If-1-x-1-1-x-1-x-1-1-x-1-find-x-

Question Number 92605 by som(math1967) last updated on 08/May/20

I f \frac{1 + x}{1 + \sqrt{1 + x}} + \frac{1 - x}{1 - \sqrt{1 - x}} = 1

f i n d x

Commented by behi83417@gmail.com last updated on 08/May/20

\frac{(1 + x) (1 - \sqrt{1 + x})}{- x} + \frac{(1 - x) (1 + \sqrt{1 - x})}{x} = 1

\Rightarrow (1 + x) \sqrt{1 + x} - (1 + x) + (1 - x) \sqrt{1 - x} + (1 - x) = x

\Rightarrow (1 + x) \sqrt{1 + x} + (1 - x) \sqrt{1 - x} = 3 x

(- 1 < x < 1, x \neq 0, let : x = \cos 2 t) \Rightarrow

{2 c}^{2} . \sqrt{2} c + {2 s}^{2} . \sqrt{2} s = 3 (c^{2} - s^{2})

\Rightarrow 2 \sqrt{2} (c^{3} + s^{3}) = 3 (c^{2} - s^{2})

\Rightarrow 2 \sqrt{2} (c + s) (c^{2} + s^{2} - cs) = 3 (c + s) (c - s)

1 . c + s = 0 \Rightarrow \sqrt{2} \cos (t - \frac{π}{4}) = 0 \Rightarrow t = \frac{π}{4} (\equiv x = 0 : not ok)

\Rightarrow 2 \sqrt{2} (1 - cs) = 3 (c - s)

\Rightarrow 8 (1 - 2 cs + c^{2} s^{2}) = 9 (c^{2} + s^{2} - 2 cs)

\Rightarrow {8 c}^{2} s^{2} + 2 cs - 1 = 0

\Rightarrow cs = \frac{- 1 \pm \sqrt{1 + 8}}{8} = \frac{1}{4}, - \frac{1}{2}

1 . cs = \frac{1}{4} \Rightarrow \sin 2 t = \frac{1}{2} \Rightarrow x = \cos 2 t = \frac{\sqrt{3}}{2}

2 . cs = - \frac{1}{2} \Rightarrow \sin 2 t = - 1 \Rightarrow x = \cos 2 t = 0 [not ok]

Commented by Tony Lin last updated on 08/May/20

l e t \sqrt{1 + x} = a, \sqrt{1 - x} = b, a^{2} + b^{2} = 2

\frac{a^{2}}{1 + a} + \frac{b^{2}}{1 - b} = 1

a^{2} - a^{2} b + b^{2} + {a b}^{2} = 1 + a - b - a b

1 - a b (a - b) - (a - b) + a b = 0

(1 + a b) [1 - (a - b)] = 0

a - b = 1 o r a b = - 1 (f a l s e)

\sqrt{1 + x} - \sqrt{1 - x} = 1

1 + x = {(1 + \sqrt{1 - x})}^{2}

2 x - 1 = 2 \sqrt{1 - x}

4 x^{2} - 4 x + 1 = 4 - 4 x

x^{2} = \frac{3}{4}

x = \pm \frac{\sqrt{3}}{2}

b u t \frac{1 - \frac{\sqrt{3}}{2}}{1 + \sqrt{1 - \frac{\sqrt{3}}{2}}} + \frac{1 + \frac{\sqrt{3}}{2}}{1 - \sqrt{1 + \frac{\sqrt{3}}{2}}} < 0 (- 5)

∴ x = \frac{\sqrt{3}}{2}

Commented by som(math1967) last updated on 08/May/20

T h a n k s s i r

Commented by som(math1967) last updated on 08/May/20

T h a n k s s i r

Answered by MJS last updated on 08/May/20

- 1 ⩽ x ⩽ 1 \land x \neq 0

\frac{(1 + x) (1 - \sqrt{1 + x})}{x} + \frac{(1 - x) (1 + \sqrt{1 - x})}{x} = 1

{(1 + x)}^{3 / 2} + {(1 - x)}^{3 / 2} = 3 x

0 < x ⩽ 1

x = \sin^{2} t

\cos^{3} t + {(1 + \sin^{2} t)}^{3 / 2} = {3 \sin}^{2} t

{(1 + \sin^{2} t)}^{3} = {({3 \sin}^{2} t - \cos^{3} t)}^{2}

\sin t = s \land \cos t = c \land s = \sqrt{1 - c^{2}}

c^{6} + 3 c^{5} + \frac{3}{2} c^{4} - 3 c^{3} - 3 c^{2} + \frac{1}{2} = 0

c = z - 1

z^{6} - 3 z^{5} + \frac{3}{2} z^{4} + z^{3} = 0

z^{3} (z - 2) (z^{2} - z - \frac{1}{2}) = 0

(c - 1) {(c + 1)}^{3} (c^{2} + c - \frac{1}{2}) = 0

\cos t = \pm 1 \lor \cos t = - \frac{1}{2} \pm \frac{\sqrt{3}}{2}

\Rightarrow only solution is

\cos t = - \frac{1}{2} + \frac{\sqrt{3}}{2} \Rightarrow \sin^{2} t = x = \frac{\sqrt{3}}{2}

Commented by som(math1967) last updated on 08/May/20

T h a n k y o u s i r

Commented by MJS last updated on 08/May/20

{you}^{'} re welcome

{it}^{'} s also possible to square it :

{(1 - x)}^{3 / 2} + {(1 + x)}^{3 / 2} = 3 x

6 x^{2} + 2 + 2 {(1 - x)}^{3 / 2} {(1 + x)}^{3 / 2} = 9 x^{2}

2 {(1 - x)}^{3 / 2} {(1 + x)}^{3 / 2} = 3 x^{2} - 2

4 {(1 - x)}^{3} {(1 + x)}^{3} = {(3 x^{2} - 2)}^{2}

4 x^{6} - 3 x^{4} = 0

x^{4} (4 x^{2} - 3) = 0

0 < x ⩽ 1 \Rightarrow x = \frac{\sqrt{3}}{2}

Commented by som(math1967) last updated on 08/May/20

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