Question

$\sf \red{\bigg(x - \frac{1}{x} \bigg) ^{ \frac{1}{2} } + \bigg(1 - \frac{1}{x} \bigg)^{ \frac{1}{2} } = x}$​

Answer 1

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

Given expression is

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

We know,

$\red{\rm :\longmapsto\: \sqrt{x} + \sqrt{y} = \dfrac{x - y}{ \sqrt{x} - \sqrt{y} }}$

So, using this we get

$\rm :\longmapsto\:\dfrac{x - \dfrac{1}{x} - \bigg(1 - \dfrac{1}{x} \bigg) }{{\bigg(x - \dfrac{1}{x} \bigg) ^{ \dfrac{1}{2} } - \bigg(1 - \dfrac{1}{x} \bigg)^{ \dfrac{1}{2} }}} = x$

can be further rewritten as

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

$\rm :\longmapsto\:\dfrac{x - 1 }{{\bigg(x - \dfrac{1}{x} \bigg) ^{ \dfrac{1}{2} } - \bigg(1 - \dfrac{1}{x} \bigg)^{ \dfrac{1}{2} }}} = x$

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

can be further rewritten as

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

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

On adding equation (1) and (2), we get

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

can be rewritten as

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

On squaring both sides, we get

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

$\rm :\longmapsto\: {\bigg(x - \dfrac{1}{x} \bigg) }^{2} + 1 - 2\bigg(x - \dfrac{1}{x} \bigg) = 0$

can be reduced to

$\rm :\longmapsto\: {\bigg(x - \dfrac{1}{x} - 1 \bigg) }^{2} = 0$

$\rm :\longmapsto\:x - \dfrac{1}{x} - 1 = 0$

$\rm :\longmapsto\:\dfrac{ {x}^{2} - 1 - x}{x} = 0$

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

Now, to get the roots of this quadratic equation, we use Quadratic Formula.

So,

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

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

$\rm :\longmapsto\:x = \dfrac{1\: \pm \: \sqrt{5} }{2}$

Hence,

$\bf\implies \:\boxed{\bf{ \: x = \dfrac{1\: \pm \: \sqrt{5} }{2} \: }}$