Question Number 115312 by naka3546 last updated on 25/Sep/20
$${Solve}\:\:{for}\:\:{x}\:\in\:\mathbb{R}\:\:{that}\:\:{suitable}\:\:{on}\:\:{this} \\ $$$${inequality}\::\:\:\:\:\sqrt{\mathrm{8}−{x}^{\mathrm{2}} }\:\:>\:\:{x} \\ $$
Commented by bemath last updated on 25/Sep/20
$${intersection}\:{two}\:{graph} \\ $$$${we}\:{get}\:\mathrm{8}−{x}^{\mathrm{2}} \:=\:{x}^{\mathrm{2}} \:\Rightarrow\:\mathrm{8}=\mathrm{2}{x}^{\mathrm{2}} \\ $$$$\:{x}\:=\:\pm\:\mathrm{2}\:\Rightarrow\:{solution}\:{set}\:{is}\:\:{x}<\:\mathrm{2}. \\ $$
Commented by MJS_new last updated on 25/Sep/20
$$\mathrm{you}\:\mathrm{are}\:\mathrm{right},\:\mathrm{two}\:\mathrm{graphs} \\ $$$$\mathrm{but}\:{f}\left({x}\right)=\sqrt{\mathrm{8}−{x}^{\mathrm{2}} }\:\mathrm{is}\:\mathrm{defined}\:\mathrm{for}\:−\mathrm{2}\sqrt{\mathrm{2}}\leqslant{x}\leqslant\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\mathrm{and}\:{f}\left({x}\right)\geqslant\mathrm{0}.\:\mathrm{you}\:\mathrm{are}\:\mathrm{not}\:\mathrm{allowed}\:\mathrm{to}\:\mathrm{square} \\ $$
Answered by john santu last updated on 25/Sep/20
$$\left(\mathrm{1}\right)\:\mathrm{8}−{x}^{\mathrm{2}} \geqslant\mathrm{0}\:\Rightarrow\left(\sqrt{\mathrm{8}}+{x}\right)\left(\sqrt{\mathrm{8}}−{x}\right)\geqslant\mathrm{0} \\ $$$$−\mathrm{2}\sqrt{\mathrm{2}}\:\leqslant\:{x}\:\leqslant\:\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{8}−{x}^{\mathrm{2}} >{x}^{\mathrm{2}} \:\rightarrow\mathrm{4}−{x}^{\mathrm{2}} >\mathrm{0}\: \\ $$$$\:\:\:\:\:\:\left(\mathrm{2}+{x}\right)\left(\mathrm{2}−{x}\right)>\mathrm{0}\: \\ $$$$\:\:\:\:\:\:−\mathrm{2}<{x}<\mathrm{2} \\ $$$${solution}\:{set}\:\left(\mathrm{1}\right)\cap\left(\mathrm{2}\right)\Rightarrow−\mathrm{2}<{x}<\mathrm{2} \\ $$
Commented by naka3546 last updated on 25/Sep/20
$${but}\:\:{why}\:?\:\:{any}\:\:{other}\:\:\:{solution}? \\ $$$${x}\:=\:−\mathrm{2}\sqrt{\mathrm{2}}\:\:{is}\:\:{right}\:\:{answer}. \\ $$
Commented by MJS_new last updated on 25/Sep/20
$$\left(\mathrm{2}\right)\:\mathrm{is}\:\mathrm{wrong}\:\mathrm{for}\:{x}<\mathrm{0} \\ $$
Answered by MJS_new last updated on 25/Sep/20
$$\sqrt{\mathrm{8}−{x}^{\mathrm{2}} }>{x} \\ $$$$\sqrt{\mathrm{8}−{x}^{\mathrm{2}} }\:\mathrm{defined}\:\mathrm{for}\:−\mathrm{2}\sqrt{\mathrm{2}}\leqslant{x}\leqslant\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\sqrt{\mathrm{8}−{x}^{\mathrm{2}} }\geqslant\mathrm{0}\:\Rightarrow\:−\mathrm{2}\sqrt{\mathrm{2}}\leqslant{x}\leqslant\mathrm{0}\:\mathrm{is}\:\mathrm{a}\:\mathrm{part}\:\mathrm{of}\:\mathrm{the} \\ $$$$\:\:\:\:\:\mathrm{solution} \\ $$$$\mathrm{for}\:{x}>\mathrm{0}\:\mathrm{we}\:\mathrm{are}\:\mathrm{allowed}\:\mathrm{to}\:\mathrm{square} \\ $$$$\mathrm{8}−{x}^{\mathrm{2}} >{x}^{\mathrm{2}} \wedge{x}>\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{0}<{x}<\mathrm{2} \\ $$$$ \\ $$$$\Rightarrow\:\mathrm{solution}\:\mathrm{is}\:−\mathrm{2}\sqrt{\mathrm{2}}\leqslant{x}<\mathrm{2} \\ $$
Answered by mr W last updated on 25/Sep/20
$$\mathrm{8}−{x}^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$$\Rightarrow−\mathrm{2}\sqrt{\mathrm{2}}\leqslant{x}\leqslant\mathrm{2}\sqrt{\mathrm{2}} \\ $$$${for}\:−\mathrm{2}\sqrt{\mathrm{2}}\leqslant{x}\leqslant\mathrm{0}: \\ $$$$\sqrt{\mathrm{8}−{x}^{\mathrm{2}} }>{x}\:{is}\:{always}\:{true}. \\ $$$${for}\:\mathrm{0}<{x}\leqslant\mathrm{2}\sqrt{\mathrm{2}}: \\ $$$$\mathrm{8}−{x}^{\mathrm{2}} >{x}^{\mathrm{2}} \\ $$$$\Rightarrow{x}<\mathrm{2} \\ $$$$ \\ $$$${answer}: \\ $$$$−\mathrm{2}\sqrt{\mathrm{2}}\leqslant{x}<\mathrm{2} \\ $$