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Question-40814




Question Number 40814 by behi83417@gmail.com last updated on 27/Jul/18
Commented by MrW3 last updated on 27/Jul/18
x=0, y=1  or  x=1, y=0
$${x}=\mathrm{0},\:{y}=\mathrm{1} \\ $$$${or} \\ $$$${x}=\mathrm{1},\:{y}=\mathrm{0} \\ $$
Commented by math khazana by abdo last updated on 28/Jul/18
from the equality we have x≥0 and y≥0 ⇒  ((√x) +(√y))^2 =x+y ⇒x+y +2(√(xy))=x+y ⇒  xy=0 ⇒x=0 or y=0  x=0 ⇒y^2 =1 ⇒y=1  y=0⇒x^2 =1⇒x=0 so the solution are (1,0)and  (0,1)
$${from}\:{the}\:{equality}\:{we}\:{have}\:{x}\geqslant\mathrm{0}\:{and}\:{y}\geqslant\mathrm{0}\:\Rightarrow \\ $$$$\left(\sqrt{{x}}\:+\sqrt{{y}}\right)^{\mathrm{2}} ={x}+{y}\:\Rightarrow{x}+{y}\:+\mathrm{2}\sqrt{{xy}}={x}+{y}\:\Rightarrow \\ $$$${xy}=\mathrm{0}\:\Rightarrow{x}=\mathrm{0}\:{or}\:{y}=\mathrm{0} \\ $$$${x}=\mathrm{0}\:\Rightarrow{y}^{\mathrm{2}} =\mathrm{1}\:\Rightarrow{y}=\mathrm{1} \\ $$$${y}=\mathrm{0}\Rightarrow{x}^{\mathrm{2}} =\mathrm{1}\Rightarrow{x}=\mathrm{0}\:{so}\:{the}\:{solution}\:{are}\:\left(\mathrm{1},\mathrm{0}\right){and} \\ $$$$\left(\mathrm{0},\mathrm{1}\right) \\ $$
Commented by behi83417@gmail.com last updated on 28/Jul/18
thank you both sirs:MrW3 & mr abdo.
$${thank}\:{you}\:{both}\:{sirs}:{MrW}\mathrm{3}\:\&\:{mr}\:{abdo}. \\ $$

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