Question

Question no. 12 solve kr do n​

Answer 1

$\begin{gathered}a = 3 +2 \sqrt{2} \\ = > \frac{1}{a} = \frac{1}{3 + 2 \sqrt{2} } \\ = > \frac{1}{a} = \frac{1}{3 + 2\sqrt{2} } \times \frac{3 - 2 \sqrt{2} }{3 - 2 \sqrt{2} } \\ = > \frac{1}{a} = \frac{3 - 2 \sqrt{2} }{9 - 8} \\ = > \frac{1}{a} = 3 - 2 \sqrt{2} \\ \\ = > so \: (a + \frac{1}{a} ) = (3 + 2 \sqrt{2} + 3 - 2 \sqrt{2} ) \\ \\ = > (a + \frac{1}{a} ) = 6 \\ = > (a + \frac{1}{a} )^{2} = {6}^{2} squaring \: both \: side\\ = > {a}^{2} + ( \frac{1}{a} )^{2} + 2 \times a \times \frac{1}{a} = 36 \\ = > {a}^{2} + \frac{1}{ {a}^{2} } = 36 - 2 \\ = > {a}^{2} + \frac{1}{ {a}^{2} } = 34\end{gathered}$

Answer 2

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