(sqrt(6) -2)/(11+sqrt(6))√6−211+√6
=(sqrt(6) -2)/(11+sqrt(6)) * (11-sqrt(6))/(11-sqrt(6))=√6−211+√6⋅11−√611−√6
this is a trick to get rid of the sqrt(6)√6 of the (11+sqrt(6))(11+√6) in the denominator. We are basically constructing what is called the "conjugate" of the denominator, (11-sqrt(6))(11−√6), which is exactly the same thing as the denominator but with the opposite sign in the middle. If the denominator is (a+b)(a+b), the conjugate is (a-b)(a−b). If the denominator is (a-b)(a−b), the conjugate is (a+b)(a+b).
Why do we want to multiply by the conjugate?
because:
(a+b)*(a-b)=a^2-b^2(a+b)⋅(a−b)=a2−b2
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
(11-sqrt(6))/(11-sqrt(6)) = 111−√611−√6=1 (as long as it's not 0/000 it is fine).
that is, we're only multiplying by 1 so we are not affecting the results at all.
Then it becomes easy:
=((sqrt(6) - 2)(11-sqrt(6))) / (11^2-6)=(√6−2)(11−√6)112−6
=(11sqrt(6)-sqrt(6)sqrt(6)-22+2sqrt(6))/(121-6)=11√6−√6√6−22+2√6121−6
=(13sqrt(6)-28)/115=13√6−28115
At this stage, I don't see any more ways to simplify this further, so this must be the answer.
