Question Number 99227 by Algoritm last updated on 19/Jun/20
Commented by Algoritm last updated on 19/Jun/20
$$\mathrm{thank}\:\mathrm{you} \\ $$
Commented by MJS last updated on 19/Jun/20
$$\left(\mathrm{I}\right)\:\:{x}\sqrt{\mathrm{10}−{x}^{\mathrm{2}} }={x}^{\mathrm{2}} −\mathrm{6} \\ $$$${x}^{\mathrm{2}} \left(\mathrm{10}−{x}^{\mathrm{2}} \right)=\left({x}^{\mathrm{2}} −\mathrm{6}\right)^{\mathrm{2}} \\ $$$$−\mathrm{2}{x}^{\mathrm{4}} +\mathrm{22}{x}^{\mathrm{2}} −\mathrm{36}=\mathrm{0} \\ $$$${x}^{\mathrm{4}} −\mathrm{11}{x}^{\mathrm{2}} −\mathrm{18}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{9}\right)\left({x}^{\mathrm{2}} −\mathrm{2}\right)=\mathrm{0} \\ $$$$\mathrm{now}\:\mathrm{test}\:\mathrm{all}\:\mathrm{solutions}\:\mathrm{if}\:\mathrm{they}\:\mathrm{fit}\:\left(\mathrm{I}\right) \\ $$$$\Rightarrow\:{x}_{\mathrm{1}} =−\sqrt{\mathrm{2}}\wedge{x}_{\mathrm{2}} =\mathrm{3} \\ $$
Answered by MJS last updated on 19/Jun/20
$$−\sqrt{\mathrm{2}}<{x}<\mathrm{3} \\ $$