Question

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

Answer 1

$a=\frac{3+\sqrt{5}}{2}\therefore \frac{1}{a}=\frac{2}{3+\sqrt{5}}$

On rationaliging the denominator , we get ,

$\frac{1}{a}=\frac{2(3-\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})}=\frac{6-2\sqrt{5}}{9-5}=\frac{6-2\sqrt{5}}{4}=\frac{3-\sqrt{5}}{2}$

Also$,{(a+\frac{1}{a})}^2=a^2+\frac{1}{a^2}+2$

Substituting the values of $a\ne \frac{1}{a},$

We get ,

`${(\frac{3+\sqrt{5}}{2}+\frac{3-\sqrt{5}}{2})}^2=(a^2+\frac{1}{a^2}+2)$

$\therefore a^2+\frac{1}{a^2}+2={\left(\frac{3+\sqrt{5}+3-\sqrt{5}}{2}\right)}^2$

$=(3{)}^2=9$

$\therefore a^2+\frac{1}{a^2}=9-2=7.$