right)^{\mathrm{2}} −\cancel{\mathrm{9}}+\cancel{\mathrm{9}}=\mathrm{0} \\ $$$$\left({x}−\mathrm{2}{y}\right)^{\mathrm{2}} +\cancel{\mathrm{4}{xy}}−\cancel{\mathrm{4}{xy}}+\left({y}+\mathrm{3}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({x}−\mathrm{2}{y}\right)^{\mathrm{2}} +\left({y}+\mathrm{3}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{x}−\mathrm{2}{y}=\mathrm{0}\:\wedge\:{y}+\mathrm{3}=\mathrm{0} \\ $$$$\Rightarrow\:{x}=\mathrm{2}{y}\:\:\:\wedge\:{y}=−\mathrm{3} \\ $$$$\therefore\:\:\:{xy}=\mathrm{2}{y}\left({y}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}{y}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\left(−\mathrm{3}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\left(\mathrm{9}\right)=\mathrm{18} \\ $$

Answered by Rasheed.Sindhi last updated on 10/Jan/22

x^2 −4xy+4y^2 +y^2 +6y+9=0 (x−2y)^2 +(y+3)^2 =0 x−2y=0 ∧ y+3=0 y=−3⇒x=2y=2(−3)=−6 xy=(−6)(−3)=18

$${x}^{\mathrm{2}} −\mathrm{4}{xy}+\mathrm{4}{y}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{6}{y}+\mathrm{9}=\mathrm{0} \\ $$$$\left({x}−\mathrm{2}{y}\right)^{\mathrm{2}} +\left({y}+\mathrm{3}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${x}−\mathrm{2}{y}=\mathrm{0}\:\wedge\:{y}+\mathrm{3}=\mathrm{0} \\ $$$${y}=−\mathrm{3}\Rightarrow{x}=\mathrm{2}{y}=\mathrm{2}\left(−\mathrm{3}\right)=−\mathrm{6} \\ $$$${xy}=\left(−\mathrm{6}\right)\left(−\mathrm{3}\right)=\mathrm{18} \\ $$

Answered by kowalsky78 last updated on 10/Jan/22

x^2 +5y^2 −4xy+6y+9=(x−2y)^2 +(y+3)^2 =0 So x=2y and y=−3 xy=(−6)×(−3)=18

$$ \\ $$$${x}^{\mathrm{2}} +\mathrm{5}{y}^{\mathrm{2}} −\mathrm{4}{xy}+\mathrm{6}{y}+\mathrm{9}=\left({x}−\mathrm{2}{y}\right)^{\mathrm{2}} +\left({y}+\mathrm{3}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${So}\:{x}=\mathrm{2}{y}\:{and}\:{y}=−\mathrm{3} \\ $$$${xy}=\left(−\mathrm{6}\right)×\left(−\mathrm{3}\right)=\mathrm{18} \\ $$$$ \\ $$

Commented by HongKing last updated on 10/Jan/22

very nice dear Sir thank you

$$\mathrm{very}\:\mathrm{nice}\:\mathrm{dear}\:\mathrm{Sir}\:\mathrm{thank}\:\mathrm{you} \\ $$

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