Question Number 170781 by solomonwells last updated on 30/May/22
$$\sqrt{\left(\mathrm{2}−\mathrm{x}\right)}\:\:\:=\:\:\left(\mathrm{2}−\mathrm{x}\right)^{\mathrm{2}} \:\:\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{for}}\:\:\:\:\:\:\boldsymbol{\mathrm{x}} \\ $$
Answered by Rasheed.Sindhi last updated on 30/May/22
$$\frac{\left(\mathrm{2}−{x}\right)^{\mathrm{2}} }{\left(\mathrm{2}−{x}\right)^{\mathrm{1}/\mathrm{2}} }=\mathrm{1} \\ $$$$\left(\mathrm{2}−{x}\right)^{\mathrm{3}/\mathrm{2}} =\mathrm{1}^{\mathrm{3}/\mathrm{2}} \\ $$$$\mathrm{2}−{x}=\mathrm{1} \\ $$$${x}=\mathrm{1} \\ $$
Commented by mr W last updated on 30/May/22
$${x}=\mathrm{2}\:{is}\:{also}\:{good} \\ $$
Commented by Rasheed.Sindhi last updated on 30/May/22
$$\mathrm{Of}\:\mathrm{course}\:\mathrm{sir}: \\ $$$$\sqrt{\mathrm{2}−{x}}\:=\left(\mathrm{2}−{x}\right)^{\mathrm{2}} \\ $$$$\sqrt{\mathrm{2}−{x}}\:−\left(\mathrm{2}−{x}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\sqrt{\mathrm{2}−{x}}\:\left(\mathrm{1}−\left(\mathrm{2}−{x}\right)^{\mathrm{3}/\mathrm{2}} \right)=\mathrm{0} \\ $$$$\sqrt{\mathrm{2}−{x}}\:=\mathrm{0}\:\mid\:\left(\mathrm{2}−{x}\right)^{\mathrm{3}/\mathrm{2}} =\mathrm{1} \\ $$$${x}=\mathrm{2}\:\:\:\:\:\:\mid\:\:\:\left(\mathrm{2}−{x}\right)^{\mathrm{3}/\mathrm{2}} =\mathrm{1}^{\mathrm{3}/\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}−{x}=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}=\mathrm{1} \\ $$
Answered by MJS_new last updated on 31/May/22
$${t}=\sqrt{\mathrm{2}−{x}}\:\geqslant\mathrm{0}\:\Leftrightarrow\:{x}=\mathrm{2}−{t}^{\mathrm{2}} \\ $$$${t}={t}^{\mathrm{4}} \\ $$$${t}\left({t}^{\mathrm{3}} −\mathrm{1}\right)=\mathrm{0} \\ $$$${t}=\mathrm{0}\vee{t}=\mathrm{1}\:\Rightarrow\:{x}=\mathrm{1}\vee{x}=\mathrm{2} \\ $$