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Question-153168




Question Number 153168 by liberty last updated on 05/Sep/21
Answered by EDWIN88 last updated on 05/Sep/21
let y=u^2  & x=v^2   (1)(√(v^2 +u)) +(√(v^2 −u)) = 2  ⇒2v^2  +2(√(v^4 −u^2 )) = 4  ⇒(√(v^4 −u^2 )) = 2−v^2   ⇒v^4 −u^2 =4−4v^2 +v^4   ⇒4v^2 −u^2  = 4    (2)(√(u^2 +v)) −(√(u^2 −v)) =1  ⇒2u^2 −2(√(u^4 −v^2 )) =1  ⇒2u^2 −1 =2(√(u^4 −v^2 ))   ⇒4u^4 −4u^2 +1=4u^4 −4v^2   ⇒4u^2 −4v^2 = 1  (1)×4+(2)⇒−4u^2 +16v^2 =16                                    4u^2 −4v^2 =1  ⇒12v^2  =17 ⇒v=(√((17)/(12)))  ⇒u^2 =4v^2 −4=((17)/3)−4=(5/3); u=(√(5/3))  ⇒4x+2y=4v^2 +2u^2 =((17)/3)+((10)/3)= 9
$${let}\:{y}={u}^{\mathrm{2}} \:\&\:{x}={v}^{\mathrm{2}} \\ $$$$\left(\mathrm{1}\right)\sqrt{{v}^{\mathrm{2}} +{u}}\:+\sqrt{{v}^{\mathrm{2}} −{u}}\:=\:\mathrm{2} \\ $$$$\Rightarrow\mathrm{2}{v}^{\mathrm{2}} \:+\mathrm{2}\sqrt{{v}^{\mathrm{4}} −{u}^{\mathrm{2}} }\:=\:\mathrm{4} \\ $$$$\Rightarrow\sqrt{{v}^{\mathrm{4}} −{u}^{\mathrm{2}} }\:=\:\mathrm{2}−{v}^{\mathrm{2}} \\ $$$$\Rightarrow{v}^{\mathrm{4}} −{u}^{\mathrm{2}} =\mathrm{4}−\mathrm{4}{v}^{\mathrm{2}} +{v}^{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{4}{v}^{\mathrm{2}} −{u}^{\mathrm{2}} \:=\:\mathrm{4} \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\sqrt{{u}^{\mathrm{2}} +{v}}\:−\sqrt{{u}^{\mathrm{2}} −{v}}\:=\mathrm{1} \\ $$$$\Rightarrow\mathrm{2}{u}^{\mathrm{2}} −\mathrm{2}\sqrt{{u}^{\mathrm{4}} −{v}^{\mathrm{2}} }\:=\mathrm{1} \\ $$$$\Rightarrow\mathrm{2}{u}^{\mathrm{2}} −\mathrm{1}\:=\mathrm{2}\sqrt{{u}^{\mathrm{4}} −{v}^{\mathrm{2}} }\: \\ $$$$\Rightarrow\mathrm{4}{u}^{\mathrm{4}} −\mathrm{4}{u}^{\mathrm{2}} +\mathrm{1}=\mathrm{4}{u}^{\mathrm{4}} −\mathrm{4}{v}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{4}{u}^{\mathrm{2}} −\mathrm{4}{v}^{\mathrm{2}} =\:\mathrm{1} \\ $$$$\left(\mathrm{1}\right)×\mathrm{4}+\left(\mathrm{2}\right)\Rightarrow−\mathrm{4}{u}^{\mathrm{2}} +\mathrm{16}{v}^{\mathrm{2}} =\mathrm{16} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}{u}^{\mathrm{2}} −\mathrm{4}{v}^{\mathrm{2}} =\mathrm{1} \\ $$$$\Rightarrow\mathrm{12}{v}^{\mathrm{2}} \:=\mathrm{17}\:\Rightarrow{v}=\sqrt{\frac{\mathrm{17}}{\mathrm{12}}} \\ $$$$\Rightarrow{u}^{\mathrm{2}} =\mathrm{4}{v}^{\mathrm{2}} −\mathrm{4}=\frac{\mathrm{17}}{\mathrm{3}}−\mathrm{4}=\frac{\mathrm{5}}{\mathrm{3}};\:{u}=\sqrt{\frac{\mathrm{5}}{\mathrm{3}}} \\ $$$$\Rightarrow\mathrm{4}{x}+\mathrm{2}{y}=\mathrm{4}{v}^{\mathrm{2}} +\mathrm{2}{u}^{\mathrm{2}} =\frac{\mathrm{17}}{\mathrm{3}}+\frac{\mathrm{10}}{\mathrm{3}}=\:\mathrm{9} \\ $$

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