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Question Number 177873 by mr W last updated on 10/Oct/22

$$solveforx\in \mathbb{R}$$$$\sqrt{3+\sqrt{3+\sqrt{3+\sqrt{3+\sqrt{3+x}}}}}=x$$

Answered by Frix last updated on 10/Oct/22

$$\mathrm{for}\mathrm{given}a⩾0\mathrm{we}\mathrm{need}x\mathrm{so}\mathrm{that}$$$$\sqrt{a+x}=x\Rightarrow x=\frac{1+\sqrt{4a+1}}{2}$$$$\mathrm{in}\mathrm{this}\mathrm{case}a=3\Rightarrow x=\frac{1+\sqrt{13}}{2}$$

Commented by mr W last updated on 10/Oct/22

$$thanksforsolving!$$

Commented by Frix last updated on 10/Oct/22

$$\mathrm{you}\mathrm{are}\mathrm{welcome}$$$$\mathrm{at}\mathrm{first}\mathrm{I}\mathrm{had}\mathrm{been}\mathrm{confused}\mathrm{to}\mathrm{get}\mathrm{the}\mathrm{same}$$$$\mathrm{solution}\mathrm{for}\mathrm{all}\mathrm{these}$$$$\sqrt{3+x}=x$$$$\sqrt{3+\sqrt{3+x}}=x$$$$...$$$$\sqrt{3+\sqrt{3\underset{\left(\mathrm{n}\mathrm{times}\right)}{+...+}\sqrt{3+x}}}=x$$$$\mathrm{but}\mathrm{then}\mathrm{I}\mathrm{saw}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{obvious}$$

Answered by a.lgnaoui last updated on 11/Oct/22

$$3+\sqrt{3+\sqrt{3+\sqrt{3+\sqrt{3+x}}}}=x^2$$$$3+\sqrt{3+\sqrt{3+\sqrt{3+x}}}={(x^2-3)}^2$$$$\sqrt{3+\sqrt{3+\sqrt{3+x}}}={(x^2-3)}^2-3$$$$3+\sqrt{3+\sqrt{3+x})}={[{(x^2-3)}^2-3]}^2$$$$\sqrt{3+\sqrt{3+x}}={[{(x^2-3)}^2-3]}^2-3$$$${3+\sqrt{x+3}={[{(x^2-3)}^2-3)}^2-3]}^2$$$${x{+3={[{(x^2-3)}^2-3)}^2-3]}^2-3]}^2$$$${x^2{{={[{(x^2-3)}^2-3)}^2-3)}^2-3)}^2-3]}^2$$$${x^2-{3{={[{(x^2-3)}^2-3)}^2-3)}^2-3)}^2-3)]}^2-3$$$$posonst={(x^2-3)}^2$$$${{({t{-3)}^2-3)}^2-3)}^2-3)}^2$$$${{({t^2-6t+6)}^2-3)}^2-3)}^2-3$$$${t^4{-24t^3+48t^2-72t+30)}^2-3)}^2-3$$$${[{(t-0,57541)}^2-3)}^2]-\stackrel{}{3}$$$${x^2-3=\sqrt{t}={(t-0,57541)}^4-6(t-0,57541)]}^2+6$$$${t={[t-0,57541)}^4-6{(t-0,57541)}^2+6]}^2$$$$.......$$$$tocomplete...$$

Commented by a.lgnaoui last updated on 11/Oct/22

$${[{(t-0,56541)}^4-6{(t-0,57541)}^2+6]}^2-(t-0,57541)+0,57541=0$$$$(Z^4-6Z^2)-Z+0,57541=0$$$$$$$${(x^2-3)}^2=t{x}^2=\sqrt{t^{}}+3$$$$$$

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