Question Number 82719 by john santu last updated on 23/Feb/20
$$\mathrm{If}\:\mathrm{x},{y}\:\in\mathbb{R}\:{satisfy}\:{in}\:{equation}\: \\ $$$${x}−\mathrm{4}\sqrt{{y}}\:=\:\mathrm{2}\sqrt{{x}−{y}}\:.\:{find}\:{range}\:{of}\:{x} \\ $$
Commented by MJS last updated on 23/Feb/20
$$\mathrm{4}\leqslant{x}\leqslant\mathrm{20} \\ $$$$\mathrm{I}\:\mathrm{will}\:\mathrm{show}\:\mathrm{later} \\ $$$$\mathrm{for}\:\mathrm{4}\leqslant{x}<\mathrm{16}\:\mathrm{we}\:\mathrm{get}\:\mathrm{1}\:\mathrm{solution}\:\mathrm{for}\:{y} \\ $$$$\mathrm{for}\:\mathrm{16}\leqslant{x}\leqslant\mathrm{20}\:\mathrm{we}\:\mathrm{get}\:\mathrm{2}\:\mathrm{solutions}\:\mathrm{for}\:{y} \\ $$
Commented by TANMAY PANACEA last updated on 23/Feb/20
$${thank}\:{you}\:{sir}… \\ $$
Commented by john santu last updated on 24/Feb/20
$${thank}\:{you}\:{all} \\ $$
Answered by TANMAY PANACEA last updated on 23/Feb/20
$$\left.\mathrm{1}\right){x}\geqslant{y} \\ $$$$\left.\mathrm{2}\right){y}\geqslant\mathrm{0} \\ $$$$\bigstar{now}\:{if}\:{y}=\mathrm{0} \\ $$$${x}=\mathrm{2}\sqrt{{x}}\: \\ $$$$\sqrt{{x}}\:\left(\sqrt{{x}}\:−\mathrm{2}\right)=\mathrm{0} \\ $$$${so}\:{either}\:{x}=\mathrm{0}\:\:\:{or}\:{x}=\mathrm{4} \\ $$$${when}\:{x}=\mathrm{0}\:{then}\mathrm{0}\:\:{y}=\mathrm{0} \\ $$$$\bigstar{when}\:{x}={y} \\ $$$${x}−\mathrm{4}\sqrt{{x}}\:=\mathrm{0} \\ $$$$\sqrt{{x}}\:\left(\sqrt{{x}}\:−\mathrm{4}\right)=\mathrm{0} \\ $$$${x}=\mathrm{0}\:\:{or}\:{x}=\mathrm{16} \\ $$$${when}\:{x}=\mathrm{0}\:\:{then}\:{y}=\mathrm{0} \\ $$$${let}\:{others}\:{throw}\:{light}\:{in}\left[{this}\:{problem}\right. \\ $$
Answered by mr W last updated on 23/Feb/20
$${y}\geqslant\mathrm{0} \\ $$$${x}\geqslant{y}\geqslant\mathrm{0} \\ $$$${x}\geqslant\mathrm{4}\sqrt{{y}} \\ $$$$ \\ $$$${x}^{\mathrm{2}} −\mathrm{8}{x}\sqrt{{y}}+\mathrm{16}{y}=\mathrm{4}\left({x}−{y}\right) \\ $$$$\mathrm{20}{y}−\mathrm{8}{x}\sqrt{{y}}+{x}^{\mathrm{2}} −\mathrm{4}{x}=\mathrm{0} \\ $$$$\sqrt{{y}}=\frac{\mathrm{2}{x}\pm\sqrt{{x}\left(\mathrm{20}−{x}\right)}}{\mathrm{10}}\geqslant\mathrm{0} \\ $$$${x}\left(\mathrm{20}−{x}\right)\geqslant\mathrm{0} \\ $$$$\Rightarrow\mathrm{0}\leqslant{x}\leqslant\mathrm{20}\:\:\:…\left({i}\right) \\ $$$$ \\ $$$$\mathrm{2}{x}+\sqrt{{x}\left(\mathrm{20}−{x}\right)}\geqslant\mathrm{0}\:\Rightarrow{always}\:{true} \\ $$$$\mathrm{2}{x}−\sqrt{{x}\left(\mathrm{20}−{x}\right)}\geqslant\mathrm{0} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} \geqslant\mathrm{20}{x}−{x}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} \geqslant\mathrm{4}{x} \\ $$$$\Rightarrow{x}\geqslant\mathrm{4}\:\:\:…\left({ii}\right) \\ $$$${from}\:\left({i}\right)\:{and}\:\left({ii}\right){we}\:{get}\:{the}\:{range}\:{of}\:{x}: \\ $$$$\mathrm{4}\leqslant{x}\leqslant\mathrm{20} \\ $$$$ \\ $$$${x}^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}+\mathrm{2}\sqrt{{y}}\right){x}+\mathrm{20}{y}=\mathrm{0} \\ $$$${x}=\mathrm{2}\left(\mathrm{1}+\mathrm{2}\sqrt{{y}}\right)\pm\mathrm{2}\sqrt{\mathrm{1}+\mathrm{4}\sqrt{{y}}−{y}} \\ $$$$\mathrm{1}+\mathrm{4}\sqrt{{y}}−{y}\geqslant\mathrm{0} \\ $$$$\mathrm{5}−\left(\mathrm{2}−\sqrt{{y}}\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$$\mid\mathrm{2}−\sqrt{{y}}\mid\leqslant\sqrt{\mathrm{5}} \\ $$$$−\sqrt{\mathrm{5}}\leqslant\mathrm{2}−\sqrt{{y}}\leqslant\sqrt{\mathrm{5}} \\ $$$$\sqrt{{y}}\leqslant\mathrm{2}+\sqrt{\mathrm{5}}\:\Rightarrow{y}\leqslant\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} =\mathrm{9}+\mathrm{4}\sqrt{\mathrm{5}} \\ $$$$\sqrt{{y}}\geqslant\mathrm{2}−\sqrt{\mathrm{5}}\:\Rightarrow{always}\:{true}\:{with}\:{y}\geqslant\mathrm{0} \\ $$$${the}\:{range}\:{of}\:{y}\:{is}: \\ $$$$\mathrm{0}\leqslant{y}\leqslant\mathrm{9}+\mathrm{4}\sqrt{\mathrm{5}} \\ $$
Commented by TANMAY PANACEA last updated on 23/Feb/20
$${thank}\:{you}\:{sir}… \\ $$