How do you simplify #(sqrt(6)-2)/(11+sqrt(6))#?
this is a trick to get rid of the
Why do we want to multiply by the conjugate?
so if a or b (or both) were square-roots, the results is squared and we are thereby getting rid of the square-root.
Note that we need to multiply the top (numerator) also by the same conjugate-of-the-denominator so that
that is, we're only multiplying by 1 so we are not affecting the results at all.
Then it becomes easy:
At this stage, I don't see any more ways to simplify this further, so this must be the answer.