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Question Number 115544 by bemath last updated on 26/Sep/20

$$Given{x}^2+12\sqrt{x}=5$$$$thenx+2\sqrt{x}?$$

Commented by MJS_new last updated on 26/Sep/20

$$x^2+12\sqrt{x}=5$$$$\mathrm{has}\mathrm{got}1\mathrm{real}\mathrm{and}2\mathrm{complex}\mathrm{solutions}:$$$$x=3-2\sqrt{2}\vee x=-3\pm 4\mathrm{i}$$$$\Rightarrow x+2\sqrt{x}=1\vee x+2\sqrt{x}=-1\pm 8\mathrm{i}$$

Answered by bobhans last updated on 26/Sep/20

$$let\sqrt{x}=w;w⩾0\Rightarrow w^4+12w-5=0$$$$factoring$$$$(w^2+2w-1)(w^2-2w+5)=0$$$$for{w}^2-2w+5>0,\mathrm{\forall }w\in \mathbb{R}$$$$for{w}^2+2w-1=0$$$$w^2+2w=1\Rightarrow {\left(\sqrt{x}\right)}^2+2\sqrt{x}=1$$$$\Rightarrow x+2\sqrt{x}=1$$

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