Question

if x=2+√5 find x^2-1/x^2

Answer 1

$</p><p>\sf x = 2 + \sqrt{5} \\ \\</p><p>$

$</p><p> \sf \rightarrow \quad x - \frac{1}{x} \\ \\ \sf \rightarrow \quad 2 + \sqrt{5} - \bigg( \frac{1}{2 + \sqrt{5} } \times \frac{2 - \sqrt{5} }{2 - \sqrt{5} } \bigg) \\ \\ \sf \rightarrow \quad 2 + \sqrt{5} - \bigg( \frac{2 - \sqrt{5} }{ {2}^{2} - {( \sqrt{5} )}^{2} } \bigg) \\ \\ \sf \rightarrow \quad 2 + \sqrt{5} - ( -( 2 - \sqrt{5} )) \\ \\ \sf \rightarrow \quad 2 + \cancel{\sqrt{5}} + 2 \cancel{- \sqrt{5} } \\ \\ \sf \rightarrow \quad x - \frac{1}{x} = 4 \qquad ...(1) \\ \\ </p><p>$

Now,

$</p><p> \sf \rightarrow \quad x - \frac{1}{x} = 4 \\ \\ \sf \quad square \: both \: sides \\ \\ \sf \rightarrow \quad { \bigg( x - \frac{1}{x} \bigg)}^{2} = {4}^{2} \\ \\ \sf \rightarrow \quad {x}^{2} - \frac{1}{ {x}^{2} } + 2 \cdot \cancel{x} \cdot \frac{1}{ \cancel{x}} = 16 \\ \\ \sf \rightarrow \quad {x}^{2} - \frac{1}{ {x}^{2} } = 14</p><p></p><p>$