Question Number 212301 by RojaTaniya last updated on 09/Oct/24
$$\:\:\:\:\mathrm{3}{x}+\frac{\mathrm{2}}{\:\sqrt{{x}}}=\mathrm{1},\:{x}−\sqrt{{x}}\:=? \\ $$$$\:\: \\ $$
Answered by Sutrisno last updated on 09/Oct/24
$${misal}\:\sqrt{{x}}={p}\rightarrow{x}={p}^{\mathrm{2}} \\ $$$$\mathrm{3}{p}^{\mathrm{2}} +\frac{\mathrm{2}}{{p}}=\mathrm{1} \\ $$$$\mathrm{3}{p}^{\mathrm{3}} +\mathrm{2}={p} \\ $$$$\mathrm{3}{p}^{\mathrm{3}} −{p}+\mathrm{2}=\mathrm{0} \\ $$$$\left({p}+\mathrm{1}\right)\left(\mathrm{3}{p}^{\mathrm{2}} −\mathrm{3}{p}+\mathrm{2}\right)=\mathrm{0} \\ $$$${p}=−\mathrm{1} \\ $$$$\sqrt{{x}}=−\mathrm{1} \\ $$$$\mathrm{3}{p}^{\mathrm{2}} −\mathrm{3}{p}+\mathrm{2}=\mathrm{0} \\ $$$${D}<\mathrm{0}\: \\ $$$${tidak}\:{ada}\:{nilai}\:{yang}\:{memenuhi} \\ $$
Answered by Frix last updated on 09/Oct/24
![Without solving: 3x+(2/( (√x)))−1=0 3((√x))^3 −(√x)+2=0 ((√x))^3 −((√x)/3)+(2/3)=0 x−(√x)=y (1+(√x))(x−(√x))=y(1+(√x)) ((√x))^3 −(y+1)(√x)−y=0 y+1=(1/3)∧−y=(2/3) ⇒ y=−(3/3) x−(√x)=−(2/3) [Solving for x we get x=−(1/6)±((√(15))/6)i]](https://www.tinkutara.com/question/Q212303.png)
$$\mathrm{Without}\:\mathrm{solving}: \\ $$$$ \\ $$$$\mathrm{3}{x}+\frac{\mathrm{2}}{\:\sqrt{{x}}}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{3}\left(\sqrt{{x}}\right)^{\mathrm{3}} −\sqrt{{x}}+\mathrm{2}=\mathrm{0} \\ $$$$\left(\sqrt{{x}}\right)^{\mathrm{3}} −\frac{\sqrt{{x}}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}}=\mathrm{0} \\ $$$$ \\ $$$${x}−\sqrt{{x}}={y} \\ $$$$\left(\mathrm{1}+\sqrt{{x}}\right)\left({x}−\sqrt{{x}}\right)={y}\left(\mathrm{1}+\sqrt{{x}}\right) \\ $$$$\left(\sqrt{{x}}\right)^{\mathrm{3}} −\left({y}+\mathrm{1}\right)\sqrt{{x}}−{y}=\mathrm{0} \\ $$$$ \\ $$$${y}+\mathrm{1}=\frac{\mathrm{1}}{\mathrm{3}}\wedge−{y}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\Rightarrow\:{y}=−\frac{\mathrm{3}}{\mathrm{3}} \\ $$$$ \\ $$$${x}−\sqrt{{x}}=−\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$ \\ $$$$\left[\mathrm{Solving}\:\mathrm{for}\:{x}\:\mathrm{we}\:\mathrm{get}\:{x}=−\frac{\mathrm{1}}{\mathrm{6}}\pm\frac{\sqrt{\mathrm{15}}}{\mathrm{6}}\mathrm{i}\right] \\ $$