Question

Solve it Please:

If
$x - \frac{1}{x} = 9$
Find the value of :

${x}^{2} + \frac{1}{ {x}^{2} }$
​

Answer 1

$\huge\bold\orange{Solution!!}$

We have :

x-1/x = 9

On squaring both sides :

$(x - \frac{1}{ {x}^{2} } ) = {9}^{2}$

${x}^{2} - 2 \times x \times \frac{1}{x} + (\frac{1}{x} ) ^{2} = 81$

${x}^{2} - 2 + \frac{1}{ {x}^{2} } = 81$

${x}^{2} + \frac{1}{ {x}^{2} } = 81 + 2$

${x}^{2} + \frac{1}{ {x}^{2} } = 83$

$\bold\yellow{Glad\:to\:help!!}$

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Answer 2

Answer:

83

Step-by-step explanation:

$x - \frac{1}{x} = 9 \\ squiring \: both \: side \: we \: get \\ {(x - \frac{1}{x} })^{2} = {9}^{2} \\ {x}^{2} + { (\frac{1}{x}) }^{2} - 2 = 81 \\ {x}^{2} - \frac{1}{ {x}^{2} } = 83$