Question

if a is equal to 3 + square root 52 then find a square + 1 upon a square​

Answer 1

$a = 3 + \sqrt{52} \\ \\ \frac{1}{a} = \frac{1}{3 + \sqrt{52} } \times \frac{3 - \sqrt{52} }{3 - \sqrt{52} } \\ \\ \: \: \: \: \: = \frac{3 - \sqrt{52} }{9 - 52} = \frac{3 - \sqrt{52} }{ - 43}$

Now,

${a}^{2} + \frac{1}{ {a}^{2} } \\ \\ = {(3 + \sqrt{52} )}^{2} + {( \frac{3 - \sqrt{52} }{ - 43} )}^{2} \\ \\ = (9 + 52 + 6 \sqrt{52} )( \frac{9 + 52 - 6 \sqrt{52} }{1849} ) \\ \\ = (61 + 6 \sqrt{52} )( \frac{61 - 6 \sqrt{52} }{1849} ) \\ \\ = \frac{ {(61)}^{2} - {(6 \sqrt{52} )}^{2} }{1849} \\ \\ = \frac{3721 - 1872}{1849} \\ \\ = \frac{1849}{1849} = 1$