Math, asked by gokulsatija, 9 months ago

if x=2+√5 find x^2-1/x^2

Answers

Answered by DrNykterstein
0

</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>

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