Surds

$\frac{6\sqrt 2}{(3+\sqrt 7)} + \frac{4\sqrt 7}{(\sqrt2-1)}$


I would probably tackle the fractions one at a time, so simplify the first one, then simplify the last one. PS my answer doesn't look too pretty so... it might be wrong?

Start with

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

To rationalise the denominator, we must multiply everything by

${(3-\sqrt 7)}$ ... see below:



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

$=\frac{18\sqrt 2 - 6\sqrt 14}{9-7}$

$=9\sqrt 2 - 3\sqrt 14$

Can you do the rest?

That is how I worked it out but the book answer says √14+4√7+9√2 actually wait I see where I messed up