Question

a = (7 - √5)/(7+√5) and b = (7+√5)/(7-√5) find a^2+b^2

Answer 1

Hey friend,
Here is the answer you were looking for:
$a = \frac{7 - \sqrt{5} }{7 + \sqrt{5} } \\ \\ on \: ratonalizing \: we \: get \\ \\ \frac{7 - \sqrt{5} }{7 + \sqrt{5} } \times \frac{7 - \sqrt{5} }{7 - \sqrt{5} } \\ \\ using \: identities \\ {(a - b)}^{2} = {(a)}^{2} + {(b)}^{2} - 2ab \\ (a + b)(a - b) = {(a)}^{2} - {(b)}^{2} \\ \\ = \frac{ {(7)}^{2} + {( \sqrt{5}) }^{2} - 2 \times 7 \times \sqrt{5} }{ {(7)}^{2} - {( \sqrt{5}) }^{2} } \\ \\ = \frac{49 + 5 - 14 \sqrt{5} }{49 - 5} \\ \\ = \frac{54 - 14 \sqrt{5} }{44} \\ \\ a= \frac{27 - 7 \sqrt{5} }{22} \\ \\ b = \frac{7 + \sqrt{5} }{7 - \sqrt{5} } \times \frac{7 + \sqrt{5} }{7 + \sqrt{5} } \\ \\ = \frac{ {(7)}^{2} + {( \sqrt{5} )}^{2} + 2 \times 7 \times \sqrt{5} }{ {(7)}^{2} - {( \sqrt{5} )}^{2} } \\ \\ = \frac{49 + 5 + 14 \sqrt{5} }{49 - 5} \\ \\ = \frac{54 + 14 \sqrt{5} }{44} \\ \\ = \frac{27 + 7 \sqrt{5} }{22} \\ \\ {a}^{2} + {b}^{2} \\ \\ = {( \frac{27 - 7 \sqrt{5} }{22} )}^{2} + {( \frac{27 + 7 \sqrt{5} }{22} )}^{2} \\ \\ using \: the \: identities \\ {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab \\ {(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab \\ \\ =( {( \frac{27}{22} )}^{2} + {( \frac{7 \sqrt{5} }{22}) }^{2} - 2 \times \frac{27}{22} \times \frac{7 \sqrt{5} }{22} ) \\ + ( {( \frac{27}{22}) }^{2} + {( \frac{7 \sqrt{5} }{22} )}^{2} + 2 \times \frac{27}{22} \times \frac{7 \sqrt{5} }{22} ) \\ \\ = (\frac{729}{484} ) + \frac{245}{484} - \frac{189 \sqrt{5} }{242} ) \\ + ( \frac{729}{484} + \frac{245}{484} + \frac{189 \sqrt{5} }{242} ) \\ \\ = \frac{729}{484} + \frac{245}{484} - \frac{189 \sqrt{5} }{242} + \frac{729}{484} \\ + \frac{245}{484} + \frac{189 \sqrt{5} }{242} \\ \\ = \frac{729}{484} + \frac{245}{484} + \frac{729}{484} + \frac{245}{484} \\ \\ = \frac{729 + 729 + 245 + 245}{484} \\ \\ = \frac{1948}{484} \\ \\ = \frac{487}{121} \\ \\ = 4.02 \: (approx)$

Hope this helps!!!

@Mahak24

Thanks...
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