Question Number 193790 by pascal889 last updated on 20/Jun/23
Answered by Subhi last updated on 20/Jun/23
$$\:\mathrm{2}^{\mathrm{4}} \sqrt{{x}}\:+\frac{\mathrm{2}}{\:^{\mathrm{4}} \sqrt{{x}}}\:=\:\mathrm{5} \\ $$$$\sqrt{{x}}−\frac{\mathrm{5}}{\mathrm{2}}\:^{\mathrm{4}} \sqrt{{x}}\:+\mathrm{1}\:=\:\mathrm{0} \\ $$$$\:^{\mathrm{4}} \sqrt{{x}}\:=\:\frac{\frac{\mathrm{5}}{\mathrm{2}}\pm\sqrt{\frac{\mathrm{25}}{\mathrm{4}}−\mathrm{4}}}{\mathrm{2}}=\frac{\frac{\mathrm{5}}{\mathrm{2}}\pm\frac{\mathrm{3}}{\mathrm{2}}}{\mathrm{2}}=\:\frac{\mathrm{1}}{\mathrm{2}}\:{or}\:\mathrm{2} \\ $$$${x}\:=\:\frac{\mathrm{1}}{\mathrm{16}}\:{or}\:\mathrm{16} \\ $$
Answered by deleteduser1 last updated on 20/Jun/23
$$\mathrm{2}\sqrt[{\mathrm{4}}]{{x}}+\frac{\mathrm{2}}{\:\sqrt[{\mathrm{4}}]{{x}}}=\mathrm{5}\Rightarrow{a}+{b}=\mathrm{5};{ab}=\mathrm{4} \\ $$$${a},{b}\:{are}\:{roots}\:{of}\:{p}^{\mathrm{2}} −\mathrm{5}{p}+\mathrm{4}=\mathrm{0}\Rightarrow{a}=\mathrm{4}\:{or}\:{b}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{2}\sqrt[{\mathrm{4}}]{{x}}=\mathrm{4}\Rightarrow\sqrt[{\mathrm{4}}]{{x}}=\mathrm{2}\Rightarrow{x}=\mathrm{16};\:\:\:\:\:\mathrm{2}\sqrt[{\mathrm{4}}]{{x}}=\mathrm{1}\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{16}} \\ $$