$$\frac{\sqrt{x} + \sqrt3}{\sqrt{x+\sqrt{x+\sqrt3}}} + \frac{\sqrt{x} - \sqrt3}{\sqrt{x-\sqrt{x-\sqrt3}}} = \sqrt3$$ I equalised with $0$, let $$u(x)= {\sqrt{x+\sqrt{x+\sqrt3}}}\cdot{\sqrt{x-\sqrt{x-\sqrt3}}}$$ Since $u(x)=0$ has no real solutions I get $$ (\sqrt{x} + \sqrt3)\left(\sqrt{x-\sqrt{x-\sqrt3}}\right) + (\sqrt{x} - \sqrt3)\left(\sqrt{x+\sqrt{x+\sqrt3}}\right) - (\sqrt{x}\cdot u(x)) = 0$$ The answer should be in form $[a, b)$ Its a mess, what do I do? Anybody know?
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It appears that there is only one real solution $x \approx 5.169810325$. I'm not sure if there is a closed form for this solution. – JimmyK4542 Dec 19 '15 at 21:55
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Are you sure you copied the equation correctly? This looks like it should be a mess. – YoTengoUnLCD Dec 19 '15 at 22:16
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It might be correct. I got a second degree polynomial with a zero very, very closed to that 5.1698... but I have no way to write it here, unless I post pictures of my calculations... but I dont know if that is according to the rules here. – Frank Wan Dec 19 '15 at 22:24
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As I said, I cannot type my huge calc... but maybe showing you my strategy might help you: 1. square both sides 2. you will get "3 things" on the LHS (Left Hand Side) 3. if you look well de squared of the numerator on first summand cancels with the "third" one, except for x. 3) Now get the same denominator on the 3 summands and multiply the denominator with 3 on the RHS. 4) square again both sides and a lot of stuff cancels. You will get a nice division of second degree polynomials. 5) I checked mine and the zeros of it get around 5.25 – Frank Wan Dec 20 '15 at 02:04
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@Frank Wan can you post a pic? Just to check myself? – diredragon Dec 20 '15 at 06:42
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@FrankWan. Could you give at least the second degree polynomials ? This problem is so messy ! – Claude Leibovici Dec 20 '15 at 15:04