If x = (√2a + 1 + √2a - 1)/(√2a + 1 - √2a - 1)prove that x² -

1.	If x = (√2a + 1 + √2a - 1)/(√2a + 1 - √2a - 1)prove that x² - 4ax + 1 = 0
Answer» $\large\underline{\sf{Given- }}$ $\rm :\longmapsto\:{x=\dfrac{\sqrt{2a+1}+\sqrt{2a-1} }{\sqrt{2a+1}-\sqrt{2a-1} } }$ $\large\underline{\sf{To\:prove- }}$ $\rm :\longmapsto\: {x}^{2} - 4ax + 1 = 0$ Concept USED :- Componendo and Dividendo $\rm :\longmapsto\:If \: \dfrac{a}{b} = \dfrac{c}{d}, \: then \: componendo \: and \: dividendo \: means$ $\boxed{ \bf{ \frac{a + b}{a - b} = \frac{c + d}{c - d}}}$ $\large\underline{\sf{Solution-}}$ Given that, $\rm :\longmapsto\:{x=\dfrac{\sqrt{2a+1}+\sqrt{2a-1} }{\sqrt{2a+1}-\sqrt{2a-1} } }$ can be rewritten as $\rm :\longmapsto\:{\dfrac{x}{1} =\dfrac{\sqrt{2a+1}+\sqrt{2a-1} }{\sqrt{2a+1}-\sqrt{2a-1} } }$ On Applying Componendo and Dividendo, we GET $\rm :\longmapsto\:{\dfrac{x + 1}{x - 1} =\dfrac{(\sqrt{2a+1}+\sqrt{2a-1} ) + (\sqrt{2a+1} - \sqrt{2a-1})}{(\sqrt{2a+1}+\sqrt{2a-1}) - (\sqrt{2a+1}-\sqrt{2a-1}) } }$ $\rm :\longmapsto\:{\dfrac{x + 1}{x - 1} =\dfrac{\sqrt{2a+1}+\sqrt{2a-1} + \sqrt{2a+1} - \sqrt{2a-1}}{\sqrt{2a+1}+\sqrt{2a-1} - \sqrt{2a+1} + \sqrt{2a-1} } }$ $\rm :\longmapsto\:{\dfrac{x + 1}{x - 1} =\dfrac{2\sqrt{2a+1}}{2\sqrt{2a - 1}}}$ can be rewritten as $\rm :\longmapsto\:{\dfrac{x + 1}{x - 1} =\dfrac{\sqrt{2a+1}}{\sqrt{2a - 1}}}$ On squaring both sides, we get $\rm :\longmapsto\:{\dfrac{(x + 1) {}^{2} }{(x - 1) {}^{2} } =\dfrac{{2a+1}}{{2a - 1}}}$ On Applying Componendo and Dividendo, we get $\rm :\longmapsto\:{\dfrac{(x + 1) {}^{2} + {(x - 1)}^{2} }{(x + 1) {}^{2} - {(x - 1)}^{2} } =\dfrac{{2a+1 + 2a - 1}}{{2a + 1 - 2a + 1}}}$ We know that, $\boxed{ \rm{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}}$ and $\boxed{ \rm{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}}$ So, using these Identities, we get $\rm :\longmapsto\:\dfrac{2( {x}^{2} + 1) }{4x} = \dfrac{4a}{2}$ $\rm :\longmapsto\:\dfrac{( {x}^{2} + 1) }{2x} = \dfrac{2a}{1}$ $\rm :\longmapsto\: {x}^{2} + 1 = 4ax$ $\bf\implies \: {x}^{2} - 4ax + 1 = 0$ HENCE, Proved ADDITIONAL Information :- $\rm :\longmapsto\:If \: \dfrac{a}{b} = \dfrac{c}{d}, \: then \:$ $\boxed{ \rm{ \frac{a}{c} = \frac{b}{d} \: is \: called \: alternendo}}$ $\boxed{ \rm{ \frac{b}{a} = \frac{d}{c} \: is \: called \: invertendo}}$ $\boxed{ \rm{ \frac{a + b}{b} = \frac{c + d}{d} \: is \: called \: componedo}}$ $\boxed{ \rm{ \frac{a - b}{b} = \frac{c - d}{d} \: is \: called \: dividendo}}$

If x = (√2a + 1 + √2a - 1)/(√2a + 1 - √2a - 1)prove that x² - 4ax + 1 = 0

Answer»

$\large\underline{\sf{Given- }}$

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

$\large\underline{\sf{To\:prove- }}$

$\rm :\longmapsto\: {x}^{2} - 4ax + 1 = 0$

Concept USED :-

Componendo and Dividendo

$\rm :\longmapsto\:If \: \dfrac{a}{b} = \dfrac{c}{d}, \: then \: componendo \: and \: dividendo \: means$

$\boxed{ \bf{ \frac{a + b}{a - b} = \frac{c + d}{c - d}}}$

$\large\underline{\sf{Solution-}}$

Given that,

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

can be rewritten as

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

On Applying Componendo and Dividendo, we GET

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

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