Question

if a=3+root5upon3-root5 and b=3-root5upon3+root5 find a2-b2​

Answer 1

Step-by-step explanation:

$a = \frac{3 + \sqrt{5} }{3 - \sqrt{5} } \\ b = \frac{3 - \sqrt{5} }{3 + \sqrt{5} } \\ {a}^{2} + {b}^{2} = {(\frac{3 + \sqrt{5} }{3 - \sqrt{5} })}^{2} + {( \frac{3 - \sqrt{5} }{3 + \sqrt{5} } )}^{2} \\ = \frac{ ( {3 + \sqrt{5} )}^{2} {(3 + \sqrt{5} )}^{2} + {(3 - \sqrt{5)} }^{2} {(3 - \sqrt{5}) }^{2} }{ {(3 + \sqrt{5} )}^{2} {(3 - \sqrt{5} )}^{2} } \\ = \frac{ {(3 + \sqrt{5}) }^{4} + {(3 - \sqrt{5} )}^{4} }{(3 + \sqrt{5} )(3 - \sqrt{5})(3 + \sqrt{5} )(3 - \sqrt{5}) } \\ = \frac{ {(3 + \sqrt{5}) }^{4} + {(3 - \sqrt{5}) }^{4} }{(9 - 5)(9 - 5)} \\ = \frac{ {(3 + \sqrt{5}) }^{4} + {(3 - \sqrt{5}) }^{4} }{(4)(4)} \\ = \frac{ {(3 + \sqrt{5}) }^{4} + {(3 - \sqrt{5}) }^{4} }{16} \\ = \frac{ {3}^{4} + (4 \times {3}^{3} \times \sqrt{5} ) + (6 \times {3}^{2} \times { \sqrt{5} }^{2} ) + (4 \times 3 \times { \sqrt{5} }^{3}) + { \sqrt{5} }^{4} +{3}^{4} - (4 \times {3}^{3} \times \sqrt{5} ) + (6 \times {3}^{2} \times { \sqrt{5} }^{2} ) - (4 \times 3 \times { \sqrt{5} }^{3}) + { \sqrt{5} }^{4} }{16} \\ = \frac{ {3}^{4} + (6 \times {3}^{2} + { \sqrt{5} }^{2}) + { \sqrt{5} }^{4} + {3}^{4} + (6 \times {3}^{2} + { \sqrt{5} }^{2}) + { \sqrt{5} }^{4} }{16} \\ = \frac{(2 \times 81) + (2 \times 180) + (2 \times 25)}{16} \\ = \frac{2 \times (81 + 180 + 25)}{16} \\ = \frac{2 \times 286}{16} \\ = \frac{572}{16}$