Question

If x= √2+1/√2-1, then the value of x^2+1/x^2​

Answer 1

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

$\boxed{ \:{1. \: \tt \:  {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}}$

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

$\boxed{ \:{3. \: \tt \:  {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2} )}}$

$\boxed{ \:{4. \: \tt \:  {(x + y)}(x - y) = {x}^{2} - {y}^{2} }}$

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$\large\underline{\bold{Solution :- }}$

☆ Consider,

$\tt \: \tt \:  ⟼ x \: = \dfrac{ \sqrt{2} + 1 }{ \sqrt{2} - 1}$

☆ On rationalizing the denominator, we get

$\tt \:  ⟼ x \: = \dfrac{ \sqrt{2} + 1 }{ \sqrt{2} - 1} \times \dfrac{\sqrt{2} + 1}{\sqrt{2} + 1}$

$\tt \:  ⟼ \: x \: = \dfrac{ {(\sqrt{2} + 1)}^{2} }{ {( \sqrt{2} )}^{2} - {1}^{2} }$

$\tt \:  ⟼ \: x \: = \dfrac{ {( \sqrt{2}) }^{2} + 1 - 2 \times 1 \times \sqrt{2} }{2 - 1}$

$\tt \:  ⟼ \: x \: = 3 + 2 \sqrt{2}$

☆ Now, Consider

$\tt \:  ⟼ \: \dfrac{1}{x} = \dfrac{1}{3 + 2 \sqrt{2} }$

$\tt \:  ⟼ \: on \: rationalizing \: the \: denominator \\ \tt \:  ⟼ \: \dfrac{1}{x} \: = \dfrac{1}{3 + 2 \sqrt{2} } \: \times \dfrac{3 - 2 \sqrt{2} }{3 - 2 \sqrt{2} } \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\tt \:  ⟼ \: \dfrac{1}{x} \: = \dfrac{3 - 2 \sqrt{2} }{ {(3)}^{2} - {(2 \sqrt{2}) }^{2} }$

$\tt \:  ⟼ \: \dfrac{1}{x} \: = \dfrac{3 - 2 \sqrt{2} }{9 - 8}$

$\tt \:  ⟼ \: \dfrac{1}{x} \: = \: 3 - 2 \sqrt{2}$

$\tt \:  ⟼☆ \: Consider \: {x}^{2} + \dfrac{1}{ {x}^{2} }$

$\tt \:  = \: { \bigg(3 + 2 \sqrt{2} \bigg) }^{2} + { \bigg(3 - 2 \sqrt{2} \bigg) }^{2}$

$\tt \:  = \: 2 \bigg( {(3)}^{2} + {(2 \sqrt{2} )}^{2} \bigg)$

$\tt \:  = \: 2(9 + 8) \: = 2 \times 17 \: = 34$

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$\large{\boxed{\boxed{\tt{Hence, \: {x}^{2} + \dfrac{1}{ {x}^{2} } = 34}}}}$

Answer 2

Answer:

$\frac{\sqrt{2+1} }{\sqrt{2-1} } = x$ ⇒

if x =$x^{2} + \frac{1}{x^{2} }$ then

x = $(\frac {\sqrt{2+1} }{\sqrt{2-1} } ) ^{2}$ + 1 ÷ $(\frac {\sqrt{2+1} }{\sqrt{2-1} } ) ^{2}$ ⇒

$x^{2} + \frac{1}{x^{2} }$= $(\frac{3}{1} ) + (\frac{1}{\frac{3}{1} } )$ ⇒

the value of $x^{2} + \frac{1}{x^{2} }$ = $\frac{9+1}{ 3}$ = $\frac{10}{3}$

Step-by-step explanation: