Question

if x is equal to 2 + root 5 prove that X square + 1 by x square is equal to 18​

Answer 1

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

Given that .

$\rm :\longmapsto\:x = \sqrt{5} + 2$

So,

$\rm :\longmapsto\:\dfrac{1}{x}$

$\rm \:  =  \:  \: \dfrac{1}{ \sqrt{5} + 2} \times \dfrac{ \sqrt{5} - 2}{ \sqrt{5} - 2}$

${\bigg \{ \because \:(x + y)(x - y) = {x}^{2} - {y}^{2} \bigg \}}$

$\rm \:  =  \:  \: \dfrac{ \sqrt{5} - 2}{ {( \sqrt{5})}^{2} - {(2)}^{2} }$

$\rm \:  =  \:  \: \dfrac{ \sqrt{5} - 2}{5 - 4}$

$\rm \:  =  \:  \: \dfrac{ \sqrt{5} - 2}{1}$

$\rm \:  =  \:  \: \sqrt{5} - 2$

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

Now,

$\rm :\longmapsto\: {x}^{2} + \dfrac{1}{ {x}^{2} }$

$\rm \:  =  \:  \: {( \sqrt{5} + 2) }^{2} + {( \sqrt{5} - 2)}^{2}$

${\bigg \{ \because \: {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2}) \bigg \}}$

$\rm \:  =  \:  \: 2(5 + 4)$

$\rm \:  =  \:  \: 2 \times 9$

$\rm \:  =  \:  \: 18$

$\bf\implies \:\: {x}^{2} + \dfrac{1}{ {x}^{2} } = 18$

Proved