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

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

Answered by nurtani last updated on 10/Jan/22

$$x^2+4y^2-4xy+y^2+6y+9=0x2$$$$x^2+{\left(2y\right)}^2-4xy+{(y+3)}^2-\cancel{9}+\cancel{9}=0$$$${(x-2y)}^2+\cancel{4xy}-\cancel{4xy}+{(y+3)}^2=0$$$${(x-2y)}^2+{(y+3)}^2=0$$$$\Rightarrow x-2y=0\wedge y+3=0$$$$\Rightarrow x=2y\wedge y=-3$$$$\therefore xy=2y\left(y\right)$$$$=2y^2$$$$=2{(-3)}^2$$$$=2\left(9\right)=18$$

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

$$x^2-4xy+4y^2+y^2+6y+9=0$$$${(x-2y)}^2+{(y+3)}^2=0$$$$x-2y=0\wedge y+3=0$$$$y=-3\Rightarrow x=2y=2(-3)=-6$$$$xy=(-6)(-3)=18$$

Answered by kowalsky78 last updated on 10/Jan/22

$$$$$$x^2+5y^2-4xy+6y+9={(x-2y)}^2+{(y+3)}^2=0$$$$Sox=2yandy=-3$$$$xy=(-6)\times (-3)=18$$$$$$

Commented by HongKing last updated on 10/Jan/22

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

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