Question

$\frac{x + 1}{x} = 11 \: find \: the \: value \: of \: {x }^{2} + \frac{1}{ {x}^{2} }$
Please give the answer fast​

Answer 1

$\underline{ \boxed{ \mathcal{ \dag \: APPROPRIATE \: QUESTION \: \dag}}}$

$\sf \frac{x²+1}{x} = 11 \: then \: find \: the \: value \: of \: {x}^{2} + \frac{1}{ {x}^{2} }$

$\underline{ \boxed{ \mathcal{ \dag \: FORMULA \: \: USED \: \dag }}}$

$\sf \leadsto {x}^{2} + \frac{1}{ {x}^{2} } = {(x + \frac{1}{x}) }^{2} - 2$

$\underline{ \boxed{ \mathcal{ \dag \: SOLUTION \: \dag}}}$

$\sf \rarr \frac{ {x}^{2} + 1 }{x} \: can \: be \: written \: as \: \: x + \frac{1}{x}$

$\sf \therefore x + \frac{1}{x} = 11$

• Using the formula according to the question we get

$\sf \rarr {x}^{2} + \frac{1}{ {x}^{2} } = {(11)}^{2} - 2$

$\sf \rarr {x}^{2} + \frac{1}{ {x}^{2} } = 121 - 2$

$\sf \rarr {x}^{2} + \frac{1}{ {x}^{2} } = 119$

So the answer is 119

$\underline{ \boxed{ \mathcal{ \dag \: KNOW \: MORE \: \dag}}}$

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

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

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

$\sf \looparrowright {a}^{2} + {b}^{2} = {(a - b)}^{2} + 2$

$\sf \looparrowright(a - b)(a + b) = {a}^{2} - {b}^{2}$

$\sf \looparrowright {(a + b)}^{2} + {(a - b)}^{2} = 2( {a}^{2} + {b}^{2} )$

$\sf \looparrowright {(a + b)}^{2} - {(a - b)}^{2} = 4ab$

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