Question Number 170849 by mathlove last updated on 01/Jun/22
$$\mathrm{2}{x}+\sqrt{{x}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{8}{x}+\frac{\mathrm{1}}{\:\sqrt{{x}}}=? \\ $$
Answered by MJS_new last updated on 01/Jun/22
$$\sqrt{{x}}\in\mathbb{R}\:\Rightarrow\:{x}\leqslant\mathrm{0} \\ $$$$\mathrm{let}\:\sqrt{{x}}={t}\geqslant\mathrm{0} \\ $$$$\mathrm{2}{t}^{\mathrm{2}} +{t}−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0}\:\Rightarrow\:{t}=\sqrt{{x}}=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}\:\Rightarrow\:{x}=\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{8}} \\ $$$$\Rightarrow \\ $$$$\mathrm{8}{x}+\frac{\mathrm{1}}{\:\sqrt{{x}}}=\mathrm{4} \\ $$