Question Number 154148 by mathdanisur last updated on 14/Sep/21
$$\boldsymbol{{z}}^{\mathrm{4}} \:-\:\frac{\mathrm{50}}{\mathrm{2}\boldsymbol{{z}}^{\mathrm{4}} \:-\:\mathrm{7}}\:=\:\mathrm{14}\:\:\:\Rightarrow\:\:\:\boldsymbol{{z}}\:=\:? \\ $$
Answered by puissant last updated on 14/Sep/21
$$\Rightarrow\:\mathrm{2}{z}^{\mathrm{8}} −\mathrm{7}{z}^{\mathrm{4}} −\mathrm{50}=\mathrm{28}{z}^{\mathrm{4}} −\mathrm{98} \\ $$$$\Rightarrow\:\mathrm{2}{z}^{\mathrm{8}} −\mathrm{35}{z}^{\mathrm{4}} +\mathrm{48}=\mathrm{0} \\ $$$${z}^{\mathrm{4}} ={x}\:\rightarrow\:\mathrm{2}{x}^{\mathrm{2}} −\mathrm{35}{x}+\mathrm{48}=\mathrm{0} \\ $$$$\Delta=\mathrm{841}\:\rightarrow\:\sqrt{\Delta}=\mathrm{29}.. \\ $$$${x}=\frac{\mathrm{3}}{\mathrm{2}}\:{or}\:{x}=\mathrm{16} \\ $$$${z}^{\mathrm{4}} =\frac{\mathrm{3}}{\mathrm{2}}\:\Rightarrow\:{z}=\pm\sqrt[{\mathrm{4}}]{\frac{\mathrm{3}}{\mathrm{2}}}.. \\ $$$${or}\:{z}^{\mathrm{4}} =\mathrm{16}\:\Rightarrow\:{z}=\pm\mathrm{2}… \\ $$
Commented by mathdanisur last updated on 14/Sep/21
$$\mathrm{very}\:\mathrm{nice}\:\mathrm{Ser}\:\mathrm{thank}\:\mathrm{you} \\ $$