Question

If $x=5+2\sqrt{3}$, find $x^2+\frac{1}{x^2}$

Answer 1

$x=5+2\sqrt{3}$

$x^2={(5+2\sqrt{3})}^2=25+12+20\sqrt{3}=37+20\sqrt{3}$

$\frac{1}{x^2}=\frac{1}{37+20\sqrt{3}}\times \frac{37-20\sqrt{3}}{37-20\sqrt{3}}$

$=\frac{37-20\sqrt{3}}{1369‬-1200}$

$=\frac{37-20\sqrt{3}}{196}$

$\therefore x^2+\frac{1}{x^2}=37+20\sqrt{3}+\frac{37-20\sqrt{3}}{196}$

$=\frac{7252+3920\sqrt{3}‬+37-20\sqrt{3}}{196}$

$=\frac{7289+3900\sqrt{3}‬}{196}$

$\therefore x^2+\frac{1}{x^2}=\frac{7289+3900\sqrt{3}‬}{196}$