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Question Number 87886 by TawaTawa1 last updated on 06/Apr/20

Commented by jagoll last updated on 07/Apr/20

$${\mathrm{x}}^2-4\mathrm{x}+1=0$$$${(\mathrm{x}-2)}^2-3=0$$$$\mathrm{x}=2\pm \sqrt{3}$$$${\mathrm{y}}^2-\mathrm{y}\sqrt{5}-1=0$$$$\mathrm{y}=\frac{\sqrt{5}\pm 3}{2}$$$$\Rightarrow \mathrm{xy}+\frac{1}{\mathrm{xy}}=\frac{{\left(\mathrm{xy}\right)}^2+1}{\mathrm{xy}}$$$$$$

Commented by TawaTawa1 last updated on 09/Apr/20

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by $@ty@m123 last updated on 07/Apr/20

$$Ifx+\frac{1}{x}=4andy-\frac{1}{y}=\sqrt{5}$$$$thenxy+\frac{1}{xy}-\sqrt{15}=...$$$$\left(a\right)8\left(b\right)4\left(c\right)5\left(d\right)6$$$$(x+\frac{1}{x}).(y-\frac{1}{y})=4\sqrt{5}$$$$xy-\frac{x}{y}+\frac{y}{x}-\frac{1}{xy}=4\sqrt{5}$$$$xy+(\sqrt{5}x-xy)+(4y-xy)-\frac{1}{xy}=4\sqrt{5}$$$$(\sqrt{5}x+4y)-xy-\frac{1}{xy}=4\sqrt{5}$$$$(\sqrt{5}x+4y)-(xy+\frac{1}{xy})=4\sqrt{5}$$$$(xy+\frac{1}{xy})-\sqrt{15}=4\sqrt{5}-\sqrt{15}-(\sqrt{5}x+4y)...\left(1\right)$$$$x+\frac{1}{x}=4\Rightarrow x^2-4x+1=0$$$$\Rightarrow x=\frac{4\pm \sqrt{12}}{2}=2\pm \sqrt{3}...\left(2\right)$$$$y-\frac{1}{y}=\sqrt{5}$$$$\Rightarrow y^2-\sqrt{5}y-1=0$$$$y=\frac{\sqrt{5}\pm \sqrt{9}}{2}=\frac{1}{2}(\sqrt{5}\pm 3)...\left(3\right)$$$$\therefore \sqrt{5}x+4y=2\sqrt{5}-\sqrt{15}+2\sqrt{5}-6\left(takingnegativesurds\right)$$$$\therefore \sqrt{5}x+4y=4\sqrt{5}-\sqrt{15}-6...\left(4\right)$$$$Substitutingitin\left(1\right),$$$$(xy+\frac{1}{xy})-\sqrt{15}=6$$

Commented by TawaTawa1 last updated on 07/Apr/20

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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