SOLUTION: 2÷[sqrt(2+sqrt(2))] simplified/rationalized

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Question 280919: 2÷[sqrt(2+sqrt(2))] simplified/rationalized
Answer by jsmallt9(3759) About Me  (Show Source):
You can put this solution on YOUR website!
2%2Fsqrt%282%2Bsqrt%282%29%29
First we will eliminate the "big" square root from the denominator. To do this we just have to multiply the numerator and denominator by it:

This gives us:
%282%28sqrt%282%2Bsqrt%282%29%29%29%29%2F%28%28sqrt%282%2Bsqrt%282%29%29%29%5E2%29
which simplifies further to:
%282sqrt%282%2Bsqrt%282%29%29%29%2F%282%2Bsqrt%282%29%29
Now we will address the remaining square root in the denominator. Since the denominator has two terms we cannot use the same technique we used on the single "big" square root. If we were to try the same technique, multiplying the denominator by itself, then we wound not eliminate the square root! This is so because
%282%2Bsqrt%282%29%29%5E2 is not 2%5E2+%2B+%28sqrt%282%29%29%5E2. Exponents do not distribute!. We can use FOIL or the pattern %28a%2Bb%29%28a%2Bb%29+=+a%5E2+%2B2ab+%2B+b%5E2 to multiply it correctly:

which, as you can see, still has a square root in it.

The proper way to rationalize a two-term denominator is to take advantage of another pattern: %28a%2Bb%29%28a-b%29+=+a%5E2+-+b%5E2. This pattern shows us how two-term expressions can be multiplied and get a result that contains only perfect squares! In your problem, your denominator is 2%2Bsqrt%282%29. This matches the a+b in the pattern. So we need to multiply by a-b to get perfect squares. If 2%2Bsqrt%282%29 is a+b then 2-sqrt%282%29 would be a-b. So this is what we will use to multiply the numerator and denominator:

In the denominator, we can use the pattern to simplify matters:

The denominator simplifies easily:
%28%282sqrt%282%2Bsqrt%282%29%29%29%282-sqrt%282%29%29%29%2F%284-2%29
%28%282sqrt%282%2Bsqrt%282%29%29%29%282-sqrt%282%29%29%29%2F2
On top we can use the Distributive Property:

%284sqrt%282%2Bsqrt%282%29%29-2sqrt%284%2B2sqrt%282%29%29%29%2F2
The numerator is a big mess, but the denominator is rational!