How do you solve #sqrt(3 + sqrt5) -sqrt(2 + sqrt3)#?

1 Answer
Gió
Jan 7, 2015

Here I can give you only a hint (I have to go to work!).

I would use a technique called "denesting";
You can write:
#sqrt(3+sqrt(5))=sqrt(a)+sqrt(b)#
You square both sides:
#3+sqrt(5)=a+2sqrt(ab)+b#
And you put:
#3=a+b#
#sqrt(5)=2sqrt(ab)# that can be squared again giving:
#5=4ab#
You must now solve this system in #a# and #b# (for both square roots in your original expression).
You should get:
#sqrt(3+sqrt(5))=sqrt(1/2)+sqrt(5/2)# and
#sqrt(2+sqrt(3))=sqrt(1/2)+sqrt(3/2)#.
Try substituring in your original expression and eventually rationalize.

PLEASE check my numbers and operations because I did them in a hurry!!! :-l

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