1-1-x-x-1-3-

Question Number 223474 by fantastic last updated on 26/Jul/25

$$\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+{x}}}=\sqrt[{\mathrm{3}}]{{x}} \\ $$

Commented by fantastic last updated on 26/Jul/25

(√(a+(√(a+x))))=(x)^(1/3) let t=(x)^(1/3) So the equation becomes (√(a+(√(a+t^3 ))))=t a+(√(a+t^3 ))=t^2 a+t^3 =t^4 +a^2 −2at^2

$$\sqrt{{a}+\sqrt{{a}+{x}}}=\sqrt[{\mathrm{3}}]{{x}} \\ $$$${let}\:{t}=\sqrt[{\mathrm{3}}]{{x}} \\ $$$${So}\:{the}\:{equation}\:{becomes} \\ $$$$\sqrt{{a}+\sqrt{{a}+{t}^{\mathrm{3}} }}={t} \\ $$$${a}+\sqrt{{a}+{t}^{\mathrm{3}} }={t}^{\mathrm{2}} \\ $$$${a}+{t}^{\mathrm{3}} ={t}^{\mathrm{4}} +{a}^{\mathrm{2}} −\mathrm{2}{at}^{\mathrm{2}} \\ $$$$ \\ $$

Answered by Ghisom last updated on 26/Jul/25

$$\mathrm{obviously}\:{x}=\mathrm{8} \\ $$$$\mathrm{no}\:\mathrm{other}\:\mathrm{solution} \\ $$

Answered by fantastic last updated on 26/Jul/25

let u=(x)^(1/3) so x=u^3 The equation becomes (√(1+(√(1+u^3 ))))=u or 1+(√(1+u^3 ))=u^2 or u^2 −1=(√(1+u^3 )) or (u^2 −1)^2 =1+u^3 or u^4 +1−2u^2 =1+u^3 or u^4 −2u^2 =u^3 or u^2 (u^2 −2)=u^2 ×u or u^2 −u−2=0 u=((1±(√9))/2)=((1±3)/2) u= 2 or −1 x=u^3 ⇒ 8 or −1 when x=8 (√(1+(√(1+8))))=(√(1+(√9)))=(√(1+3))=(√4)=2=(8)^(1/3) when x=−1 (√(1+(√(1−1))))=(√1)=1≠(8)^(1/3) So x=8✓

$${let}\:{u}=\sqrt[{\mathrm{3}}]{{x}} \\ $$$${so}\:{x}={u}^{\mathrm{3}} \\ $$$${The}\:{equation}\:{becomes} \\ $$$$\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+{u}^{\mathrm{3}} }}={u} \\ $$$${or}\:\mathrm{1}+\sqrt{\mathrm{1}+{u}^{\mathrm{3}} }={u}^{\mathrm{2}} \\ $$$${or}\:{u}^{\mathrm{2}} −\mathrm{1}=\sqrt{\mathrm{1}+{u}^{\mathrm{3}} } \\ $$$${or}\:\left({u}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} =\mathrm{1}+{u}^{\mathrm{3}} \\ $$$${or}\:{u}^{\mathrm{4}} +\cancel{\mathrm{1}}−\mathrm{2}{u}^{\mathrm{2}} =\cancel{\mathrm{1}}+{u}^{\mathrm{3}} \\ $$$${or}\:{u}^{\mathrm{4}} −\mathrm{2}{u}^{\mathrm{2}} ={u}^{\mathrm{3}} \\ $$$${or}\:\cancel{{u}^{\mathrm{2}} }\left({u}^{\mathrm{2}} −\mathrm{2}\right)=\cancel{{u}^{\mathrm{2}} }×{u} \\ $$$${or}\:{u}^{\mathrm{2}} −{u}−\mathrm{2}=\mathrm{0} \\ $$$${u}=\frac{\mathrm{1}\pm\sqrt{\mathrm{9}}}{\mathrm{2}}=\frac{\mathrm{1}\pm\mathrm{3}}{\mathrm{2}} \\ $$$${u}=\:\mathrm{2}\:{or}\:−\mathrm{1} \\ $$$${x}={u}^{\mathrm{3}} \Rightarrow\:\mathrm{8}\:{or}\:−\mathrm{1} \\ $$$${when}\:{x}=\mathrm{8} \\ $$$$\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{8}}}=\sqrt{\mathrm{1}+\sqrt{\mathrm{9}}}=\sqrt{\mathrm{1}+\mathrm{3}}=\sqrt{\mathrm{4}}=\mathrm{2}=\sqrt[{\mathrm{3}}]{\mathrm{8}} \\ $$$${when}\:{x}=−\mathrm{1} \\ $$$$\sqrt{\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{1}}}=\sqrt{\mathrm{1}}=\mathrm{1}\neq\sqrt[{\mathrm{3}}]{\mathrm{8}} \\ $$$${So}\:{x}=\mathrm{8}\checkmark \\ $$

Answered by mr W last updated on 26/Jul/25

say x=t^3 >0 (√(1+(√(1+t^3 ))))=t (√(1+t^3 ))=t^2 −1 1+t^3 =(1+t)^2 (t−1)^2 (1+t)(t^2 −t+1)=(1+t)^2 (t−1)^2 (t+1)t^2 (t−2)=0 t=−1 ⇒rejected t=0 ⇒ rejected t=2 ⇒x=2^3 =8 ✓

$${say}\:{x}={t}^{\mathrm{3}} >\mathrm{0} \\ $$$$\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+{t}^{\mathrm{3}} }}={t} \\ $$$$\sqrt{\mathrm{1}+{t}^{\mathrm{3}} }={t}^{\mathrm{2}} −\mathrm{1} \\ $$$$\mathrm{1}+{t}^{\mathrm{3}} =\left(\mathrm{1}+{t}\right)^{\mathrm{2}} \left({t}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{1}+{t}\right)\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right)=\left(\mathrm{1}+{t}\right)^{\mathrm{2}} \left({t}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\left({t}+\mathrm{1}\right){t}^{\mathrm{2}} \left({t}−\mathrm{2}\right)=\mathrm{0} \\ $$$${t}=−\mathrm{1}\:\Rightarrow{rejected} \\ $$$${t}=\mathrm{0}\:\Rightarrow\:{rejected} \\ $$$${t}=\mathrm{2}\:\Rightarrow{x}=\mathrm{2}^{\mathrm{3}} =\mathrm{8}\:\checkmark \\ $$

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