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

Question

Eva Abramson · Accepted Answer

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

#(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.

Kennedy Adams · Answer

undefined To simplify the expression (sqrt(6)-2)/(11+sqrt(6)), we can use the conjugate of the denominator to eliminate the square root.

The conjugate of 11+sqrt(6) is 11-sqrt(6).

To simplify, we multiply both the numerator and denominator by the conjugate:

[(sqrt(6)-2)(11-sqrt(6))] / [(11+sqrt(6))(11-sqrt(6))]

Expanding the numerator and denominator:

[11sqrt(6) - 2sqrt(6) - 22 + 2sqrt(6)] / [121 - (sqrt(6))^2]

Combining like terms in the numerator:

[9sqrt(6) - 22] / [121 - 6]

Simplifying further:

[9sqrt(6) - 22] / 115

Therefore, the simplified expression is (9sqrt(6) - 22) / 115.