Solve for x in terms of a (√(a−(√(a+x )))) + (√(a+(√(a−x)))) = 2x Please sir i request you to solve this question =


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Question Number 61211 by alphaprime last updated on 30/May/19

$S o l v e f o r x i n t e r m s o f a$ $\sqrt{a - \sqrt{a + x}} + \sqrt{a + \sqrt{a - x}} = 2 x$ $P l e a s e s i r i r e q u e s t y o u t o s o l v e t h i s$ $q u e s t i o n =_=$
Commented by maxmathsup by imad last updated on 30/May/19

$f i r s t x \in [0, a] a - \sqrt{a + x} = u a n d a + \sqrt{a - x} = v \Rightarrow$ $\sqrt{a - x} - \sqrt{a + x} = v - u \Rightarrow a - x + a + x - 2 \sqrt{a^{2} - x^{2}} = {(u - v)}^{2} \Rightarrow$ $2 a - 2 \sqrt{a^{2} - x^{2}} = {(u - v)}^{2} \Rightarrow 2 \sqrt{a^{2} - x^{2}} = 2 a - {(u - v)}^{2} \Rightarrow$ $\sqrt{a^{2} - x^{2}} = a - \frac{{(u - v)}^{2}}{2} \Rightarrow a^{2} - x^{2} = {(\frac{2 a - {(u - v)}^{2}}{2})}^{2} \Rightarrow$ $x^{2} = a^{2} - \frac{1}{4} {(2 a - {(u - v)}^{2})}^{2} = \frac{4 a^{2} - (4 a^{2} - 4 a {(u - v)}^{2} + {(u - v)}^{4})}{4}$ $= \frac{4 a {(u - v)}^{2} - {(u - v)}^{4}}{4} \Rightarrow x = \frac{1}{2} \sqrt{4 a {(u - v)}^{2} - {(u - v)}^{4}}$ $= \frac{∣ u - v ∣}{2} \sqrt{4 a - {(u - v)}^{2}}$ $(e) \Rightarrow \sqrt{u} + \sqrt{v} =∣ u - v ∣ \sqrt{4 a - {(u - v)}^{2}} \Rightarrow$ $u + v + 2 u v = (u^{2} - 2 u v + v^{2}) (4 a - {(u - v)}^{2}) \Rightarrow$ $u + v + 2 u v = (u^{2} - 2 u v + v^{2}) (4 a - u^{2} - v^{2} + 2 u v . . . . . b e c o n t i n u e d . . .$
Commented by maxmathsup by imad last updated on 30/May/19

$p e r h a p s s i r m j s o r s i r m r w c a n f i n d s o m e t h i n g w i t h t h i s . . .$
Commented by behi83417@gmail.com last updated on 31/May/19

$sir! i have some treis on this Q .$ $You can't use 'macro parameter character #' in math mode$
	Answered by MJS last updated on 30/May/19

	$squaring a few times leads to$ $[t = x^{2}]$ $t^{6} - 4 {a t}^{5} + \frac{a (12 a - 1)}{2} t^{4} - \frac{128 a^{3} - 16 a^{2} - 17}{32} t^{3} + \frac{a (16 a^{3} + 8 a^{2} - 7 a - 5)}{16} t^{2} - \frac{a (64 a^{3} - 48 a^{2} - 4 a - 7)}{128} t + \frac{256 a^{4} - 256 a^{3} - 32 a^{2} + 1}{4096} = 0$ $which cannot be solved exactly for t$ $or$ $a^{4} - \frac{16 t^{2} + 2 t - 1}{4 t - 1} a^{3} + \frac{768 t^{4} + 64 t^{3} - 56 t^{2} + 4 t - 1}{8 (16 t^{2} - 8 t + 1)} a^{2} - \frac{t (512 t^{4} + 64 t^{3} + 40 t - 7)}{8 (16 t^{2} - 8 t + 1)} a + \frac{4096 t^{6} + 2176 t^{3} + 1}{256 (16 t^{2} - 8 t + 1)} = 0$ $which can be solved exactly for a using$ ${Ferrari}^{'} s formula with a hugh amount of$ $devotion$ $in both cases we get invalid solutions because$ $of squaring . in fact, for given a only one t$ $exists if that . . .$
	Commented by alphaprime last updated on 30/May/19
	Thank you both sir at least you shown interest in the question and I hope we one day get solutions to it . completely.
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