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Solve the Equation in real/Complex numbers:

Solve $$\left(\sqrt{\sqrt{2}-4-x}\right)+x^{\frac{1}{4}}=2^{\frac{-1}{4}}$$

My try:

Letting $x=t^4$ we get

We get

$$\left(\sqrt{\sqrt{2}-4-t^4}\right)+|t|=2^{\frac{-1}{4}}$$

Then:

$$\left(\sqrt{\sqrt{2}-4-t^4}\right)=2^{\frac{-1}{4}}-|t|$$

Squaring we get:

$$\sqrt{2}-4-t^4=t^2+\frac{1}{\sqrt{2}}-|t|2^{\frac{3}{4}}$$

$\implies$

we get

$$t^4+t^2+4=\frac{1}{\sqrt{2}}+|t|2^{\frac{3}{4}}$$

Any clue here?

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    $\begingroup$ Wolfram Alpha gives me awful-looking solutions. Are you sure the problem is correctly stated? $\endgroup$ – greenturtle3141 Apr 27 at 4:16
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One solution is given by $$x=\frac{1}{16} \left(8 \sqrt{2}-56\right)+\frac{1}{2} \sqrt{-5-\frac{14 \sqrt{2}}{3}+\frac{713}{3 \sqrt[3]{16569-3994 \sqrt{2}+2 i \sqrt{40035168-28289088 \sqrt{2}}}}-\frac{52 \sqrt{2}}{\sqrt[3]{16569-3994 \sqrt{2}+2 i \sqrt{40035168-28289088 \sqrt{2}}}}+\frac{1}{3} \sqrt[3]{16569-3994 \sqrt{2}+2 i \sqrt{40035168-28289088 \sqrt{2}}}}-\frac{1}{2} \sqrt{5+\frac{14 \sqrt{2}}{3}+\frac{3}{64} \left(56-8 \sqrt{2}\right)^2+\frac{1}{2} \left(56 \sqrt{2}-336\right)-\frac{713}{3 \sqrt[3]{16569-3994 \sqrt{2}+2 i \sqrt{40035168-28289088 \sqrt{2}}}}+\frac{52 \sqrt{2}}{\sqrt[3]{16569-3994 \sqrt{2}+2 i \sqrt{40035168-28289088 \sqrt{2}}}}-\frac{1}{3} \sqrt[3]{16569-3994 \sqrt{2}+2 i \sqrt{40035168-28289088 \sqrt{2}}}+\frac{\frac{1}{4} \left(56-8 \sqrt{2}\right) \left(336-56 \sqrt{2}-\frac{1}{16} \left(56-8 \sqrt{2}\right)^2\right)-2 \left(924-228 \sqrt{2}\right)}{4 \sqrt{-5-\frac{14 \sqrt{2}}{3}+\frac{713}{3 \sqrt[3]{16569-3994 \sqrt{2}+2 i \sqrt{40035168-28289088 \sqrt{2}}}}-\frac{52 \sqrt{2}}{\sqrt[3]{16569-3994 \sqrt{2}+2 i \sqrt{40035168-28289088 \sqrt{2}}}}+\frac{1}{3} \sqrt[3]{16569-3994 \sqrt{2}+2 i \sqrt{40035168-28289088 \sqrt{2}}}}}}$$

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  • $\begingroup$ Too long is not the best qualifier. $\endgroup$ – Claude Leibovici Apr 27 at 16:07
  • $\begingroup$ Yes thank you, i mean too long! $\endgroup$ – Dr. Sonnhard Graubner Apr 27 at 16:09
  • $\begingroup$ To all downvoters The term is too long for posting as a comment! $\endgroup$ – Dr. Sonnhard Graubner Apr 27 at 16:09
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As said in comments, I seriously wonder it the problem is correctly stated.

This equation shows two complex roots. Using Newton method, we have the awful iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & +1.00000+1.00000\, i \\ 1 & -3.58944+1.23135\, i \\ 2 & -0.75253+0.37882\, i \\ 3 & -1.89384+0.00982\, i \\ 4 & -1.89949+0.01512\, i \\ 5 & -1.89949+0.01510\, i \end{array} \right)$$ $$\left( \begin{array}{cc} n & x_n \\ 0 & +1.00000-1.00000\, i \\ 1 & -3.58944-1.23135\, i \\ 2 & -0.75253-0.37882\, i \\ 3 & -1.89384-0.00982\,i \\ 4 & -1.89949-0.01512\, i \\ 5 & -1.89949-0.01510\, i \end{array} \right)$$

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