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Question Number 191947 by universe last updated on 04/May/23

p r o v e t h a t

\sqrt{a + b \sqrt{a - b \sqrt{a + b \sqrt{a - b \sqrt{\dots}}}}} = \frac{\sqrt{4 a - 3 b^{2}} + b}{2}

Answered by ajfour last updated on 05/May/23

L e t s = \sqrt{a + b \sqrt{a - b s}} = \frac{b + \sqrt{4 a - 3 b^{2}}}{2}

\Rightarrow s^{2} = a + b \sqrt{a - b s}

{(s^{2} - a)}^{2} = b^{2} (a - b s) \dots (i)

& {(2 s - b)}^{2} = 4 a - 3 b^{2}

\Rightarrow 4 s^{2} + b^{2} - 4 b s = 4 a - 3 b^{2}

o r {(s^{2} - a)}^{2} = b^{2} {(s - b)}^{2} . . (i i)

c o m p a r i n g (i) & (i i)

\Rightarrow {(s - b)}^{2} = a - b s

{(s - b)}^{2} + b (s - b) - (a - b^{2}) = 0

s - b = \frac{- b \pm \sqrt{4 a - 3 b^{2}}}{2}

s = \frac{b \pm \sqrt{4 a - 3 b^{2}}}{2}

Commented by mr W last updated on 04/May/23

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Commented by mehdee42 last updated on 05/May/23

d e a r f r i e n d, w h a t i s t h e r e a s o n f o r u s i n g e q u a t i o n {(2 s - b)}^{2} = 4 a - 3 b^{2} ?

Answered by universe last updated on 05/May/23

\sqrt{a + b \sqrt{a - b \sqrt{a + b \sqrt{a - b \sqrt{\dots}}}}} = x (l e t) \dots (1)

a n d

\sqrt{a - b \sqrt{a + b \sqrt{a - b \sqrt{a + b \sqrt{\dots}}}}} = y (l e t) \dots (2)

b y (1)

\sqrt{a + b y} = x \Rightarrow a + b y = x^{2} \dots . (3)

n o w b y (2)

\sqrt{a - b x} = y \Rightarrow a - b x = y^{2} \dots . (4)

{e q}^{n} (3) - {e q}^{n} (4)

b (x + y) = (x - y) (x + y) \Rightarrow x - y = b

y = x - b

b y (3)

a + b x - b^{2} = x^{2} \Rightarrow x^{2} - b x + b^{2} - a = 0

x = \frac{b + \sqrt{b^{2} - 4 (b^{2} - a)}}{2}

x = \frac{b + \sqrt{4 a - 3 b^{2}}}{2}

Commented by mehdee42 last updated on 05/May/23

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