Question

tell fast its emergency ​

Answer 1

Solution

Please find the answer in attachment

$x\: = \: \dfrac{\sqrt{a+1}\: +\: \sqrt{a-1}}{\sqrt{a+1}\: - \: \sqrt{a-1}}$

$x\: = \: \dfrac{\sqrt{a+1}\: +\: \sqrt{a-1}}{\sqrt{a+1}\: - \: \sqrt{a-1}}\times\dfrac{\sqrt{a+1}\: +\: \sqrt{a-1}}{\sqrt{a+1}\: +\: \sqrt{a-1}}$

$x\: = \: \dfrac{(\sqrt{a+1}\: +\: \sqrt{a-1})^{2}}{(\sqrt{a+1})^{2}\: - \: (\sqrt{a-1})^{2}}$

$x\:=\: \dfrac{(\sqrt{a+1})^{2}\:+\:(\sqrt{a-1})^{2}\:+\: 2\sqrt{a^{2}\:-\:1^{2}}}{(a\:+\:1)\:-\:(a\:-\:1)}$

$x\:=\: \dfrac{a\:+\:1\:+\:a\:-\:1\:+\: 2\sqrt{a^{2}\:-\:1^{2}}}{(a\:+\:1\:-\:a\:+\:1)}$

$x\: =\: \dfrac{2a\: + \: 2\sqrt{a^2-1}}{2}$

$x\:=\: a\: +\: \sqrt{a^2-1}$

$(x\:-\:a)^2\:=\:a^2\: - \:1$

$x^{2}\:+\:a^2\: - \:2ax\: =\: a^2\: - \: 1$

$\boxed{x^{2}\:-\:2ax\:+\:1\:=\:0}$

HENCE PROVED

Answer 2

Solution:

$\sf x = \dfrac{ \sqrt{a + 1} + \sqrt{a - 1} }{ \sqrt{a + 1} - \sqrt{a - 1} }$

Using componendo Dividendo theorem i.e If a/b = c/d then (a + b)/(a - b) = (c + d)/(c - d)

$\sf \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{ \sqrt{a + 1} + \sqrt{a - 1} + \sqrt{a + 1} - \sqrt{a - 1} }{ \sqrt{a + 1} + \sqrt{a - 1} - (\sqrt{a + 1} - \sqrt{a - 1} )}$

$\sf \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{ 2 \sqrt{a + 1} }{ 2\sqrt{a - 1} }$

$\sf \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{ \sqrt{a + 1} }{ \sqrt{a - 1} }$

Squaring on both sides

$\sf \Rightarrow \dfrac{(x + 1) ^{2} }{(x - 1)^{2} } = \dfrac{( \sqrt{a + 1})^{2} }{( \sqrt{a - 1}) ^{2} }$

$\sf \Rightarrow \dfrac{ {x}^{2} + 2x + 1 }{ {x}^{2} - 2x + 1 } = \dfrac{a + 1 }{a - 1 }$

Again using Componedo Dividendo theorem

$\sf \Rightarrow \dfrac{ {x}^{2} + 2x + 1 + {x}^{2} - 2x + 1}{ {x}^{2} + 2x + 1 - ( {x}^{2} - 2x + 1 )} = \dfrac{a + 1 + a - 1 }{a + 1 - (a - 1 ) }$

$\sf \Rightarrow \dfrac{2 {x}^{2} + 2}{ 4x } = \dfrac{2a }{2 }$

$\sf \Rightarrow \dfrac{{x}^{2} + 1}{ 2x } = a$

$\sf \Rightarrow {x}^{2} + 1= 2ax$

$\sf \Rightarrow {x}^{2} - 2ax+ 1= 0$

Hence shown.