Question

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

Answer 1

$\Rightarrow$ We are given, $a=\frac{3+\sqrt{5}}{2}$

we have to find $a^2+\frac{2}{a^2}$

now, $(a^2+\frac{1}{a^2})=(a^2+2a^2.\frac{1}{a^2}+\frac{1}{a^2})-2$

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

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

$={\left[\frac{(3+\sqrt{5}{)}^2+(2{)}^2}{\left(2\right)(3+\sqrt{5})}\right]}^2-2$

$={\left[\frac{9+6\sqrt{5}+5+4}{6+2\sqrt{5}}\right]}^2-2$

$={\left(\frac{18+6\sqrt{5}}{6+2\sqrt{5}}\right)}^2-2$

$={\left[\frac{3(6+2\sqrt{5})}{6+2\sqrt{3}}\right]}^2-2$

$={\left[3\right(1\left)\right]}^2-2$

$=3^2-2$

$=9-2$

$(a^2+\frac{1}{a^2})=7$