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Question Number 11748 by tawa last updated on 30/Mar/17

$$\mathrm{Solve}\mathrm{simultaneously}$$$${\mathrm{x}}^2+{4\mathrm{y}}^2+\mathrm{xy}=6..............\mathrm{equation}\left(\mathrm{i}\right)$$$${3\mathrm{x}}^2+{8\mathrm{y}}^2=14..............\mathrm{equation}\left(\mathrm{ii}\right)$$

Answered by mrW1 last updated on 30/Mar/17

$$\left(ii\right)-\left(i\right)\times 2:$$$$x^2-2xy=2$$$$\Rightarrow y=\frac{x^2-2}{2x}$$$$3x^2+8\times \frac{{(x^2-2)}^2}{4x^2}=14$$$$3x^4+2(x^4-4x^2+4)=14x^2$$$$5x^4-22x^2+8=0$$$$x^2=\frac{22\pm \sqrt{{22}^2-4\times 5\times 8}}{10}=\frac{22\pm 18}{10}=4,\frac{2}{5}$$$$x_{1,2}=\pm 2$$$$x_{3,4}=\pm \sqrt{\frac{2}{5}}$$$$y_{1,2}=\frac{x^2-2}{2x}=\pm 2\times \frac{4-2}{2\times 4}=\pm \frac{1}{2}$$$$y_{3,4}=\frac{x^2-2}{2x}=\pm \sqrt{\frac{2}{5}}\times \frac{\frac{2}{5}-2}{2\times \frac{2}{5}}=\mp 2\sqrt{\frac{2}{5}}$$

Commented by tawa last updated on 30/Mar/17

$$\mathrm{Wow},\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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