Question

if a is equal to root 3 minus root 2 upon root 3 + root 2 and b is equal to root 3 + root 2 upon root 3 minus root 2 find the value of a square + b square minus 5 ab...​

Answer 1

$a = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ b = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ \\ ab = \frac{( \sqrt{3} - \sqrt{2} )}{ (\sqrt{ 3} + \sqrt{2} )} \frac{( \sqrt{3} + \sqrt{2}) }{( \sqrt{3} - \sqrt{2} ) } \\ \\ ab = 1 \\ \\ now \\ \\ a = \frac{( \sqrt{3} - \sqrt{2}) }{( \sqrt{3} + \sqrt{2} )} \\ \\ a = \frac{ (\sqrt{3} - \sqrt{2} ) }{ (\sqrt{3} + \sqrt{2}) } \frac{ (\sqrt{3} - \sqrt{2} ) }{( \sqrt{3} - \sqrt{2} )} \\ \\ a = {( \sqrt{3} - \sqrt{2} ) }^{2} \\ \\ a = 5 - 2 \sqrt{6} \\ \\ b = \frac{( \sqrt{3} + \sqrt{2} )}{( \sqrt{3} - \sqrt{2} )} \\ \\ b = \frac{( \sqrt{3} + \sqrt{2} )}{( \sqrt{3} - \sqrt{2}) } \frac{( \sqrt{3} + \sqrt{2} )}{( \sqrt{3} + \sqrt{2}) } \\ \\ b = {( \sqrt{3} + \sqrt{2}) }^{2} \\ \\ b = 5 + 2 \sqrt{6} \\ \\ {a}^{2} = 49 - 20 \sqrt{6} \\ \\ {b}^{2} = 49 + 20 \sqrt{6} \\ \\ now \\ \\ {a}^{2} + {b}^{2} - 5ab \\ \\ = 93$

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