Question

If $a=6-\sqrt{3}5$. find the value of $a^2+\frac{1}{a^2}$.

Answer 1

$a=6-\sqrt{35}a+\frac{1}{a}=6-\sqrt{35}+\frac{6+\sqrt{35}}{(6-\sqrt{35})(6+\sqrt{35})}=(6-\sqrt{35}+6+\sqrt{35})=12{(a+\frac{1}{a})}^2=a^2+\frac{1}{a^2}+2\Rightarrow {12}^2=a^2+\frac{1}{a^2}+2\Rightarrow a^2+\frac{1}{a^2}=142$