Question

If
$x + \frac{1}{x} = 8$
find the value of
${x}^{2} + \frac{ {1} }{ {x}^{2} }$
​

Answer 1

Given :-

$\large \bf x + \frac{1}{x} = 8$

To find :-

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

Calculation :-

$\huge \sf (x + \frac{1}{x} )^{2} = {(8)}^{2} \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } + 2( \cancel x)( \frac{1}{ \cancel x} ) = 64 \\ \\ \sf {x}^{2} + \frac{1}{ {x}^{2} } + 2 = 64 \\ \\ \sf \: \: \: \: \: \: \: \: {x}^{2} + \frac{1}{ {x}^{2} } = 64 - 2 \\ \\ \sf \: \: \: \: \: \: \: \: {x}^{2} + \frac{1}{ {x}^{2} } = \red{62}$

#BAL

Answer 2

$\huge{\fbox{\fbox {\mathfrak{Solution :}}}}$

Given:

• $x + \frac{1}{x} = 8$

To Find :

• The value of $x^2 + \frac{1}{x^2 }$

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

Squaring on both sides :

$\implies {(x + \frac{1}{x} )}^{2} = 8^2 \\\implies (x)^2 + ( \frac{1}{x} )^2 + 2 \times x \times \frac{1}{x} = 64 \\ \implies x^2 + \frac{1}{x^2 } + 2 = 64 \\ \implies x^2 + \frac{1}{x^2 } = 64 - 2 \\\implies x^2 + \frac{1}{x^2 } = 62$

$\boxed {\green {\therefore{x^2 + \frac{1}{x^2 } = 62}}}$