How do you simplify #(sqrt(6)-2)/(11+sqrt(6))#?

1 Answer
Sep 23, 2015

Answer:

#(13sqrt(6)-28)/115#

Explanation:

#(sqrt(6) -2)/(11+sqrt(6))#
#=(sqrt(6) -2)/(11+sqrt(6)) * (11-sqrt(6))/(11-sqrt(6))#

this is a trick to get rid of the #sqrt(6)# of the #(11+sqrt(6))# in the denominator. We are basically constructing what is called the "conjugate" of the denominator, #(11-sqrt(6))#, which is exactly the same thing as the denominator but with the opposite sign in the middle. If the denominator is #(a+b)#, the conjugate is #(a-b)#. If the denominator is #(a-b)#, the conjugate is #(a+b)#.

Why do we want to multiply by the conjugate?
because:
#(a+b)*(a-b)=a^2-b^2#
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)) = 1# (as long as it's not #0/0# 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)#
#=(11sqrt(6)-sqrt(6)sqrt(6)-22+2sqrt(6))/(121-6)#
#=(13sqrt(6)-28)/115#

At this stage, I don't see any more ways to simplify this further, so this must be the answer.