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Question Number 158065 by zainaltanjung last updated on 30/Oct/21

Answered by Rasheed.Sindhi last updated on 30/Oct/21

(√(2+2(√(3/4)))) =(√(2+(√3))) =p+q(√3) =2+(√3) =p^2 +3q^2 +2pq(√3) p^2 +3q^2 =2 ∧ 2pq=1 ((1/(2q)))^2 +3q^2 −2=0 12q^4 −8q^2 +1=0 (2q^2 −1)(6q^2 −1)=0 q=±(1/( (√2))),±(1/( (√6))) =±((√2)/2),±((√6)/6) ⇒p=(1/(2q))=(1/(2(±((√2)/2)))),(1/(2(±((√6)/6)))) p=±((√2)/2),±((3(√6))/( 6)) p+q(√3)=±((√2)/2)±((√2)/2)((√3) ) =±((√2)/2)±((√6)/2)=±(((√2) +(√6))/2) p+q(√3) =±((3(√6))/( 6))±((√6)/6)((√3) )=±((3(√6) +3(√2))/6) =±(((√6) +(√2))/2) (same)

$$\:\:\:\:\:\:\:\sqrt{\mathrm{2}+\mathrm{2}\sqrt{\frac{\mathrm{3}}{\mathrm{4}}}} \\ $$$$=\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\:={p}+{q}\sqrt{\mathrm{3}} \\ $$$$=\mathrm{2}+\sqrt{\mathrm{3}}\:={p}^{\mathrm{2}} +\mathrm{3}{q}^{\mathrm{2}} +\mathrm{2}{pq}\sqrt{\mathrm{3}} \\ $$$$\:\:\:\:{p}^{\mathrm{2}} +\mathrm{3}{q}^{\mathrm{2}} =\mathrm{2}\:\wedge\:\mathrm{2}{pq}=\mathrm{1} \\ $$$$\:\:\:\:\left(\frac{\mathrm{1}}{\mathrm{2}{q}}\right)^{\mathrm{2}} +\mathrm{3}{q}^{\mathrm{2}} −\mathrm{2}=\mathrm{0} \\ $$$$\:\:\:\:\mathrm{12}{q}^{\mathrm{4}} −\mathrm{8}{q}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\left(\mathrm{2}{q}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{6}{q}^{\mathrm{2}} −\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\:\:\:{q}=\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}},\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}} \\ $$$$\:\:\:\:\:\:=\pm\frac{\sqrt{\mathrm{2}}}{\mathrm{2}},\pm\frac{\sqrt{\mathrm{6}}}{\mathrm{6}} \\ $$$$\Rightarrow{p}=\frac{\mathrm{1}}{\mathrm{2}{q}}=\frac{\mathrm{1}}{\mathrm{2}\left(\pm\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)},\frac{\mathrm{1}}{\mathrm{2}\left(\pm\frac{\sqrt{\mathrm{6}}}{\mathrm{6}}\right)} \\ $$$$\:\:\:\:\:{p}=\pm\frac{\sqrt{\mathrm{2}}}{\mathrm{2}},\pm\frac{\mathrm{3}\sqrt{\mathrm{6}}}{\:\mathrm{6}} \\ $$$$\:\:\:{p}+{q}\sqrt{\mathrm{3}}=\pm\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\pm\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}=\pm\frac{\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{6}}}{\mathrm{2}} \\ $$$${p}+{q}\sqrt{\mathrm{3}}\:=\pm\frac{\mathrm{3}\sqrt{\mathrm{6}}}{\:\mathrm{6}}\pm\frac{\sqrt{\mathrm{6}}}{\mathrm{6}}\left(\sqrt{\mathrm{3}}\:\right)=\pm\frac{\mathrm{3}\sqrt{\mathrm{6}}\:+\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{6}} \\ $$$$\:\:\:\:\:=\pm\frac{\sqrt{\mathrm{6}}\:+\sqrt{\mathrm{2}}}{\mathrm{2}}\:\left({same}\right) \\ $$

Commented by zainaltanjung last updated on 31/Oct/21

Ok. you are right brotber

$$\mathrm{Ok}.\:\mathrm{you}\:\mathrm{are}\:\mathrm{right}\:\mathrm{brotber} \\ $$

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