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Question Number 33764 by .none. last updated on 23/Apr/18

$$2x-y=\frac{1}{\sqrt{5}-2}$$$$x+2y=\frac{1}{\sqrt{5}+2}$$$$[3x^2-8xy-3y^2]?$$

Answered by Rasheed.Sindhi last updated on 23/Apr/18

$${3\mathrm{x}}^2-8\mathrm{xy}-{3\mathrm{y}}^2=(\mathrm{x}-3\mathrm{y})(3\mathrm{x}+\mathrm{y})$$$$---------$$$$2x-y=\frac{1}{\sqrt{5}-2}............\mathrm{A}$$$$x+2y=\frac{1}{\sqrt{5}+2}..............\mathrm{B}$$$$\mathrm{A}+\mathrm{B}:3\mathrm{x}+\mathrm{y}=\frac{1}{\sqrt{5}-2}+\frac{1}{\sqrt{5}+2}=\frac{\sqrt{5}+2+\sqrt{5}-2}{5-4}$$$$=2\sqrt{5}$$$$\mathrm{A}-\mathrm{B}:\mathrm{x}-3\mathrm{y}=\frac{1}{\sqrt{5}-2}-\frac{1}{\sqrt{5}+2}=\frac{\sqrt{5}+2-\sqrt{5}+2}{5-4}$$$$=4$$$$------------$$$${3\mathrm{x}}^2-8\mathrm{xy}-{3\mathrm{y}}^2=(\mathrm{x}-3\mathrm{y})(3\mathrm{x}+\mathrm{y})$$$$=\left(4\right)\left(2\sqrt{5}\right)=8\sqrt{5}$$$$$$$$$$

Commented by .none. last updated on 23/Apr/18

$$Thankstoyou,icansolvethisnow$$

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