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Question Number 28288 by Mr eaay last updated on 23/Jan/18

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

$${}^{\bullet }{(2+\sqrt{3})}^{-1}=\frac{1}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}$$$$=2-\sqrt{3}$$$${}^{\bullet }{(2+\sqrt{3})}^2={\left(2\right)}^2+2\left(2\right)\left(\sqrt{3}\right)+{\left(\sqrt{3}\right)}^2$$$$=7+4\sqrt{\mathrm{x}}$$$${}^{\bullet }{(2+\sqrt{3})}^3={\left(2\right)}^3+{\left(\sqrt{3}\right)}^3+3{\left(2\right)}^2\left(\sqrt{3}\right)+3\left(2\right){\left(\sqrt{3}\right)}^2$$$$=8+3\sqrt{3}+12\sqrt{3}+18$$$$=26+15\sqrt{3}$$$$\mathrm{The}\mathrm{given}\mathrm{equation}\Rightarrow$$$${\left({(2+\sqrt{3})}^3\right)}^{\mathrm{x}}-5{\left({(2+\sqrt{3})}^2\right)}^{\mathrm{x}}+6{(2+\sqrt{3})}^{\mathrm{x}}+{\left({(2+\sqrt{3})}^{-1}\right)}^{\mathrm{x}}=5$$$${\left({(2+\sqrt{3})}^{\mathrm{x}}\right)}^3-5{\left({(2+\sqrt{3})}^{\mathrm{x}}\right)}^2+6{(2+\sqrt{3})}^{\mathrm{x}}+{\left({(2+\sqrt{3})}^{\mathrm{x}}\right)}^{-1}=5$$$$\mathrm{Let}{(2+\sqrt{3})}^{\mathrm{x}}=\mathrm{y}$$$$\Rightarrow {\mathrm{y}}^3-{5\mathrm{y}}^2+6\mathrm{y}+{\mathrm{y}}^{-1}=5$$$${\mathrm{y}}^4-{5\mathrm{y}}^3+{6\mathrm{y}}^2-5\mathrm{y}+1=0$$$$⋮$$

Commented by Rasheed.Sindhi last updated on 23/Jan/18

��şίŕ ţħάήķş ғόŕ ςόŕŕέςţίόή!��

Answered by ajfour last updated on 23/Jan/18

$$x=\pm 1$$

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