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Question Number 166921 by cortano1 last updated on 02/Mar/22

Commented by MJS_new last updated on 03/Mar/22

$$\left(2\right)$$$$\sqrt{x}+\sqrt{y}=2u>0[\mathrm{let}u=\frac{1}{4}(\sqrt{5}-1)]$$$$\Rightarrow \sqrt{x}=u-v\mathrm{i}\wedge \sqrt{y}=u+v\mathrm{i}\wedge u,v\in \mathbb{R}$$$$[\mathrm{we}\mathrm{can}\mathrm{let}v⩾0\mathrm{because}\mathrm{of}\mathrm{symmetry}]$$$$\Rightarrow$$$$\left(1\right)$$$$v^4-6u^2{v}^2+u^2(u^2-1)=0$$$$\mathrm{to}\mathrm{get}v⩾0\mathrm{only}\mathrm{one}\mathrm{solution}\mathrm{fits}:$$$$v=\sqrt{3u^2+u\sqrt{8u^2+1}}$$$$\mathrm{now}\mathrm{insert}u=\frac{1}{4}(\sqrt{5}-1)$$$$\mathrm{the}\mathrm{exact}\mathrm{values}\mathrm{are}\mathrm{possible}\mathrm{to}\mathrm{get}\mathrm{but}\mathrm{not}$$$$\mathrm{very}\mathrm{easy}\mathrm{to}\mathrm{handle}...$$$$x\approx .309016994375\pm .834799117309\mathrm{i}$$$$y\approx .309016994375\mp .834799117309\mathrm{i}$$

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