Question Number 97479 by Hassen_Timol last updated on 08/Jun/20
Commented by Hassen_Timol last updated on 08/Jun/20
$$\mathrm{How}\:\mathrm{can}\:\mathrm{I}\:\mathrm{prove}\:\mathrm{these}\:\mathrm{equalities}\:\mathrm{please}? \\ $$
Commented by john santu last updated on 08/Jun/20
$$\mathrm{in}\:\mathrm{generally}\: \\ $$$$\sqrt{\mathrm{a}+\mathrm{b}+\mathrm{2}\sqrt{\mathrm{ab}}\:}\:=\:\sqrt{\mathrm{a}}\:+\:\sqrt{\mathrm{b}}\: \\ $$$$\sqrt{\mathrm{a}+\mathrm{b}−\mathrm{2}\sqrt{\mathrm{ab}}\:}\:=\:\sqrt{\mathrm{a}}\:−\sqrt{\mathrm{b}}\:;\:\mathrm{a}\:>\:\mathrm{b}\: \\ $$
Answered by Aziztisffola last updated on 08/Jun/20
$$\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}=\mathrm{1}+\mathrm{2}×\mathrm{1}×\sqrt{\mathrm{3}}+\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} =\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$$\sqrt{\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}}=\sqrt{\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }=\mathrm{1}+\sqrt{\mathrm{3}} \\ $$
Commented by Hassen_Timol last updated on 08/Jun/20
$$\mathrm{Thanks}\:\mathrm{a}\:\mathrm{lot}.\:\mathrm{I}\:\mathrm{didn}'\mathrm{t}\:\mathrm{think}\:\mathrm{about}\:\mathrm{that}… \\ $$
Answered by Aziztisffola last updated on 08/Jun/20
$$\left(\sqrt{\mathrm{3}+\sqrt{\mathrm{5}}}−\sqrt{\mathrm{3}−\sqrt{\mathrm{5}}}\right)^{\mathrm{2}} \\ $$$$\:=\mathrm{3}+\sqrt{\mathrm{5}}+\mathrm{3}−\sqrt{\mathrm{5}}−\mathrm{2}\sqrt{\left(\mathrm{3}+\sqrt{\mathrm{5}}\right)\left(\mathrm{3}−\sqrt{\mathrm{5}}\right)} \\ $$$$\:=\mathrm{6}−\mathrm{2}\sqrt{\mathrm{9}−\mathrm{5}}=\mathrm{6}−\mathrm{2}×\mathrm{2}=\mathrm{2} \\ $$$$\sqrt{\mathrm{3}+\sqrt{\mathrm{5}}}−\sqrt{\mathrm{3}−\sqrt{\mathrm{5}}}=\sqrt{\mathrm{2}} \\ $$$$ \\ $$