Question

Question 4
If $a=\frac{3+\sqrt{5}}{2}$, then find the value of $a^2+\frac{1}{a^2}$.

Answer 1

Given, $a=\frac{3+\sqrt{5}}{2}$
Now, $\frac{1}{a}=\frac{2}{3+\sqrt{5}}=\frac{2}{3+\sqrt{5}}\times \frac{3-\sqrt{5}}{3-\sqrt{5}}$
[Multiplying numerator and denominator by $3-\sqrt{5}]$
$=\frac{6-2\sqrt{5}}{3^2-(\sqrt{5}{)}^2}$ [Using identity, $(a-b)(a+b)=a^2-b^2$]
$=\frac{6-2\sqrt{5}}{9-5}=\frac{6-2\sqrt{5}}{4}$
$\Rightarrow \frac{1}{a}=\frac{2(3-\sqrt{5})}{4}=\frac{3-\sqrt{5}}{2}$
$\therefore a^2+\frac{1}{a^2}=a^2+\frac{1}{a^2}+2-2={(a+\frac{1}{a})}^2-2$
[adding and subtracting 2]
$={(\frac{3+\sqrt{5}}{2}+\frac{3-\sqrt{5}}{2})}^2-2$
$={\left(\frac{6}{2}\right)}^2-2=(3{)}^2-2=9-2=7$