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Question Number 212301 by RojaTaniya last updated on 09/Oct/24

$$3x+\frac{2}{\sqrt{x}}=1,x-\sqrt{x}=?$$$$$$

Answered by Sutrisno last updated on 09/Oct/24

$$misal\sqrt{x}=p\to x=p^2$$$$3p^2+\frac{2}{p}=1$$$$3p^3+2=p$$$$3p^3-p+2=0$$$$(p+1)(3p^2-3p+2)=0$$$$p=-1$$$$\sqrt{x}=-1$$$$3p^2-3p+2=0$$$$D<0$$$$tidakadanilaiyangmemenuhi$$

Answered by Frix last updated on 09/Oct/24

$$\mathrm{Without}\mathrm{solving}:$$$$$$$$3x+\frac{2}{\sqrt{x}}-1=0$$$$3{\left(\sqrt{x}\right)}^3-\sqrt{x}+2=0$$$${\left(\sqrt{x}\right)}^3-\frac{\sqrt{x}}{3}+\frac{2}{3}=0$$$$$$$$x-\sqrt{x}=y$$$$(1+\sqrt{x})(x-\sqrt{x})=y(1+\sqrt{x})$$$${\left(\sqrt{x}\right)}^3-(y+1)\sqrt{x}-y=0$$$$$$$$y+1=\frac{1}{3}\wedge -y=\frac{2}{3}$$$$\Rightarrow y=-\frac{3}{3}$$$$$$$$x-\sqrt{x}=-\frac{2}{3}$$$$$$$$[\mathrm{Solving}\mathrm{for}x\mathrm{we}\mathrm{get}x=-\frac{1}{6}\pm \frac{\sqrt{15}}{6}\mathrm{i}]$$

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