Question

$x + \frac{1}{x} = 6$
Find the value of
${x}^{2} + \frac{1}{ {x}^{2} } \: and \: {x}^{2} - \frac{1}{ {x}^{2} }$
​

Answer 1

Answer:

As x + 1/x = 6

square on both terms

(x + 1/x)² = (6)² {(a+b)² = a² + b² + 2ab)}

x² + 1/x² + 2×x×1/x = 36 {x will be cancelled}

x² + 1/x² + 2 = 36

x² + 1/x² = 36 - 2 = 34

Do the same in x - 1/x by prop. a² - 2ab + b²

Answer 2

Answer:

$x + \frac{1}{x} = 6 .......(i)\\ 1 \} \: \: squaring \: both \: side \: of \: eq(i) \\ {(x + \frac{1}{x}) }^{2} = 6 {}^{2} \\ {x}^{2} + \frac{1}{ {x}^{2} } + 2x \times \frac{1}{x} = 36 \\ {x}^{2} + \frac{1}{ {x}^{2} } + 2 = 36 \\ = > {x}^{2} + \frac{1 }{ {x}^{2} } = 36 - 2 = 34$