Question Number 174269 by mr W last updated on 28/Jul/22
$${solve}\:\left({x}^{\mathrm{2}} −\mathrm{3}\right)^{\mathrm{2}} ={x}+\mathrm{3} \\ $$
Answered by Frix last updated on 28/Jul/22
$$\mathrm{obviously}\:{x}=\mathrm{1}\vee{x}=−\mathrm{2}\:\mathrm{the}\:\mathrm{other}\:\mathrm{2}\:\mathrm{are}\:\mathrm{easy} \\ $$
Answered by mr W last updated on 28/Jul/22
$$\left({x}^{\mathrm{2}} −\mathrm{3}\right)^{\mathrm{2}} −\mathrm{3}={x} \\ $$$${let}\:{x}^{\mathrm{2}} −\mathrm{3}={t}\:\:\:…\left({i}\right) \\ $$$$\Rightarrow{t}^{\mathrm{2}} −\mathrm{3}={x}\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)−\left({ii}\right): \\ $$$${x}^{\mathrm{2}} −{t}^{\mathrm{2}} ={t}−{x} \\ $$$$\left({t}−{x}\right)\left({t}+{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{t}−{x}=\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −{x}−\mathrm{3}=\mathrm{0} \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}\pm\sqrt{\mathrm{13}}}{\mathrm{2}} \\ $$$$\Rightarrow{t}+{x}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{x}−\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow{x}=−\mathrm{2}\:{or}\:\mathrm{1} \\ $$
Commented by Tawa11 last updated on 28/Jul/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$