Question

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Answer 1

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

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

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

$\rm :\longmapsto\:value \: of \: x$

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\rm :\longmapsto\:If \: \dfrac{a}{b} = \dfrac{c}{d} \: then \:$

$\red{\boxed{ \sf{ \:\dfrac{b}{a} = \dfrac{d}{c} \: is \: called \: invertendo}}}$

$\red{\boxed{ \sf{ \:\dfrac{a + b}{a - b} = \dfrac{c + d}{c - d} \: is \: called \: componendo \: and \: dividendo}}}$

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

Given that

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

On using process of invertendo, we get

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

On Applying Componendo and Dividendo, we get

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

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

$\rm :\longmapsto\:\dfrac{ 2\sqrt{x + 1} }{2\sqrt{x - 1}} = \dfrac{3}{1}$

$\rm :\longmapsto\:\dfrac{ \sqrt{x + 1} }{\sqrt{x - 1}} = \dfrac{3}{1}$

On squaring both sides, we get

$\rm :\longmapsto\:\dfrac{x + 1}{x - 1} = \dfrac{9}{1}$

$\rm :\longmapsto\:9x - 9 = x + 1$

$\rm :\longmapsto\:9x - x = 9 + 1$

$\rm :\longmapsto\:8x = 10$

$\rm :\longmapsto\:4x = 5$

$\bf\implies \:x = \dfrac{5}{4}$

Additional Information :-

$\rm :\longmapsto\:If \: \dfrac{a}{b} = \dfrac{c}{d} \: then \:$

$\red{\boxed{ \sf{ \:\dfrac{a}{c} = \dfrac{b}{d} \: is \: called \: alternendo}}}$

$\red{\boxed{ \sf{ \:\dfrac{a + b}{b} = \dfrac{c + d}{d} \: is \: called \: componendo}}}$

$\red{\boxed{ \sf{ \:\dfrac{a - b}{b} = \dfrac{c - d}{d} \: is \: called \: dividendo}}}$

Answer 2

Given:

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

Solution:

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

Applying Componendo and Dividendo we get:

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \implies \tt \: \dfrac{ \sqrt{x + 1 }- \sqrt{x - 1} + \sqrt{x + 1 }+ \sqrt{x - 1}} { \sqrt{x + 1 }- \sqrt{x - 1} - \sqrt{x + 1 } - \sqrt{x - 1} } = \dfrac{3}{ - 1} \: \\ \\ \\ \tt \implies \frac{2 \sqrt{x + 1} }{ - 2 \sqrt{x - 1} } = \frac{3}{ - 1} \\ \\ \\ \tt \implies \frac { \cancel{2} \sqrt{x + 1} }{ \cancel{2 }\sqrt{x - 1} } = 3 \\ \\ \\ \tt \implies \frac{x + 1}{x - 1} = 9 \\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt \implies9(x - 1) = x + 1 \\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt \implies9x - 9 = x + 1 \\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt \implies9x - x = 9 + 1 \\ \\ \\\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt \implies8x = 10 \\ \\ \\\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt \implies \: x = \frac{10}{8} \\ \\ \\\: \: \: \: \: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt \implies \: x = \frac{ \cancel{10}}{ \cancel{8}} \\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \ \: \: \: \ \: \: \: \: \:\:\implies \tt \: x = \frac{5}{4}$

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• Hence Solved!

Explore more:

1. Dividendo:

• a – b : b = c – d : d

2. Componendo

• a + b : b = c + d : d

3. Componendo and Dividendo:

• a + b : a – b = c + d : c – d

4. Convertendo:

• a : a – b = c : c – d