Question

$\\ \sf{if \: x = \dfrac{ \sqrt{2a + 3b} + \sqrt{2a - 3b} }{ \sqrt{2a + 3b} - \sqrt{2a - 3b}} } \: then \: prove \: that \ratio-$
$\\ \sf{ \implies{3b {x}^{2} - 4ax + 3b = 0}}$

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Answer 1

$\\ \huge{ \underline{ \underline{ \sf{ \pmb{Question}}}}}$

$\\ \sf{If \: x = \dfrac{ \sqrt{2a + 3b} + \sqrt{2a - 3b} }{ \sqrt{2a + 3b} - \sqrt{2a - 3b}} } \: then \: prove \: that \ratio-$

$\\ \sf{ \implies{3b {x}^{2} - 4ax + 3b = 0}}$

$\\ \huge{ \underline{ \underline{ \sf{ \pmb{Required \: Answer}}}}}$

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

• $\\ \sf{x = \dfrac{ \sqrt{2a + 3b} + \sqrt{2a - 3b} }{ \sqrt{2a + 3b} - \sqrt{2a - 3b}} }$

$\\ \large{ \underline{ \underline{ \sf{ \pmb{To \: Do}}}}}$

• $\\ \sf{Prove \: that \ratio-3b {x}^{2} - 4ax + 3b = 0}$

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

$\\ \sf{ \bold{{x = \dfrac{ \sqrt{2a + 3b} + \sqrt{2a - 3b} }{ \sqrt{2a + 3b} - \sqrt{2a - 3b}}}}}$

By Componendo-Divinendo :-

$\\ \sf{ \implies{\dfrac{x + 1}{x - 1} = \dfrac{ \sqrt{2a + 3b} + \cancel{\sqrt{2a - 3b}} + \sqrt{2a + 3b} - \cancel{ \sqrt{2a - 3b}} }{ \cancel{\sqrt{2a + 3b}} + \sqrt{2a - 3b} - \cancel{ \sqrt{2a + 3b}} + \sqrt{2a - 3b}} }}$

$\\ \sf{ \implies{ \dfrac{x + 1}{x - 1} = \dfrac{2( \sqrt{2a + 3b} )}{2( \sqrt{2a - 3b} )}}}$

$\\ \sf{ \implies{ \dfrac{x + 1}{x - 1} = \dfrac{ \sqrt{2a + 3b} }{\sqrt{2a - 3b} }}}$

Squaring in both sides:-

$\\ \sf{ \implies{ \dfrac{(x + 1) {}^{2} }{(x - 1) {}^{2} } = \dfrac{ (\sqrt{2a + 3b} ) {}^{2} }{(\sqrt{2a - 3b}) {}^{2} }}}$

Applying formula of (a+b)^2 , (a-b)^2 and removing the roots:-

$\\ \sf{ \implies{ \dfrac{ {x}^{2} + 2x + 1 }{ {x}^{2} - 2x + 1} = \dfrac{2a + 3b}{2a - 3b} }}$

Again Componendo-Divinendo :-

$\\ \sf{ \implies{ \dfrac{ {x}^{2} + \cancel{ 2x} + 1 + {x}^{2} - \cancel{ 2x} + 1 }{{x}^{2} + 2x+ \cancel{1} - {x}^{2} + 2x - \cancel{1}} = \dfrac{2a + \cancel{3b} + 2a - \cancel{ 3b}}{ \cancel{2a} + 3b - \cancel{ 2a} + 3b}}}$

$\\ \sf{ \implies{ \dfrac{2 {x}^{2} + 2 }{4x} = \dfrac{4a}{6b} }}$

$\\ \sf{ \implies{ \dfrac{2( {x}^{2} + 1) }{2 \times 2x} = \dfrac{2 \times 2a}{2 \times 3b}}}$

$\\ \sf{ \implies{ \dfrac{ {x}^{2} + 1}{2x} = \dfrac{2a}{3b}}}$

$\\ \sf{ \implies{3b {x}^{2} + 3b = 4ax}}$

$\\ \sf{ \implies{3b {x}^{2} + 3b - 4ax = 0}}$

$\\ \sf{ \implies{ \red{ \bold{3b {x}^{2} - 4ax + 3b = 0}}}}$

Hence Proved!!

Answer 2

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