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Question Number 206155 by dimentri last updated on 08/Apr/24
x = 2+(2/(2+(2/(2+(2/(2+(2/(2+(2/x))))))))) find x.
$$\:\:\:\:\mathrm{x}\:=\:\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{2}}{\mathrm{x}}}}}} \\ $$$$\:\:\:\mathrm{find}\:\mathrm{x}.\: \\ $$
Answered by MM42 last updated on 08/Apr/24
2+(2/x)=((2x+2)/x)=a 2+(2/a)=2+(x/(x+1))=((3x+2)/(x+1))=b 2+(2/b)=2+((2x+2)/(3x+2))=((8x+6)/(3x+2))=c 2+(2/c)=2+((6x+4)/(8x+6))=((22x+16)/(8x+6))=((11x+8)/(4x+3))=d x=2+(2/d)=2+((8x+6)/(11x+8))=((30x+22)/(11x+8)) ⇒11x^2 +8x=30x+22 11x^2 −22x−22=0 x=1±(√(3 )) ✓
$$\mathrm{2}+\frac{\mathrm{2}}{{x}}=\frac{\mathrm{2}{x}+\mathrm{2}}{{x}}={a} \\ $$$$\mathrm{2}+\frac{\mathrm{2}}{{a}}=\mathrm{2}+\frac{{x}}{{x}+\mathrm{1}}=\frac{\mathrm{3}{x}+\mathrm{2}}{{x}+\mathrm{1}}={b} \\ $$$$\mathrm{2}+\frac{\mathrm{2}}{{b}}=\mathrm{2}+\frac{\mathrm{2}{x}+\mathrm{2}}{\mathrm{3}{x}+\mathrm{2}}=\frac{\mathrm{8}{x}+\mathrm{6}}{\mathrm{3}{x}+\mathrm{2}}={c} \\ $$$$\mathrm{2}+\frac{\mathrm{2}}{{c}}=\mathrm{2}+\frac{\mathrm{6}{x}+\mathrm{4}}{\mathrm{8}{x}+\mathrm{6}}=\frac{\mathrm{22}{x}+\mathrm{16}}{\mathrm{8}{x}+\mathrm{6}}=\frac{\mathrm{11}{x}+\mathrm{8}}{\mathrm{4}{x}+\mathrm{3}}={d} \\ $$$${x}=\mathrm{2}+\frac{\mathrm{2}}{{d}}=\mathrm{2}+\frac{\mathrm{8}{x}+\mathrm{6}}{\mathrm{11}{x}+\mathrm{8}}=\frac{\mathrm{30}{x}+\mathrm{22}}{\mathrm{11}{x}+\mathrm{8}} \\ $$$$\Rightarrow\mathrm{11}{x}^{\mathrm{2}} +\mathrm{8}{x}=\mathrm{30}{x}+\mathrm{22} \\ $$$$\mathrm{11}{x}^{\mathrm{2}} −\mathrm{22}{x}−\mathrm{22}=\mathrm{0} \\ $$$${x}=\mathrm{1}\pm\sqrt{\mathrm{3}\:}\:\:\checkmark \\ $$$$ \\ $$
Answered by Rasheed.Sindhi last updated on 08/Apr/24
x=2+(2/x)⇒x^2 −2x−2=0 x_(>0) =((2+(√(4+8)))/2) =1+(√3)
$${x}=\mathrm{2}+\frac{\mathrm{2}}{{x}}\Rightarrow{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{2}=\mathrm{0} \\ $$$${x}_{>\mathrm{0}} =\frac{\mathrm{2}+\sqrt{\mathrm{4}+\mathrm{8}}}{\mathrm{2}}\:=\mathrm{1}+\sqrt{\mathrm{3}}\: \\ $$
Commented by cortano21 last updated on 08/Apr/24
no. it not x=2+(2/(2+(2/(2+(2/(2+...))))))
$${no}. \\ $$$${it}\:{not}\:{x}=\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}+…}}} \\ $$
Commented by Frix last updated on 08/Apr/24
x=2+(2/x) insert x=2+(2/x) on the rhs ⇒ x=2+(2/(2+(2/x))) insert x=2+(2/x) on the rhs ⇒ x=2+(2/(2+(2/(2+(2/x))))) ...
$${x}=\mathrm{2}+\frac{\mathrm{2}}{{x}} \\ $$$$\mathrm{insert}\:{x}=\mathrm{2}+\frac{\mathrm{2}}{{x}}\:\mathrm{on}\:\mathrm{the}\:\mathrm{rhs}\:\Rightarrow \\ $$$${x}=\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{2}}{{x}}} \\ $$$$\mathrm{insert}\:{x}=\mathrm{2}+\frac{\mathrm{2}}{{x}}\:\mathrm{on}\:\mathrm{the}\:\mathrm{rhs}\:\Rightarrow \\ $$$${x}=\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{2}}{{x}}}} \\ $$$$… \\ $$
Answered by MATHEMATICSAM last updated on 08/Apr/24
Answered by Rasheed.Sindhi last updated on 08/Apr/24
x = 2+(2/(2+(2/(2+(2/(2+(2/((2x+2)/x)))))))) x = 2+(2/(2+(2/(2+(2/(2+((2x)/(2x+2)))))))) x = 2+(2/(2+(2/(2+(2/(2+(x/(x+1)))))))) x = 2+(2/(2+(2/(2+(2/(2+((3x+2)/(x+1)))))))) x = 2+(2/(2+(2/(2+((2x+2)/(3x+2)))))) x = 2+(2/(2+(2/((8x+6)/(3x+2))))) x = 2+(2/(2+((3x+2)/(4x+3)))) x = 2+(2/((11x+8)/(4x+3))) x = ((22x+16+8x+6)/(11x+8)) x = ((30x+22)/(11x+8)) 11x^2 +8x−30x−22=0 11x^2 −22x−22=0 x^2 −2x−2=0 x=((2+(√(4+8)))/2)=1+(√3)
$$\:\:\:\:\mathrm{x}\:=\:\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{2}}{\frac{\mathrm{2}{x}+\mathrm{2}}{\mathrm{x}}}}}} \\ $$$$\:\:\:\:\mathrm{x}\:=\:\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{2}{x}}{\mathrm{2}{x}+\mathrm{2}}}}} \\ $$$$\:\:\:\:\mathrm{x}\:=\:\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}+\frac{{x}}{{x}+\mathrm{1}}}}} \\ $$$$\:\:\:\:\mathrm{x}\:=\:\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{3}{x}+\mathrm{2}}{{x}+\mathrm{1}}}}} \\ $$$$\:\:\:\:\mathrm{x}\:=\:\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{2}{x}+\mathrm{2}}{\mathrm{3}{x}+\mathrm{2}}}} \\ $$$$\:\:\:\:\mathrm{x}\:=\:\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{2}}{\frac{\mathrm{8}{x}+\mathrm{6}}{\mathrm{3}{x}+\mathrm{2}}}} \\ $$$$\:\:\:\:\mathrm{x}\:=\:\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{3}{x}+\mathrm{2}}{\mathrm{4}{x}+\mathrm{3}}} \\ $$$$\:\:\:\:\mathrm{x}\:=\:\mathrm{2}+\frac{\mathrm{2}}{\frac{\mathrm{11}{x}+\mathrm{8}}{\mathrm{4}{x}+\mathrm{3}}} \\ $$$$\:\:\:\:\mathrm{x}\:=\:\frac{\mathrm{22}{x}+\mathrm{16}+\mathrm{8}{x}+\mathrm{6}}{\mathrm{11}{x}+\mathrm{8}} \\ $$$$\:\:\:\:\mathrm{x}\:=\:\frac{\mathrm{30}{x}+\mathrm{22}}{\mathrm{11}{x}+\mathrm{8}} \\ $$$$\mathrm{11}{x}^{\mathrm{2}} +\mathrm{8}{x}−\mathrm{30}{x}−\mathrm{22}=\mathrm{0} \\ $$$$\mathrm{11}{x}^{\mathrm{2}} −\mathrm{22}{x}−\mathrm{22}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{2}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{2}+\sqrt{\mathrm{4}+\mathrm{8}}}{\mathrm{2}}=\mathrm{1}+\sqrt{\mathrm{3}} \\ $$$$ \\ $$$$ \\ $$
Commented by MM42 last updated on 08/Apr/24
yes.thanks
$${yes}.{thanks}\: \\ $$

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