Question Number 184305 by HeferH last updated on 05/Jan/23
$${if}:\:{f}\left({x}\right)\:=\:{x}^{\mathrm{2}} \:+\:\mathrm{2}{x} \\ $$$${find}\:{x}: \\ $$$$\:{f}\left({f}\left({f}\left({x}\:+\:\mathrm{2}\right)\right)\right)\:=\:\mathrm{99}\:\mathrm{999}\:\mathrm{999}\: \\ $$
Answered by Frix last updated on 05/Jan/23
![x=t−3 f(x)=(t−3)(t−1) f(x+2)=t^2 −1 f(f(x+2))=t^4 −1 f(f(f(x+2)))=t^8 −1 t^8 =10^8 t=±10∨t=±10i∨t=±5(√2)(1±i)_([4 solutions])](https://www.tinkutara.com/question/Q184309.png)
$${x}={t}−\mathrm{3} \\ $$$${f}\left({x}\right)=\left({t}−\mathrm{3}\right)\left({t}−\mathrm{1}\right) \\ $$$${f}\left({x}+\mathrm{2}\right)={t}^{\mathrm{2}} −\mathrm{1} \\ $$$${f}\left({f}\left({x}+\mathrm{2}\right)\right)={t}^{\mathrm{4}} −\mathrm{1} \\ $$$${f}\left({f}\left({f}\left({x}+\mathrm{2}\right)\right)\right)={t}^{\mathrm{8}} −\mathrm{1} \\ $$$${t}^{\mathrm{8}} =\mathrm{10}^{\mathrm{8}} \\ $$$${t}=\pm\mathrm{10}\vee{t}=\pm\mathrm{10i}\vee\underset{\left[\mathrm{4}\:\mathrm{solutions}\right]} {{t}=\pm\mathrm{5}\sqrt{\mathrm{2}}\left(\mathrm{1}\pm\mathrm{i}\right)} \\ $$