Question-149043

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Question Number 149043 by Tawa11 last updated on 02/Aug/21

Commented by EDWIN88 last updated on 02/Aug/21

$${x}=\mathrm{2} \\ $$

Commented by Tawa11 last updated on 02/Aug/21

$$\mathrm{Workings}\:\mathrm{sir}. \\ $$

Commented by Tawa11 last updated on 02/Aug/21

$$\mathrm{Solve}\:\mathrm{for}\:\:\mathrm{x} \\ $$

Answered by EDWIN88 last updated on 02/Aug/21

2(√x) (u−v)=4(√(3 )) where { ((u=(√x) +(√(x+(√3))))),((v=(√x)−(√(x−(√3))))) :} ⇔2(√x) ((√(x+(√3))) +(√(x−(√3))) )=4(√3) ⇔(√(x^2 +x(√3))) +(√(x^2 −x(√3))) = 2(√3) ⇒2x^2 +2(√(x^4 −3x^2 )) = 12 ⇒(√(x^4 −3x^2 )) =6−x^2 (√(u^2 −3u)) = 6−u ; u=x^2 u^2 −3u=36−12u+u^2 ⇒9u=36⇒u=4 then x = 2 , since x=−2 rejected

$$\mathrm{2}\sqrt{{x}}\:\left({u}−{v}\right)=\mathrm{4}\sqrt{\mathrm{3}\:} \\ $$$${where}\:\begin{cases}{{u}=\sqrt{{x}}\:+\sqrt{{x}+\sqrt{\mathrm{3}}}}\\{{v}=\sqrt{{x}}−\sqrt{{x}−\sqrt{\mathrm{3}}}}\end{cases} \\ $$$$\Leftrightarrow\mathrm{2}\sqrt{{x}}\:\left(\sqrt{{x}+\sqrt{\mathrm{3}}}\:+\sqrt{{x}−\sqrt{\mathrm{3}}}\:\right)=\mathrm{4}\sqrt{\mathrm{3}} \\ $$$$\Leftrightarrow\sqrt{{x}^{\mathrm{2}} +{x}\sqrt{\mathrm{3}}}\:+\sqrt{{x}^{\mathrm{2}} −{x}\sqrt{\mathrm{3}}}\:=\:\mathrm{2}\sqrt{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}\sqrt{{x}^{\mathrm{4}} −\mathrm{3}{x}^{\mathrm{2}} }\:=\:\mathrm{12} \\ $$$$\Rightarrow\sqrt{{x}^{\mathrm{4}} −\mathrm{3}{x}^{\mathrm{2}} }\:=\mathrm{6}−{x}^{\mathrm{2}} \\ $$$$\sqrt{{u}^{\mathrm{2}} −\mathrm{3}{u}}\:=\:\mathrm{6}−{u}\:;\:{u}={x}^{\mathrm{2}} \\ $$$${u}^{\mathrm{2}} −\mathrm{3}{u}=\mathrm{36}−\mathrm{12}{u}+{u}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{9}{u}=\mathrm{36}\Rightarrow{u}=\mathrm{4} \\ $$$${then}\:{x}\:=\:\mathrm{2}\:,\:{since}\:{x}=−\mathrm{2}\:{rejected} \\ $$

Commented by bramlexs22 last updated on 02/Aug/21

$$\mathrm{nice} \\ $$

Commented by Tawa11 last updated on 02/Aug/21

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you} \\ $$

Commented by Tawa11 last updated on 22/Aug/21

Sir. I could not get the first part since. That is: 2(√x) (u − v) = 4(√3) But I understand other steps very well.

$$\mathrm{Sir}.\:\mathrm{I}\:\mathrm{could}\:\mathrm{not}\:\mathrm{get}\:\mathrm{the}\:\mathrm{first}\:\mathrm{part}\:\mathrm{since}. \\ $$$$\mathrm{That}\:\mathrm{is}:\:\:\:\:\:\:\mathrm{2}\sqrt{\mathrm{x}}\:\left(\mathrm{u}\:\:−\:\:\mathrm{v}\right)\:\:=\:\:\mathrm{4}\sqrt{\mathrm{3}} \\ $$$$\mathrm{But}\:\mathrm{I}\:\mathrm{understand}\:\mathrm{other}\:\mathrm{steps}\:\mathrm{very}\:\mathrm{well}. \\ $$

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