Question

if a=3+√5/3-√5 b=3-√5/3+√5 find a square plus b square

Answer 1

Answer:

$a = \frac{1}{b} \\ \\ {a}^{2} + {b}^{2} = {(a + b)}^{2} - 2 \\ \\ = {( \frac{3 + \sqrt{5} }{3 - \sqrt{5} } + \frac{3 - \sqrt{5} }{3 + \sqrt{5} } )}^{2} - 2 \\ \\ = {( \frac{9 + 5 + 9 + 5}{9-5} )}^{2} - 2 \\ \\ = {( \frac{28}{4} )}^{2} - 2 \\ \\ = {(7)}^{2} - 2 = 49 - 2 = 47$

Answer 2

$\huge \bold{ \color{blue}{{Solution}}}$

$a = \frac{1}{b}$

$a {}^{2} + b{}^{2} = (a + b) {}^{2} - 2$

$= ( \frac{3 + \sqrt{5} }{3 - \sqrt{5} } + \frac{3 - \sqrt{5 {}^{2} } }{3 + \sqrt{5} }) - 2$

$= \frac{(9 + 5 + 9 + 5 {}^{2}) }{9 - 5} - 2$

$= ( \frac{28 {}^{2} }{4}) - 2$

$= (7) {}^{2} - 2$

$= 49 - 2$

$= 47$