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Question Number 95159 by i jagooll last updated on 23/May/20

$$\sqrt[3]{8-\mathrm{x}}+\sqrt{\mathrm{x}}=2$$

Commented by hknkrc46 last updated on 23/May/20

$$★\sqrt[2n+1]{f\left(x\right)}\Rightarrow \mathrm{\forall }f\left(x\right)\in \mathbb{R}\to f\left(x\right)⩾0\vee f\left(x\right)⩽0$$$$★\sqrt[2n]{f\left(x\right)}\Rightarrow \mathrm{\forall }f\left(x\right)\in {\mathbb{R}}^{+}\to f\left(x\right)⩾0$$$$★\sqrt[2]{f\left(x\right)}=\sqrt{f\left(x\right)}$$$$\bullet x⩾0$$$$★m\ne k,h=ekok(m,k)=lcm(m,k)$$$$\sqrt[m]{f\left(x\right)}+\sqrt[k]{g\left(x\right)}=t\Rightarrow f\left(x\right)=q^h$$$$\bullet h=ekok(3,2)=6$$$$\bullet f\left(x\right)=8-x=q^6$$$$\bullet \sqrt[3]{8-x}+\sqrt{x}=q^2+\sqrt{8-q^6}=2$$$$\bullet \sqrt{8-q^6}=2-q^2$$$$\bullet 8-q^6={(2-q^2)}^2=q^4-4q^2+4$$$$\bullet q^6+q^4-4q^2-4=0$$$$\bullet q^2(q^4-4)+q^4-4=(q^2+1)(q^4-4)$$$$=(q^2+1)(q^2-2)(q^2+2)\Rightarrow q=\pm \sqrt{2},\pm i\sqrt{2},\pm i$$$$\bullet 8-x={(\pm \sqrt{2})}^6=8\Rightarrow x=0$$$$\bullet 8-x={(\pm i\sqrt{2})}^6=-8\Rightarrow x=16$$$$\bullet 8-x={(\pm i)}^6=-1\Rightarrow x=9$$$$$$

Answered by bobhans last updated on 23/May/20

$$\sqrt[3]{8-x}=2-\sqrt{x}\Rightarrow 8-x=8-12\sqrt{x}+6x-x\sqrt{x}$$$$x\sqrt{x}+12\sqrt{x}-7x=0$$$$\sqrt{x}(x+12-7\sqrt{x})=0$$$$\sqrt{x}(\sqrt{x}-3)(\sqrt{x}-4)=0\Rightarrow \{\begin{array}{l}x=0\\ x=9\\ x=16\end{array}$$

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