Question

if x =√3+√2/√3-√2 , find x`4+1/x`4​

Answer 1

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

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

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

$\rm :\longmapsto\: {x}^{4} + \dfrac{1}{ {x}^{4} }$

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\rm :\longmapsto\: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy$

$\rm :\longmapsto\: {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy$

$\rm :\longmapsto\: {(x + y)}^{2} + {(x - y)}^{2} =2({x}^{2} + {y}^{2})$

$\rm :\longmapsto\:(x + y)(x - y) = {x}^{2} - {y}^{2}$

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

Given that,

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

On rationalizing the denominator, we get

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

$\rm \: = \: \: \dfrac{ {( \sqrt{3} + \sqrt{2}) }^{2} }{ {( \sqrt{3} )}^{2} - {( \sqrt{2} )}^{2} }$

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

$\rm \: = \: \: 5 + 2 \sqrt{6}$

$\bf\implies \:x = \: \: 5 + 2 \sqrt{6}$

Now,

Consider,

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

$\rm \: = \: \: \dfrac{1}{5 + 2 \sqrt{6} }$

On rationalizing the denominator, we get

$\rm \: = \: \: \dfrac{1}{5 + 2 \sqrt{6} } \times \dfrac{5 - 2 \sqrt{6} }{5 - 2 \sqrt{6} }$

$\rm \: = \: \: \dfrac{5 - 2 \sqrt{6} }{ {(5)}^{2} - {(2 \sqrt{6}) }^{2} }$

$\rm \: = \: \: \dfrac{5 - 2 \sqrt{6} }{ 25 - 24}$

$\rm \: = \: \: 5 - 2 \sqrt{6}$

$\bf\implies \:\dfrac{1}{x} = \: \: 5 - 2 \sqrt{6}$

Therefore,

$\bf\implies \:x + \dfrac{1}{x} = \:5 + 2 \sqrt{6} + \: 5 - 2 \sqrt{6} = 10$

Hence,

$\bf\implies \:x + \dfrac{1}{x} = 10$

On squaring both sides, we get

$\rm :\longmapsto\: {\bigg(x + \dfrac{1}{x} \bigg) }^{2} = {(10)}^{2}$

$\rm :\longmapsto\: {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 \times x \times \dfrac{1}{x} = 100$

$\rm :\longmapsto\: {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 = 100$

$\rm :\longmapsto\: {x}^{2} + \dfrac{1}{ {x}^{2} } = 100 - 2$

$\rm :\longmapsto\: {x}^{2} + \dfrac{1}{ {x}^{2} } = 98$

Again, On squaring both sides, we get

$\rm :\longmapsto\: {\bigg( {x}^{2} + \dfrac{1}{ {x}^{2} } \bigg) }^{2} = {(98)}^{2}$

$\rm :\longmapsto\: {x}^{4} + \dfrac{1}{ {x}^{4} } + 2 \times {x}^{2} \times \dfrac{1}{ {x}^{2} } = 9604$

$\rm :\longmapsto\: {x}^{4} + \dfrac{1}{ {x}^{4} } + 2 = 9604$

$\rm :\longmapsto\: {x}^{4} + \dfrac{1}{ {x}^{4} } = 9604 - 2$

$\rm :\longmapsto\: {x}^{4} + \dfrac{1}{ {x}^{4} } = 9602$

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)