Question

If a=1-√2 find value of a+1/a and a^2+1/a^2

Answer 1

Heya there !!!
Here is the answer you were looking for:
$a = 1 - \sqrt{2} \\ \\ \frac{1}{a} = \frac{1}{1 - \sqrt{2} }$

On rationalizing the denominator we get,

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

Squaring both the sides

${(a + \frac{1}{a}) }^{2} = {( - 2 \sqrt{2}) }^{2} \\ \\ {(a)}^{2} + {( \frac{1}{a}) }^{2} + 2 \times a \times \frac{1}{a} = 8 \\ \\ {a}^{2} + \frac{1}{ {a}^{2} } + 2 = 8 \\ \\ {a}^{2} + \frac{1}{ {a}^{2} } = 8 - 2 \\ \\ {a}^{2} + \frac{1}{ {a}^{2} } = 6$

Hope this helps!!!

Feel free to ask in the comment section if you have any doubt regarding to my answer...

@Mahak24

Thanks...
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