Question Number 208681 by Noorzai last updated on 20/Jun/24
Answered by Rasheed.Sindhi last updated on 21/Jun/24
$$\sqrt{\mathrm{7}+\sqrt{\mathrm{48}}}\:=\sqrt{\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}}\:={a}+{b}\sqrt{\mathrm{3}}\:\left(>\mathrm{0}\right)\left({let}\right) \\ $$$$\:{Where}\:{a},{b}\in\mathbb{Z} \\ $$$$\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}\:={a}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} +\mathrm{2}{ab}\sqrt{\mathrm{3}}\: \\ $$$${a}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} =\mathrm{7}\:\&\:\mathrm{2}{ab}=\mathrm{4}\Rightarrow{b}=\frac{\mathrm{2}}{{a}} \\ $$$${a}^{\mathrm{2}} +\mathrm{3}\left(\frac{\mathrm{2}}{{a}}\right)^{\mathrm{2}} =\mathrm{7} \\ $$$${a}^{\mathrm{2}} +\frac{\mathrm{12}}{{a}^{\mathrm{2}} }=\mathrm{7} \\ $$$${a}^{\mathrm{4}} −\mathrm{7}{a}^{\mathrm{2}} +\mathrm{12}=\mathrm{0} \\ $$$$\left({a}^{\mathrm{2}} −\mathrm{3}\right)\left({a}^{\mathrm{2}} −\mathrm{4}\right)=\mathrm{0} \\ $$$${a}^{\mathrm{2}} =\mathrm{4}\Rightarrow{a}=\pm\mathrm{2} \\ $$$${b}=\frac{\mathrm{2}}{\pm\mathrm{2}}=\pm\mathrm{1} \\ $$$${a}+{b}\sqrt{\mathrm{3}}\:=\mathrm{2}+\sqrt{\mathrm{3}}\:\:,\:−\mathrm{2}−\sqrt{\mathrm{3}}\:\left(<\mathrm{0}\right) \\ $$$$\therefore\:\:{a}+{b}\sqrt{\mathrm{3}}\:=\mathrm{2}+\sqrt{\mathrm{3}}\: \\ $$
Answered by a.lgnaoui last updated on 20/Jun/24
$$=\sqrt{\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}} \\ $$$$=\sqrt{\left(\mathrm{4}+\mathrm{3}\right)+\mathrm{4}\sqrt{\mathrm{3}}} \\ $$$$=\sqrt{\left(\mathrm{2}+\sqrt{\mathrm{3}}\:\right)^{\mathrm{2}} } \\ $$$$=\mathrm{2}+\sqrt{\mathrm{3}} \\ $$