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If-x-1-5-2-find-x-12-

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Question Number 203264 by hardmath last updated on 13/Jan/24
If x = ((1 + (√5))/2) find: x^(12) = ?
$$\mathrm{If}\:\:\:\:\:\mathrm{x}\:=\:\frac{\mathrm{1}\:+\:\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:\:\:\:\mathrm{find}:\:\mathrm{x}^{\mathrm{12}} \:=\:? \\ $$
Answered by MM42 last updated on 13/Jan/24
{ ((x^2 =((3+(√5))/2))),((x^4 =((7+3(√5))/2))) :} ⇒x^6 =9+4(√5) ⇒ x^(12) =161+72(√5)
$$\begin{cases}{{x}^{\mathrm{2}} =\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}}\\{{x}^{\mathrm{4}} =\frac{\mathrm{7}+\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{2}}}\end{cases}\:\:\Rightarrow{x}^{\mathrm{6}} =\mathrm{9}+\mathrm{4}\sqrt{\mathrm{5}} \\ $$$$\Rightarrow\:{x}^{\mathrm{12}} =\mathrm{161}+\mathrm{72}\sqrt{\mathrm{5}} \\ $$
Commented by hardmath last updated on 13/Jan/24
▲ x = ((1 + (√5))/2) = ϕ ▲ ϕ^n = F_n ϕ + F_(n−1) ϕ^(12) = F_(12) ϕ + F_(11) = 144ϕ + 89 = 144(((1 + (√5))/2)) + 89 = 161 + 72 (√5) ✓
$$\blacktriangle\:\mathrm{x}\:=\:\frac{\mathrm{1}\:+\:\sqrt{\mathrm{5}}}{\mathrm{2}}\:=\:\varphi \\ $$$$\blacktriangle\:\varphi^{\boldsymbol{\mathrm{n}}} \:=\:\mathrm{F}_{\boldsymbol{\mathrm{n}}} \varphi\:+\:\mathrm{F}_{\boldsymbol{\mathrm{n}}−\mathrm{1}} \\ $$$$\varphi^{\mathrm{12}} \:=\:\mathrm{F}_{\mathrm{12}} \varphi\:+\:\mathrm{F}_{\mathrm{11}} \\ $$$$\:\:\:\:\:\:\:\:=\:\mathrm{144}\varphi\:+\:\mathrm{89} \\ $$$$\:\:\:\:\:\:\:\:=\:\mathrm{144}\left(\frac{\mathrm{1}\:+\:\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\:+\:\mathrm{89} \\ $$$$\:\:\:\:\:\:\:\:=\:\mathrm{161}\:+\:\mathrm{72}\:\sqrt{\mathrm{5}}\:\:\:\checkmark \\ $$
Answered by Rasheed.Sindhi last updated on 13/Jan/24
(2x−1)^2 =5 4x^2 −4x−4=0 x^2 =x+1⇒x^3 =x^2 +x=2x+1 ⇒x^4 =2x^2 +x=3x+2 ⇒x^5 =3x^2 +2x=5x+3 ⇒x^6 =5x^2 +3x=8x+5 ⇒x^(12) =64x^2 +80x+25=144x+89 =144(((1+(√5) )/2))+89 =72+72(√5) +89 =161+72(√5)
$$\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{5} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{4}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} ={x}+\mathrm{1}\Rightarrow{x}^{\mathrm{3}} ={x}^{\mathrm{2}} +{x}=\mathrm{2}{x}+\mathrm{1} \\ $$$$\Rightarrow{x}^{\mathrm{4}} =\mathrm{2}{x}^{\mathrm{2}} +{x}=\mathrm{3}{x}+\mathrm{2} \\ $$$$\Rightarrow{x}^{\mathrm{5}} =\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}=\mathrm{5}{x}+\mathrm{3} \\ $$$$\Rightarrow{x}^{\mathrm{6}} =\mathrm{5}{x}^{\mathrm{2}} +\mathrm{3}{x}=\mathrm{8}{x}+\mathrm{5} \\ $$$$\Rightarrow{x}^{\mathrm{12}} =\mathrm{64}{x}^{\mathrm{2}} +\mathrm{80}{x}+\mathrm{25}=\mathrm{144}{x}+\mathrm{89} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{144}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}\:}{\mathrm{2}}\right)+\mathrm{89} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{72}+\mathrm{72}\sqrt{\mathrm{5}}\:+\mathrm{89} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{161}+\mathrm{72}\sqrt{\mathrm{5}}\: \\ $$

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