Question

$\sf\red{If \: (x + \frac{1}{x}) = 5 \: then \: find \: ( {x}^{2} + \frac{1}{ {x}^{2} } and \: ( {x}^{4} + \frac{1}{ {x}^{4} }. }$
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Answer 1

We are given,

$\bigg(x + \frac{1}{x} \bigg ) = 5$

Let's figure out,

$\longrightarrow \bigg(x + \frac{1}{x} \bigg ) = 5$

$\longrightarrow {\bigg(x + \frac{1}{x} \bigg ) }^{2} = {5}^{2}$

$\longrightarrow \bigg( {x}^{2} + \frac{1}{ {x}^{2}} + 2 \times x \times \frac{1}{x} \bigg ) = 25$

$\longrightarrow \bigg( {x}^{2} + \frac{1}{ {x}^{2}} + 2 \times \cancel{x} \times \frac{1}{ \cancel{x}} \bigg ) = 25$

$\longrightarrow \bigg( {x}^{2} + \frac{1}{ {x}^{2}} \bigg) = 25 - 2$

$\longrightarrow \bf\bigg( {x}^{2} + \frac{1}{ {x}^{2}} \bigg) = 23 \: \: \: Ans. no ( 1 )$

$\: \: \mapsto \ \: Let's \: \: square \: \: both \: \: sides$

$\longrightarrow {\bigg( {x}^{2} + \frac{1}{ {x}^{2}} \bigg) }^{2} = {23}^{2}$

$\longrightarrow {\bigg( {x}^{4} + \frac{1}{ {x}^{4}} + 2 \times {x}^{2} \times \frac{1}{ {x}^{2} } \bigg) }^{} = 529$

$\longrightarrow {\bigg( {x}^{4} + \frac{1}{ {x}^{4}} + 2 \times \cancel{ {x}^{2}} \times \frac{1}{ \cancel{{x}^{2} }} \bigg) }^{} = 529$

$\longrightarrow \bf {\bigg( {x}^{4} + \frac{1}{ {x}^{4}} \bigg) }^{} = 527 \: \: Ans. no ( 2)$

Answer 2

αղsաҽɾ :

• $\sf x^2 + \frac{1}{x^2} = 23$

• $\sf x^4 + \frac{1}{x^4} = 527$

ҽxթlαղαԵíօղ :

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$\longrightarrow$ In equation $\sf x + \frac{1}{x} = 5$

$\longrightarrow$ Now take square from both side In given equation ,

• $\sf \bigg( x + \frac{1}{x} \bigg)^{2} = (5)²$

$\longrightarrow$ Here, We can apply Identity of (a + b)² = a² + b² + 2ab

• $\sf (x )^2 + \bigg(\frac{1}{x}\bigg)^2 + 2(x) \bigg(\frac{1}{x}\bigg) = 25$

• $\sf (x)^2 + \bigg(\frac{1}{x}\bigg)^2 + 2(\cancel{x}) \bigg(\frac{1}{\cancel{x}}\bigg) = 25$

• $\sf x^2 + \frac{1}{x^2} + 2 = 25$

• $\sf {x}^{2} + \frac{1}{ {x}^{2} } = 25 - 2$

• $\sf {x}^{2} + \frac{1}{ {x}^{2} } = 23$

$\longrightarrow$ Hence,

• $\purple{\underline{\boxed{\bf {x}^{2} + \frac{1}{ {x}^{2} } = 23 }}}$

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$\longrightarrow$ Now, For $\sf x^4 + \frac{1}{x^4}$

$\longrightarrow$ Here, we had already Find out that,

• $\sf {x}^{2} + \frac{1}{ {x}^{2} } = 23$

$\longrightarrow$ So, Now Again We have to take square from both sides in above equation

$\longrightarrow$ Hence,

• $\sf \bigg( {x}^{2} + \frac{1}{ {x}^{2} }\bigg)^{2} = ({23})^{2}$

$\longrightarrow$ Here, Again We can apply Identity of (a + b)² = a² + b² + 2ab

• $\sf ( {x}^{2})^2 + \bigg(\frac{1}{ {x}^{2} }\bigg)^{2} + 2( {x}^{2}) \bigg( \frac{1}{ {x}^{2} } \bigg) = ({23})^{2}$

• $\sf ( {x}^{2} )^{2} + \bigg(\frac{1}{ {x}^{2} }\bigg)^{2} + 2( \cancel{{x}^{2}}) \bigg( \frac{1}{\cancel{ {x}^{2}} } \bigg) = ({23})^{2}$

• $\sf {x}^{4} + \frac{1}{ {x}^{4} } + 2 = 529$

• $\sf {x}^{4} + \frac{1}{ {x}^{4} } = 529 - 2$

• $\sf {x}^{4} + \frac{1}{ {x}^{4} } = 527$

$\longrightarrow$ Hence,

• $\purple { \underline{\boxed{\bf {x}^{4} + \frac{1}{ {x}^{4} } = 527 }}}$

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I hope it helps you ❤️✔️