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Question Number 186662 by pascal889 last updated on 08/Feb/23

Answered by Frix last updated on 08/Feb/23

$$u=\sqrt{x+1}⩾0\iff x=u^2-1$$$$\frac{u^2}{\sqrt{u+2}}=2u^2-1$$$$v=\sqrt{u+2}⩾\sqrt{2}\iff u=v^2-1$$$$\frac{{(v^2-2)}^2}{v}=2v^4-8v^2+7$$$$v^5-\frac{v^4}{2}-4v^3+2v^2+\frac{7v}{2}-2=0$$$$(v-1)(v^2+v-1)(v^2-\frac{v}{2}-2)=0$$$$v⩾\sqrt{2}\Rightarrow v=\frac{1+\sqrt{33}}{4}\Rightarrow u=\frac{1+\sqrt{33}}{8}\Rightarrow$$$$x=\frac{-15+\sqrt{33}}{32}$$

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