Question Number 172282 by BaliramSingh last updated on 25/Jun/22
$${Find}\:{the}\:{all}\:{valid}\:{value}\:{of}\:\:\:{x} \\ $$$${x}^{\mathrm{2}} \:−\:\sqrt{\mathrm{4}+{x}}\:=\:\mathrm{4} \\ $$
Answered by dumitrel last updated on 25/Jun/22
$${x}^{\mathrm{2}} −\mathrm{4}\geqslant\mathrm{0};{x}+\mathrm{4}\geqslant\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\mathrm{4}=\sqrt{\mathrm{4}+{x}}={y} \\ $$$${x}^{\mathrm{2}} −{y}=\mathrm{4} \\ $$$${y}^{\mathrm{2}} −{x}=\mathrm{4}\Rightarrow{x}^{\mathrm{2}} −{y}^{\mathrm{2}} +\left({x}−{y}\right)=\mathrm{0} \\ $$$$\left({x}−{y}\right)\left({x}+{y}+\mathrm{1}\right)=\mathrm{0} \\ $$$${x}={y}\Rightarrow{x}^{\mathrm{2}} −{x}−\mathrm{4}=\mathrm{0}\Rightarrow{x}=\frac{\mathrm{1}+\sqrt{\mathrm{17}}}{\mathrm{2}} \\ $$$${x}+{y}+\mathrm{1}=\mathrm{0}\Rightarrow{x}^{\mathrm{2}} +{x}−\mathrm{3}=\mathrm{0} \\ $$$${x}=\frac{−\mathrm{1}−\sqrt{\mathrm{13}}}{\mathrm{2}} \\ $$
Commented by BaliramSingh last updated on 25/Jun/22
$${Thanks}\:{sir} \\ $$