Question

if x=2+root2, find the value of X^2 + 1/x^2

Answer 1

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

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

$\large\underline{\sf{To\:Find - }}$

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

Formula Used :-

$\boxed{ \sf{ \: {(x)}^{2} - {(y)}^{2} = (x + y)(x - y)}}$

$\boxed{ \sf{ \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}}$

$\boxed{ \sf{ \: {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy}}$

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

Given that,

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

Consider,

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

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

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

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

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

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

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

Consider,

$\rm :\longmapsto\: {x}^{2}$

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

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

$\rm \: = \: \: 4 + 2 + 4 \sqrt{2}$

$\rm \: = \: \: 6+ 4 \sqrt{2}$

$\bf\implies \: {x}^{2} = 6 + 4 \sqrt{2}$

Now,

Consider,

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

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

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

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

$\rm \: = \: \: \dfrac{6 - 4 \sqrt{2} }{4}$

$\rm \: = \: \: \dfrac{3 - 2 \sqrt{2} }{2}$

$\bf\implies \:\dfrac{1}{ {x}^{2} } = \: \: \dfrac{3 - 2 \sqrt{2} }{2}$

Hence,

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

$\rm \: = \: \: 6 + 4 \sqrt{2} + \dfrac{3 - 2 \sqrt{2} }{2}$

$\rm \: = \: \: \dfrac{12 + 8 \sqrt{2} + 3 - 2 \sqrt{2} }{2}$

$\rm \: = \: \: \dfrac{15+ 6\sqrt{2} }{2}$

$\rm \: = \: \: \dfrac{15}{2} + 3 \sqrt{2}$

Additional Information :-

More Identities to know:

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

a² - b² = (a + b)(a - b)

(a + b)² = (a - b)² + 4ab

(a - b)² = (a + b)² - 4ab

(a + b)² + (a - b)² = 2(a² + b²)

(a + b)³ = a³ + b³ + 3ab(a + b)

(a - b)³ = a³ - b³ - 3ab(a - b)