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

Commented by EDWIN88 last updated on 02/Aug/21

$x = 2$
Commented by Tawa11 last updated on 02/Aug/21

$Workings sir .$
Commented by Tawa11 last updated on 02/Aug/21

$Solve for x$
	Answered by EDWIN88 last updated on 02/Aug/21
	$2(√x) (u−v)=4(√(3 )) where { ((u=(√x) +(√(x+(√3))))),((v=(√x)−(√(x−(√3))))) :} ⇔2(√x) ((√(x+(√3))) +(√(x−(√3))) )=4(√3) ⇔(√(x^2 +x(√3))) +(√(x^2 −x(√3))) = 2(√3) ⇒2x^2 +2(√(x^4 −3x^2 )) = 12 ⇒(√(x^4 −3x^2 )) =6−x^2 (√(u^2 −3u)) = 6−u ; u=x^2 u^2 −3u=36−12u+u^2 ⇒9u=36⇒u=4 then x = 2 , since x=−2 rejected$
	$2 \sqrt{x} (u - v) = 4 \sqrt{3}$ $w h e r e {\begin{cases} u = \sqrt{x} + \sqrt{x + \sqrt{3}} \\ v = \sqrt{x} - \sqrt{x - \sqrt{3}} \end{cases}$ $\Leftrightarrow 2 \sqrt{x} (\sqrt{x + \sqrt{3}} + \sqrt{x - \sqrt{3}}) = 4 \sqrt{3}$ $\Leftrightarrow \sqrt{x^{2} + x \sqrt{3}} + \sqrt{x^{2} - x \sqrt{3}} = 2 \sqrt{3}$ $\Rightarrow 2 x^{2} + 2 \sqrt{x^{4} - 3 x^{2}} = 12$ $\Rightarrow \sqrt{x^{4} - 3 x^{2}} = 6 - x^{2}$ $\sqrt{u^{2} - 3 u} = 6 - u; u = x^{2}$ $u^{2} - 3 u = 36 - 12 u + u^{2}$ $\Rightarrow 9 u = 36 \Rightarrow u = 4$ $t h e n x = 2, s i n c e x = - 2 r e j e c t e d$
	Commented by bramlexs22 last updated on 02/Aug/21

	$nice$
	Commented by Tawa11 last updated on 02/Aug/21

	$Thank you sir . God bless you$
	Commented by Tawa11 last updated on 22/Aug/21

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