Question-48367

Question Number 48367 by behi83417@gmail.com last updated on 22/Nov/18

Answered by tanmay.chaudhury50@gmail.com last updated on 23/Nov/18

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

\frac{1}{\sqrt{3}} = \frac{x + y}{x y - 1}

\frac{x + y}{1 - x y} = - \frac{1}{\sqrt{3}}

t a n a = x t a n b = y t a n c = z

t a n (a + b) = - \frac{1}{\sqrt{3}}

t a n (b + c) = - \frac{1}{\sqrt{3}}

t a n (a + c) = - \frac{1}{\sqrt{3}}

t a n (a + b) = t a n (b + c) = t a n (a + c) s o

a = b = c s o

t a n a = t a n b = t a n c

x = y = z

\frac{x + x}{1 - x \times x} = - \frac{1}{\sqrt{3}}

2 \sqrt{3} x = - 1 (1 - x^{2})

2 \sqrt{3} x + 1 - x^{2} = 0

x^{2} - 2 \sqrt{3} x - 1 = 0

x = \frac{2 \sqrt{3} \pm \sqrt{12 + 4}}{2} = \sqrt{3} \pm 2

x = y = z = \sqrt{3} \pm 2

Commented by behi83417@gmail.com last updated on 23/Nov/18

v e r y b e a u t i f u l m e t h o d! g o d b l e s s y o u s i r .

Commented by tanmay.chaudhury50@gmail.com last updated on 23/Nov/18

t h a n k y o u s i r \dots

Answered by MJS last updated on 22/Nov/18

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

(2) z = \frac{1 + y \sqrt{3}}{y - \sqrt{3}} = x

(3) \frac{{(1 + y \sqrt{3})}^{2}}{{(y - \sqrt{3})}^{2}} = 2 \sqrt{3} \frac{1 + y \sqrt{3}}{y - \sqrt{3}} + 1

\Rightarrow y^{2} - 2 \sqrt{3} y - 1 = 0 \Rightarrow y = \pm 2 + \sqrt{3}

x = y = x = \pm 2 + \sqrt{3}

Commented by behi83417@gmail.com last updated on 23/Nov/18

s o f a s t! t h a n k s i n a d v a n c e s i r .

Commented by MJS last updated on 23/Nov/18

if we see that x = y = z at start we can directly

go to x^{2} = 2 \sqrt{3} x + 1

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