Question and Answers Forum

All Questions      Topic List

Operation Research Questions

Previous in All Question      Next in All Question      

Previous in Operation Research      Next in Operation Research      

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

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

how x=4 y=9

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

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

bah darun excellent....

$${bah}\:{darun}\:{excellent}.... \\ $$

Commented by Joel578 last updated on 23/Aug/18

I can′t imagine how you get this idea

$$\mathrm{I}\:\mathrm{can}'\mathrm{t}\:\mathrm{imagine}\:\mathrm{how}\:\mathrm{you}\:\mathrm{get}\:\mathrm{this}\:\mathrm{idea} \\ $$

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

i think sir behi have played a match criket with this equation...

$${i}\:{think}\:{sir}\:{behi}\:{have}\:{played}\:{a}\:{match}\:{criket}\:\:{with}\:{this}\:{equation}... \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com