Question Number 178251 by mr W last updated on 14/Oct/22
$${solve} \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{5}\right)^{\mathrm{2}} −{x}=\mathrm{5} \\ $$
Answered by BaliramKumar last updated on 14/Oct/22
$$\left({x}^{\mathrm{2}} −\mathrm{5}\right)^{\mathrm{2}} −{x}=\mathrm{5} \\ $$$$ \\ $$$${x}^{\mathrm{2}} −\mathrm{5}\:=\:\sqrt{\mathrm{5}+{x}} \\ $$$${x}^{\mathrm{2}} \:=\:\mathrm{5}+\sqrt{\mathrm{5}+{x}} \\ $$$${x}\:=\:\sqrt{\mathrm{5}+\sqrt{\mathrm{5}+{x}}} \\ $$$${x}\:=\:\sqrt{\mathrm{5}+\sqrt{\mathrm{5}+\sqrt{\mathrm{5}+………..\infty}}} \\ $$$${x}\:=\:\sqrt{\mathrm{5}+{x}} \\ $$$${x}^{\mathrm{2}} −{x}−\mathrm{5}\:=\:\mathrm{0} \\ $$$${x}\:=\:\frac{\mathrm{1}\pm\sqrt{\mathrm{21}}}{\mathrm{2}} \\ $$
Commented by mr W last updated on 14/Oct/22
$${thanks}! \\ $$
Answered by Frix last updated on 14/Oct/22
$$\left({x}^{\mathrm{2}} −\mathrm{5}\right)^{\mathrm{2}} −{x}−\mathrm{5}=\mathrm{0} \\ $$$${x}^{\mathrm{4}} −\mathrm{10}{x}^{\mathrm{2}} −{x}+\mathrm{20}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} +{x}−\mathrm{4}\right)\left({x}^{\mathrm{2}} −{x}−\mathrm{5}\right)=\mathrm{0} \\ $$$${x}=−\frac{\mathrm{1}\pm\sqrt{\mathrm{17}}}{\mathrm{2}}\vee{x}=\frac{\mathrm{1}\pm\sqrt{\mathrm{21}}}{\mathrm{2}} \\ $$
Commented by mr W last updated on 14/Oct/22
$${thanks}! \\ $$
Answered by mr W last updated on 14/Oct/22
$$\left({x}^{\mathrm{2}} −\mathrm{5}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} =\mathrm{5}+{x}−{x}^{\mathrm{2}} \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{5}+{x}\right)\left({x}^{\mathrm{2}} −\mathrm{5}−{x}\right)=−\left({x}^{\mathrm{2}} −\mathrm{5}−{x}\right) \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{4}+{x}\right)\left({x}^{\mathrm{2}} −\mathrm{5}−{x}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −\mathrm{4}+{x}=\mathrm{0}\:\Rightarrow{x}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{17}}}{\mathrm{2}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −\mathrm{5}−{x}=\mathrm{0}\:\Rightarrow{x}=\frac{\mathrm{1}\pm\sqrt{\mathrm{21}}}{\mathrm{2}} \\ $$