Question

Using properties of proportions ,solve for x given that x is positive:- *

Answer 1

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

Given that

$\rm :\longmapsto\:\dfrac{2x + \sqrt{ {4x}^{2} - 1} }{2x - \sqrt{ {4x}^{2} - 1 } } = \dfrac{4}{1}$

We know,

Componendo and Dividendo

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

$\red{\rm :\longmapsto\:\dfrac{a + b}{a - b} = \dfrac{c + d}{c - d}}$

So, on applying Componendo and Dividendo in equation, we get

$\rm :\longmapsto\:\dfrac{2x + \sqrt{ {4x}^{2} - 1} + 2x - \sqrt{ {4x}^{2} - 1 }}{2x + \sqrt{ {4x}^{2} - 1} - 2x + \sqrt{ {4x}^{2} - 1}} = \dfrac{4 + 1}{4 - 1}$

$\rm :\longmapsto\:\dfrac{4x}{2 \sqrt{ {4x}^{2} - 1} } = \dfrac{5}{3}$

$\rm :\longmapsto\:\dfrac{2x}{ \sqrt{ {4x}^{2} - 1} } = \dfrac{5}{3}$

On squaring both sides, we get

$\rm :\longmapsto\:\dfrac{ {4x}^{2} }{{4x}^{2} - 1} = \dfrac{25}{9}$

$\rm :\longmapsto\:36 {x}^{2} = 25( {4x}^{2} - 1)$

$\rm :\longmapsto\:36 {x}^{2} = {100x}^{2} -25$

$\rm :\longmapsto\:36 {x}^{2} - {100x}^{2} = -25$

$\rm :\longmapsto\: - {64x}^{2} = -25$

$\rm :\longmapsto\: {64x}^{2} = 25$

$\rm :\longmapsto\: {x}^{2} = \dfrac{25}{64}$

$\rm :\implies\:x = \: \pm \: \dfrac{5}{8}$

But it is given that x > 0

$\bf\implies \:x = \dfrac{5}{8}$

Hence,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underbrace{ \boxed{ \bf \: Option \: (a) \: is \: correct}}$

Additional Information :-

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

$\red{\rm :\longmapsto\: \: \dfrac{a}{c} = \dfrac{b}{d} \: is \: called \: alternendo}$

$\red{\rm :\longmapsto\: \: \dfrac{b}{a} = \dfrac{d}{c} \: is \: called \: invertendo}$

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

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