Question Number 171815 by Mikenice last updated on 21/Jun/22
$${solve}\:{for}\:{x}: \\ $$$$\left(\mathrm{2}−{x}^{\mathrm{2}} \right)^{{x}^{\mathrm{2}} −\mathrm{3}\sqrt{\mathrm{2}\:}\:{x}+\mathrm{4}} =\mathrm{1} \\ $$
Answered by Rasheed.Sindhi last updated on 21/Jun/22
$$\left(\mathrm{2}−{x}^{\mathrm{2}} \right)^{{x}^{\mathrm{2}} −\mathrm{3}\sqrt{\mathrm{2}\:}\:{x}+\mathrm{4}} =\mathrm{1} \\ $$$${case}\mathrm{1}:\:\mathrm{2}−{x}^{\mathrm{2}} =\mathrm{1}\Rightarrow{x}=\pm\mathrm{1} \\ $$$${case}\mathrm{2}:{x}^{\mathrm{2}} −\mathrm{3}\sqrt{\mathrm{2}}\:{x}+\mathrm{4}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:{x}=\frac{\mathrm{3}\sqrt{\mathrm{2}}\:\pm\sqrt{\left(−\mathrm{3}\sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}\right)\left(\mathrm{4}\right)}}{\mathrm{2}\left(\mathrm{1}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:{x}=\frac{\mathrm{3}\sqrt{\mathrm{2}}\:\pm\sqrt{\mathrm{18}−\mathrm{16}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:{x}=\frac{\mathrm{3}\sqrt{\mathrm{2}}\:\pm\sqrt{\mathrm{2}}}{\mathrm{2}}=\mathrm{2}\sqrt{\mathrm{2}}\:,\:\sqrt{\mathrm{2}} \\ $$
Commented by Mikenice last updated on 21/Jun/22
$${please}\:{continue}\:{your}\:{solution} \\ $$
Commented by Mikenice last updated on 21/Jun/22
$${please}\:{continue}\:{your}\:{solution} \\ $$