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Question Number 108309 by bemath last updated on 16/Aug/20

((△BeMath△)/∴) Given (√(5+(√(9+2(√(15)))))) +(√(5−(√(9+2(√(15)))))) = x find the value of (x−(1/x))^2

$$\:\:\:\:\:\frac{\bigtriangleup\mathcal{B}{e}\mathcal{M}{ath}\bigtriangleup}{\therefore} \\ $$$${Given}\:\sqrt{\mathrm{5}+\sqrt{\mathrm{9}+\mathrm{2}\sqrt{\mathrm{15}}}}\:+\sqrt{\mathrm{5}−\sqrt{\mathrm{9}+\mathrm{2}\sqrt{\mathrm{15}}}}\:=\:{x} \\ $$$${find}\:{the}\:{value}\:{of}\:\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} \\ $$

Answered by bobhans last updated on 16/Aug/20

((∈bobhans∈)/≎) ⇒(x−(1/x))^2 =x^2 +(1/x^2 )−2 from the equation ⇒set (√(9+2(√(15))))= ♭ (√(5+♭)) +(√(5−♭)) = x ⇒5+♭+5−♭+2(√(25−♭^2 ))=x^2 10+2(√(25−(9+2(√(15))))) = x^2 10+2(√(16−2(√(15)))) = x^2 10+2((√(15))−1)=x^2 ; x^2 =8+2(√(15)) and (1/x^2 ) = (1/(8+2(√(15)))) ×((8−2(√(15)))/(8−2(√(15)))) = ((8−2(√(15)))/4) (1/x^2 ) = 2−((√(15))/2) . Finally we got (x−(1/x))^2 = 8+2(√(15)) + 2−((√(15))/2) −2 = 8+((3(√(15)))/2) = ((16+3(√(15)))/2)

$$\:\:\frac{\in{bobhans}\in}{\Bumpeq} \\ $$$$\Rightarrow\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\mathrm{2} \\ $$$${from}\:{the}\:{equation}\:\Rightarrow{set}\:\sqrt{\mathrm{9}+\mathrm{2}\sqrt{\mathrm{15}}}=\:\flat \\ $$$$\sqrt{\mathrm{5}+\flat}\:+\sqrt{\mathrm{5}−\flat}\:=\:{x}\:\Rightarrow\mathrm{5}+\flat+\mathrm{5}−\flat+\mathrm{2}\sqrt{\mathrm{25}−\flat^{\mathrm{2}} }={x}^{\mathrm{2}} \\ $$$$\mathrm{10}+\mathrm{2}\sqrt{\mathrm{25}−\left(\mathrm{9}+\mathrm{2}\sqrt{\mathrm{15}}\right)}\:=\:{x}^{\mathrm{2}} \\ $$$$\mathrm{10}+\mathrm{2}\sqrt{\mathrm{16}−\mathrm{2}\sqrt{\mathrm{15}}}\:=\:{x}^{\mathrm{2}} \\ $$$$\mathrm{10}+\mathrm{2}\left(\sqrt{\mathrm{15}}−\mathrm{1}\right)={x}^{\mathrm{2}} \:;\:{x}^{\mathrm{2}} =\mathrm{8}+\mathrm{2}\sqrt{\mathrm{15}} \\ $$$${and}\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{8}+\mathrm{2}\sqrt{\mathrm{15}}}\:×\frac{\mathrm{8}−\mathrm{2}\sqrt{\mathrm{15}}}{\mathrm{8}−\mathrm{2}\sqrt{\mathrm{15}}}\:=\:\frac{\mathrm{8}−\mathrm{2}\sqrt{\mathrm{15}}}{\mathrm{4}} \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:=\:\mathrm{2}−\frac{\sqrt{\mathrm{15}}}{\mathrm{2}}\:.\:{Finally}\:{we}\:{got} \\ $$$$\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} =\:\mathrm{8}+\mathrm{2}\sqrt{\mathrm{15}}\:+\:\mathrm{2}−\frac{\sqrt{\mathrm{15}}}{\mathrm{2}}\:−\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{8}+\frac{\mathrm{3}\sqrt{\mathrm{15}}}{\mathrm{2}}\:=\:\frac{\mathrm{16}+\mathrm{3}\sqrt{\mathrm{15}}}{\mathrm{2}} \\ $$

Answered by 1549442205PVT last updated on 16/Aug/20

Put (√(5+(√(9+2(√(15)))) )) =a,(√(5−(√(9+2(√(15)) )) )) =b Then x=a+b and ab=(√(25−(9+2(√(15))))) =(√(16−2(√(15))))= (√(((√(15))−1)^2 )) =(√(15))−1⇒2ab=2(√(15)) −2 a^2 +b^2 =10⇒x^2 =(a+b)^2 =a^2 +b^2 +2ab =10+2(√(15)) −2=8+2(√(15)) =((√5) +(√3) )^2 ⇒x=a+b=(√5) +(√3) ⇒(1/x)=(1/( (√5)+(√3))) =(((√5) −(√3))/2)⇒(x−(1/x))^2 =((√5)+(√3)−(((√5)−(√3))/2))^2 =((((√5)+3(√3))/2))^2 =((32+6(√(15)))/4)=((16+3(√(15)))/2) Thus,(x−(1/x))^2 =((16+3(√(15)))/2)

$$\mathrm{Put}\:\sqrt{\mathrm{5}+\sqrt{\mathrm{9}+\mathrm{2}\sqrt{\mathrm{15}}}\:}\:=\mathrm{a},\sqrt{\mathrm{5}−\sqrt{\mathrm{9}+\mathrm{2}\sqrt{\mathrm{15}}\:\:}\:}\:=\mathrm{b} \\ $$$$\mathrm{Then}\:\mathrm{x}=\mathrm{a}+\mathrm{b}\:\mathrm{and}\: \\ $$$$\mathrm{ab}=\sqrt{\mathrm{25}−\left(\mathrm{9}+\mathrm{2}\sqrt{\mathrm{15}}\right)}\:=\sqrt{\mathrm{16}−\mathrm{2}\sqrt{\mathrm{15}}}= \\ $$$$\sqrt{\left(\sqrt{\mathrm{15}}−\mathrm{1}\right)^{\mathrm{2}} }\:=\sqrt{\mathrm{15}}−\mathrm{1}\Rightarrow\mathrm{2ab}=\mathrm{2}\sqrt{\mathrm{15}}\:−\mathrm{2} \\ $$$$\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} =\mathrm{10}\Rightarrow\mathrm{x}^{\mathrm{2}} =\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{2}} =\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{2ab} \\ $$$$=\mathrm{10}+\mathrm{2}\sqrt{\mathrm{15}}\:−\mathrm{2}=\mathrm{8}+\mathrm{2}\sqrt{\mathrm{15}}\:=\left(\sqrt{\mathrm{5}}\:+\sqrt{\mathrm{3}}\:\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{a}+\mathrm{b}=\sqrt{\mathrm{5}}\:+\sqrt{\mathrm{3}}\:\Rightarrow\frac{\mathrm{1}}{\mathrm{x}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}+\sqrt{\mathrm{3}}} \\ $$$$=\frac{\sqrt{\mathrm{5}}\:−\sqrt{\mathrm{3}}}{\mathrm{2}}\Rightarrow\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{2}} =\left(\sqrt{\mathrm{5}}+\sqrt{\mathrm{3}}−\frac{\sqrt{\mathrm{5}}−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$=\left(\frac{\sqrt{\mathrm{5}}+\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{32}+\mathrm{6}\sqrt{\mathrm{15}}}{\mathrm{4}}=\frac{\mathrm{16}+\mathrm{3}\sqrt{\mathrm{15}}}{\mathrm{2}} \\ $$$$\mathrm{Thus},\left(\boldsymbol{\mathrm{x}}−\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\right)^{\mathrm{2}} =\frac{\mathrm{16}+\mathrm{3}\sqrt{\mathrm{15}}}{\mathrm{2}}\: \\ $$

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