Question

if
$x + \frac{1}{x} = 7 \: then find \: the \: value \: of \: x {}^{2} + \frac{1}{x {}^{2} }$
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Answer 1

$\huge\red{SOLUTION}$

Given :-

$x + \frac{1}{x} = 7$

Squaring on both the sides :-

$(x + \frac{1}{x} )^{2} = ( {7)}^{2}$

${x}^{2} + ( \frac{1}{x})^{2} + 2(x)( \frac{1}{x} ) = 49$

$( \: \: ( {a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab \: \: )$

${x}^{2} + \frac{1}{ {(x)}^{2} } + 2 = 49$

${x}^{2} + \frac{1}{ {x}^{2} }$

$49 - 2$

Therefore,

${x}^{2} + \frac{1}{ {x}^{2} } = 47$

Answer 2

${ \tt{ \huge \underline{Given}}}$

${ \rm{ \large{x + \dfrac{1}{x} = 7}}}$

${ \tt{ \huge \underline{Solution}}}$

${ \rm{ \large{ {x}^{2} + \dfrac{1}{ {x}^{2} }}} }$

${ \rm{ \large{ = ( {x + \dfrac{1}{x}) }^{2} - 2.x. \dfrac{1}{x} }} }$

${ \rm{ \large{ = ( {x + \dfrac{1}{ x}) }^{2} - 2. \not x. \dfrac{1}{ \not x} }} }$

${ \rm{ \large{ = 49 - 2} }}$

${ \rm{ \large{ = 47} }}$

${ \tt{Ans \colon The \: value \: of \: {x}^{2} + \dfrac{1}{ {x}^{2} } \: is = 47}}$

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More Info

$\implies{(a+b)^2 = a^2 + b^2 + 2ab}$

$\implies{(a-b)^2 = a^2 + b^2 - 2ab}$

$\implies{(a+b)^3 = a^3 + b^3 + 3ab(a + b)}$

$\implies{(a-b)^3 = a^3 - b^3 - 3ab(a-b)}$

$\implies{(a^3+b^3)= (a+b)(a^2 - ab + b^2)}$

$\implies{(a^3-b^3)= (a-b)(a^2 + ab + b^2)}$

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