Question-42316

Question Number 42316 by Raj Singh last updated on 23/Aug/18

Commented by Raj Singh last updated on 23/Aug/18

$${how}\:{x}=\mathrm{4}\:{y}=\mathrm{9} \\ $$

Commented by maxmathsup by imad last updated on 23/Aug/18

let put (√x)=u and (√y) =v ⇒u +v^2 =11 and u^2 +v =7 ⇒ u =11−v^2 and (11−v^2 )^2 +v =7 ⇒v^4 −22v^2 +121 +v−7 =0 ⇒ v^4 −22v^2 +v + 114 =0 the solution of this equation are v_1 ∼−3,7793 (real) v_2 ∼3,5844(real) v_3 =3 (real) v_4 ∼−2,8051 (real) but v must be +ve ⇒ v=v_2 or v_3 v =3,7793 ⇒y =(3,7793)^2 and u =11−(3,7793)^2 =... v =3 ⇒y =9 and u =11−3^2 =2 =(√x) ⇒x =4 .

$${let}\:{put}\:\sqrt{{x}}={u}\:{and}\:\sqrt{{y}}\:={v}\:\Rightarrow{u}\:+{v}^{\mathrm{2}} \:\:=\mathrm{11}\:{and}\:{u}^{\mathrm{2}} \:+{v}\:=\mathrm{7}\:\Rightarrow \\ $$$${u}\:=\mathrm{11}−{v}^{\mathrm{2}} \:{and}\:\left(\mathrm{11}−{v}^{\mathrm{2}} \right)^{\mathrm{2}} \:+{v}\:=\mathrm{7}\:\Rightarrow{v}^{\mathrm{4}} \:−\mathrm{22}{v}^{\mathrm{2}} \:+\mathrm{121}\:+{v}−\mathrm{7}\:=\mathrm{0}\:\Rightarrow \\ $$$${v}^{\mathrm{4}} \:−\mathrm{22}{v}^{\mathrm{2}} \:\:+{v}\:\:+\:\:\:\:\:\mathrm{114}\:=\mathrm{0}\:\:{the}\:{solution}\:{of}\:{this}\:{equation}\:{are} \\ $$$${v}_{\mathrm{1}} \sim−\mathrm{3},\mathrm{7793}\:\:\left({real}\right) \\ $$$${v}_{\mathrm{2}} \sim\mathrm{3},\mathrm{5844}\left({real}\right) \\ $$$${v}_{\mathrm{3}} \:=\mathrm{3}\:\:\left({real}\right) \\ $$$${v}_{\mathrm{4}} \:\sim−\mathrm{2},\mathrm{8051}\:\left({real}\right)\:\:{but}\:{v}\:{must}\:{be}\:+{ve}\:\Rightarrow\:{v}={v}_{\mathrm{2}} \:\:{or}\:{v}_{\mathrm{3}} \\ $$$${v}\:=\mathrm{3},\mathrm{7793}\:\Rightarrow{y}\:=\left(\mathrm{3},\mathrm{7793}\right)^{\mathrm{2}} \:\:\:{and}\:{u}\:=\mathrm{11}−\left(\mathrm{3},\mathrm{7793}\right)^{\mathrm{2}} \:=… \\ $$$${v}\:=\mathrm{3}\:\Rightarrow{y}\:=\mathrm{9}\:{and}\:{u}\:=\mathrm{11}−\mathrm{3}^{\mathrm{2}} \:=\mathrm{2}\:=\sqrt{{x}}\:\Rightarrow{x}\:=\mathrm{4}\:\:. \\ $$

Answered by behi83417@gmail.com last updated on 23/Aug/18

((√x)−2)+((√y)−3)((√y)+3)=0 ((√x)−2)((√x)+2)+((√y)−3)=0 ((√x)−2)−((√x)−2)((√x)+2)((√y)+3)=0 ((√x)−2)(1−((√x)+2)((√y)+3))=0 ⇒(√x)−2=0⇒x=4 x+(√y)=7⇒(√y)=7−4=3⇒y=9.■

$$\left(\sqrt{{x}}−\mathrm{2}\right)+\left(\sqrt{{y}}−\mathrm{3}\right)\left(\sqrt{{y}}+\mathrm{3}\right)=\mathrm{0} \\ $$$$\left(\sqrt{{x}}−\mathrm{2}\right)\left(\sqrt{{x}}+\mathrm{2}\right)+\left(\sqrt{{y}}−\mathrm{3}\right)=\mathrm{0} \\ $$$$\left(\sqrt{{x}}−\mathrm{2}\right)−\left(\sqrt{{x}}−\mathrm{2}\right)\left(\sqrt{{x}}+\mathrm{2}\right)\left(\sqrt{{y}}+\mathrm{3}\right)=\mathrm{0} \\ $$$$\left(\sqrt{{x}}−\mathrm{2}\right)\left(\mathrm{1}−\left(\sqrt{{x}}+\mathrm{2}\right)\left(\sqrt{{y}}+\mathrm{3}\right)\right)=\mathrm{0} \\ $$$$\Rightarrow\sqrt{{x}}−\mathrm{2}=\mathrm{0}\Rightarrow{x}=\mathrm{4} \\ $$$${x}+\sqrt{{y}}=\mathrm{7}\Rightarrow\sqrt{{y}}=\mathrm{7}−\mathrm{4}=\mathrm{3}\Rightarrow{y}=\mathrm{9}.\blacksquare \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 23/Aug/18