Question

$if \: x - \frac{1}{x} = 2 \: \\ \\ find \: {x}^{2} + \frac{1}{ {x}^{2} }$
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Answer 1

$\huge\underline\mathfrak\pink{Answer}$

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

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$\huge\underline\mathfrak\pink{Explanation}$

Given :

$x - \frac{1}{x} = 6$

To find :

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

Solution :

We know that,

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

Let,

$a = x$

$b = \frac{1}{x}$

Put the value of a and b in the above identity,

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

Now, Put the given values in above identity.

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

$\rightarrow$ $4 + 2 = x^{2} + \frac{1}{x^{2} }$

$\rightarrow$ $x^{2} + \frac{1}{x^{2} } = 6$