Question

If a = 5 + 2√6, then find the value of a² + 1/a²

P. S. - (a+b)²= a² + 2ab +b²​

Answer 1

Identities used :

$\begin{gathered} {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab \\ \\ {(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab \\ \\ (a + b)(a - b) = {a}^{2} - {b}^{2}\end{gathered}$

Given,

a = 5+ 2√6

So,

$\begin{gathered} \frac{1}{a} = \frac{1}{5 + 2 \sqrt{6} } \\ \end{gathered}$

On rationalizing the denominator we get,

$\begin{gathered} \frac{1}{a} = \frac{1}{5 + 2 \sqrt{6} } \times \frac{5 - 2 \sqrt{6} }{5 - 2 \sqrt{6} } \\ \\ \frac{1}{a} = \frac{5 - 2 \sqrt{6} }{ {(5)}^{2} - {(2 \sqrt{6} )}^{2} } \\ \\ \frac{1}{a} = \frac{5 - 2 \sqrt{6} }{25 - 24} \\ \\ \frac{1}{a} = 5 - 2 \sqrt{6} \end{gathered}$

Now,

$\begin{gathered}a + \frac{1}{a} = (5 + 2 \sqrt{6} ) + (5 - 2 \sqrt{6} ) \\ \\ a + \frac{1}{a} = 5 + 2 \sqrt{6} + 5 - 2 \sqrt{6} \\ \\ a + \frac{1}{a} = 10\end{gathered}$

Squaring both the sides we get,

$\begin{gathered} {(a + \frac{1}{a} )}^{2} = {(10)}^{2} \\ \\ {a}^{2} + {( \frac{1}{a} )}^{2} + 2 \times a \times \frac{1}{a} = 100 \\ \\ {a}^{2} + \frac{1}{ {a}^{2} } = 100 - 2 \\ \\ { a }^{2} + \frac{1}{ {a}^{2} } = 98\end{gathered}$