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Question Number 97227 by bemath last updated on 07/Jun/20

Commented by bemath last updated on 07/Jun/20

$$\mathrm{thank}\mathrm{you}\mathrm{both}$$

Answered by som(math1967) last updated on 07/Jun/20

$${\mathrm{x}}^2+\mathrm{xy}+{\mathrm{y}}^2=84$$$$\Rightarrow (\mathrm{x}+\sqrt{\mathrm{xy}}+\mathrm{y})(\mathrm{x}-\sqrt{\mathrm{xy}}+\mathrm{y})=84$$$$\therefore \mathrm{x}-\sqrt{\mathrm{xy}}+\mathrm{y}=\frac{84}{14}=6$$$$\therefore 2(\mathrm{x}+\mathrm{y})=20$$$$\mathrm{x}+\mathrm{y}=10$$$$2\sqrt{\mathrm{xy}}=8$$$$\mathrm{xy}=16$$$$\mathrm{x}(10-\mathrm{x})=16$$$${\mathrm{x}}^2-10\mathrm{x}+16=0$$$$(\mathrm{x}-8)(\mathrm{x}-2)=0$$$$\mathrm{x}=8\mathrm{or}2$$$$\mathrm{when}\mathrm{x}=8\mathrm{y}=2$$$$\mathrm{when}\mathrm{x}=2,\mathrm{y}=8\mathrm{ans}$$

Answered by john santu last updated on 07/Jun/20

$$\iff x^2+2xy+y^2-xy=84$$$${(x+y)}^2-{\left(\sqrt{xy}\right)}^2=84$$$$(x+y+\sqrt{xy})(x+y-\sqrt{xy})=84$$$$\iff 14\times (x+y-\sqrt{xy})=84$$$$x+y-\sqrt{xy}=6...\left(2\right)$$$$x+y+\sqrt{xy}=14...\left(1\right)$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}+$$$$x+y=10\wedge \sqrt{xy}=4$$$$\iff x(10-x)=16$$$$\iff x^2-10x+16=0$$$$(x-2)(x-8)=0$$$$\{\begin{array}{l}x=2\wedge \mathrm{y}=8\\ x=8\wedge \mathrm{y}=2\end{array}$$

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