Question

Rationalise
$\frac{3 + \sqrt[2]{2} }{3 - \sqrt{2} }$
a+b=
$\sqrt{2}$
find a&b​

Answer 1

$\frac{3 + \sqrt{2} }{3 - \sqrt{2} } \\ \\ = \frac{(3 + \sqrt{2} )}{(3 - \sqrt{2}) } \frac{(3 + \sqrt{2}) }{(3 + \sqrt{2}) } \\ \\ = \frac{13 + 6 \sqrt{2} }{7}$

$a + b = \sqrt{2}$

$\frac{3 + \sqrt{2} }{2} + \frac{3 - \sqrt{2} }{2} \\ \\ = \sqrt{2}$

$a = \frac{3 + \sqrt{2} }{2} \\ \\ b = \frac{3 - \sqrt{2} }{2}$

hope it helps