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Question Number 133072 by bemath last updated on 18/Feb/21

$$\mathrm{Show}\mathrm{that}\frac{5}{2-\sqrt[4]{3}}\mathrm{is}\mathrm{in}{\mathrm{F}}_2\mathrm{by}\mathrm{expressing}$$$$\mathrm{the}\mathrm{number}\mathrm{in}\mathrm{form}{a}_1+b_1\sqrt{k_1}\mathrm{where}$$$$a_1,b_1,k_{1}arein{F}_1$$

Answered by EDWIN88 last updated on 18/Feb/21

$$\frac{5}{2-\sqrt[4]{3}}\times \frac{2+\sqrt[4]{3}}{2+\sqrt[4]{3}}=\frac{10+5\sqrt[4]{3}}{4-\sqrt[4]{9}}\times \frac{4+\sqrt[4]{9}}{4+\sqrt[4]{9}}$$$$=\frac{40+10\sqrt[4]{9}+20\sqrt[4]{3}+5\sqrt[4]{27}}{13}$$$$=\frac{40+10\sqrt{3}+20\sqrt[4]{3}+5\sqrt{3}\sqrt[4]{3}}{13}$$$$=\frac{40+10\sqrt{3}}{13}+\left(\frac{20+5\sqrt{3}}{13}\right)\sqrt[4]{3}$$$$\mathrm{then}{a}_1=\frac{40+10\sqrt{3}}{13};b_1=\frac{20+5\sqrt{3}}{13};k_1=\sqrt{3}$$$$$$

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