Math, asked by Spoileralert, 1 month ago

Solve for x ,Solve for x ,Solve for x ,Solve for x ,Solve for x ,​

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Answered by mathdude500
9

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

Given expression is

\rm :\longmapsto\:\dfrac{ \sqrt{a + x}  +  \sqrt{a - x} }{ \sqrt{a + x}  -  \sqrt{a - x} }  = b

can be rewritten as

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

Apply Componendo and Dividendo, we get

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

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

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

On squaring both sides, we get

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

\rm :\longmapsto\:\dfrac{a + x}{a - x}  = \dfrac{ {b}^{2}  + 1 + 2b}{ {b}^{2} + 1 - 2b}

On applying Componendo and Dividendo, we get

\rm :\longmapsto\:\dfrac{a + x + a - x}{a + x - a + x}  = \dfrac{ {b}^{2}  + 1 + 2b +  {b}^{2}  + 1 - 2b}{ {b}^{2} + 1 + 2b -  {b}^{2} - 1 + 2b }

\rm :\longmapsto\:\dfrac{2a}{2x}  = \dfrac{2 {b}^{2}  + 2 }{ 4b }

\rm :\longmapsto\:\dfrac{a}{x}  = \dfrac{2( {b}^{2}  + 1) }{ 4b }

\rm :\longmapsto\:\dfrac{a}{x}  = \dfrac{ {b}^{2}  + 1 }{ 2b }

\bf\implies \:x = \dfrac{2ab}{ {b}^{2}  + 1}

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Additional Information :-

If a, b, c, d are positive numbers such that a : b :: c : d then

1. Invertendo

\boxed{ \tt{ \:  \frac{b}{a} =  \frac{d}{c} \: }}

2. Alternendo

\boxed{ \tt{ \:  \frac{a}{c} =  \frac{b}{d} \: }}

3. Componendo

\boxed{ \tt{ \:  \frac{a + b}{b} =  \frac{c + d}{d} \: }}

4. Dividendo

\boxed{ \tt{ \:  \frac{a  -  b}{b} =  \frac{c - d}{d} \: }}

5. Componendo and Dividendo

\boxed{ \tt{ \:  \frac{a + b}{a - b} =  \frac{c + d}{c - d} \: }}

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