Question
If $\text{a}=\frac{3+\sqrt{5}}{2}$ then find the value of $\text{a}^2+\frac{1}{\text{a}^2}.$
✓
Answer
Given, $\text{a}=\frac{3+\sqrt{5}}{2}\ ...\text{(i)}$
Now, $\frac{1}{\text{a}}=\frac{2}{3+\sqrt{5}}=\frac{2}{3+\sqrt{5}}\times\frac{3-\sqrt{5}}{3-\sqrt{5}}$
$[$multiplying numerator and denominator by${ 3}-\sqrt{5}]$
$=\frac{6-2\sqrt{5}}{3^2-(\sqrt{5})^2}$
$[\text{using identity, (a}-\text{b) (a+b)}=\text{a}^2-\text{b}^2]$
$=\frac{6-2\sqrt{5}}{9-5}=\frac{6-2\sqrt{5}}{4}$
$\Rightarrow\ \frac{1}{\text{a}}=\frac{2(3-\sqrt{5})}{4}=\frac{3-\sqrt{5}}{2}\ ...(\text{ii})$
$\therefore\ \text{a}^2+\frac{1}{\text{a}^2}=\text{a}^2+\frac{1}{\text{a}^2}+2-2=\Big(\text{a}+\frac{1}{\text{a}}\Big)^2-2$ [adding and subtracting 2] $=\Big(\frac{3+\sqrt{5}}{2}+\frac{3-\sqrt{5}}{2}\Big)^2-2$ [from Eqs. $(i)$ and $(ii)$] $=\Big(\frac{6}{2}\Big)^2-2=(3)^2-2=9-2=7$
Now, $\frac{1}{\text{a}}=\frac{2}{3+\sqrt{5}}=\frac{2}{3+\sqrt{5}}\times\frac{3-\sqrt{5}}{3-\sqrt{5}}$
$[$multiplying numerator and denominator by${ 3}-\sqrt{5}]$
$=\frac{6-2\sqrt{5}}{3^2-(\sqrt{5})^2}$
$[\text{using identity, (a}-\text{b) (a+b)}=\text{a}^2-\text{b}^2]$
$=\frac{6-2\sqrt{5}}{9-5}=\frac{6-2\sqrt{5}}{4}$
$\Rightarrow\ \frac{1}{\text{a}}=\frac{2(3-\sqrt{5})}{4}=\frac{3-\sqrt{5}}{2}\ ...(\text{ii})$
$\therefore\ \text{a}^2+\frac{1}{\text{a}^2}=\text{a}^2+\frac{1}{\text{a}^2}+2-2=\Big(\text{a}+\frac{1}{\text{a}}\Big)^2-2$ [adding and subtracting 2] $=\Big(\frac{3+\sqrt{5}}{2}+\frac{3-\sqrt{5}}{2}\Big)^2-2$ [from Eqs. $(i)$ and $(ii)$] $=\Big(\frac{6}{2}\Big)^2-2=(3)^2-2=9-2=7$
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