Question

if x=
$\frac{ \sqrt{5 } + 3 }{2}$
find the value of
${x}^{2} + \frac{1}{ {x}^{2} }$
​

Answer 1

ANSWER

$x = \frac{ \sqrt{5} + 3 }{2} \\ \frac{1}{x} = \frac{2}{ \sqrt{5} + 3} \\ = > \frac{1}{x} = \frac{2( \sqrt{5} - 3) }{( \sqrt{5} + 3)( \sqrt{5} - 3)} \\ = > \frac{1}{x} = \frac{2( \sqrt{5} - 3) }{5 - 9} \\ = > \frac{1}{x} = \frac{2( \sqrt{5} - 3) }{ - 4} \\ = > \frac{1}{x} = - \frac{ \sqrt{5} - 3}{2}$

therefore....

$x + \frac{1}{x} = \frac{ \sqrt{5} + 3}{2} + ( - \frac{ \sqrt{5} - 3}{2} ) \\ = > x + \frac{1}{x} = \frac{ \sqrt{5} + 3 - \sqrt{5} + 3}{2} \\ = > x + \frac{1}{x} = 3 \\ = > (x + \frac{1}{x} ) {}^{2} = 3 {}^{2} \\ = > x {}^{2} + \frac{1}{x {}^{2} } + 2(x)( \frac{1}{x} ) = 9 \\ = > x {}^{2} + \frac{1}{x {}^{2} } = 9$