Question Number 80433 by jagoll last updated on 03/Feb/20
$${find}\:{the}\:{solution}\:{of} \\ $$$$\sqrt{\mathrm{4}−{x}}−\mathrm{2}\leqslant{x}\mid{x}−\mathrm{3}\mid+\mathrm{4}{x} \\ $$
Commented by john santu last updated on 03/Feb/20
$$\left(\mathrm{1}\right)\:\mathrm{4}−{x}\geqslant\mathrm{0}\:\Rightarrow{x}\leqslant\mathrm{4} \\ $$$$\left(\mathrm{2}\right)\sqrt{\mathrm{4}−{x}\:}\leqslant{x}\mid{x}−\mathrm{3}\mid+\mathrm{4}{x}+\mathrm{2} \\ $$$${for}\:{x}\:\geqslant\mathrm{3}\:\Rightarrow\sqrt{\mathrm{4}−{x}}\:\leqslant{x}\left({x}−\mathrm{3}+\mathrm{4}\right)+\mathrm{2} \\ $$$$\sqrt{\mathrm{4}−{x}}\:\leqslant\:{x}\left({x}+\mathrm{1}\right)+\mathrm{2} \\ $$$$ \\ $$$$ \\ $$
Commented by john santu last updated on 03/Feb/20
$${solution}\:\mathrm{0}\leqslant{x}\leqslant\mathrm{4}\: \\ $$
Commented by jagoll last updated on 03/Feb/20
$${please}\:{explain}\:\:{me} \\ $$
Commented by jagoll last updated on 03/Feb/20
$${mister}\:{W}\:{please}\:{help}\:{me}\:{the} \\ $$$${question} \\ $$
Commented by mr W last updated on 03/Feb/20
$${solution}\:{from}\:{santu}\:{sir}\:\mathrm{0}\leqslant{x}\leqslant\mathrm{4}\:{is}\:{correct}. \\ $$
Commented by jagoll last updated on 03/Feb/20
$${please}\:{explain}\:{me}.\:{how}\:{get}\:{it}? \\ $$
Answered by mr W last updated on 03/Feb/20
$${x}\leqslant\mathrm{4} \\ $$$$ \\ $$$${if}\:\mathrm{3}\leqslant{x}\leqslant\mathrm{4}: \\ $$$$\sqrt{\mathrm{4}−{x}}−\mathrm{2}\leqslant{x}\left({x}−\mathrm{3}\right)+\mathrm{4}{x} \\ $$$$\sqrt{\mathrm{4}−{x}}\leqslant{x}^{\mathrm{2}} +{x}+\mathrm{2} \\ $$$${LHS}\leqslant\mathrm{1} \\ $$$${RHS}\geqslant\mathrm{3}^{\mathrm{2}} +\mathrm{3}+\mathrm{2}=\mathrm{14} \\ $$$$\Rightarrow{LHS}\leqslant{RHS}\:{is}\:{true} \\ $$$$ \\ $$$${if}\:{x}<\mathrm{3}: \\ $$$$\sqrt{\mathrm{4}−{x}}−\mathrm{2}\leqslant{x}\left(\mathrm{3}−{x}\right)+\mathrm{4}{x} \\ $$$$\sqrt{\mathrm{4}−{x}}−\mathrm{2}\leqslant{x}\left(\mathrm{7}−{x}\right) \\ $$$${LHS}\leqslant\mathrm{0}\:{when}\:{x}\geqslant\mathrm{0} \\ $$$${LHS}>\mathrm{0}\:{when}\:{x}<\mathrm{0} \\ $$$${RHS}\geqslant\mathrm{0}\:{when}\:{x}\geqslant\mathrm{0}\:\left({x}<\mathrm{3}\right) \\ $$$${RHS}<\mathrm{0}\:{when}\:{x}<\mathrm{0} \\ $$$${when}\:{x}\geqslant\mathrm{0}: \\ $$$${LHS}\leqslant\mathrm{0}\:{and}\:{RHS}\geqslant\mathrm{0}\:\Rightarrow{LHS}\leqslant{RHS}\:{is}\:{true}. \\ $$$${when}\:{x}<\mathrm{0}: \\ $$$${LHS}>\mathrm{0}\:{and}\:{RHS}<\mathrm{0}\:\Rightarrow{LHS}\leqslant{RHS}\:{is}\:{not}\:{true}. \\ $$$$ \\ $$$${that}\:{means}\:{the}\:{inequality}\:{is}\:{valid},\:{when} \\ $$$$\mathrm{3}\leqslant{x}\leqslant\mathrm{4}\:{or}\:\mathrm{0}\leqslant{x}<\mathrm{3} \\ $$$$\Rightarrow\mathrm{0}\leqslant{x}\leqslant\mathrm{4} \\ $$
Commented by mr W last updated on 03/Feb/20
$${by}\:{handling}\:{inequality}\:{it}'{s}\:{not}\:{good} \\ $$$${idea}\:{to}\:{square}\:{both}\:{sides},\:{because}\:{that} \\ $$$${will}\:{change}\:{the}\:{range}\:{of}\:{validity}\:{of} \\ $$$${the}\:{inequality}! \\ $$
Commented by john santu last updated on 03/Feb/20
$${thank}\:{you}\:{sir} \\ $$
Commented by jagoll last updated on 03/Feb/20
$${thank}\:{you}\:{mister} \\ $$