Cancelling out square roots gives 2?

Question

Question:

If $$N = \frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}$$Find N ^{(This is a subset of a larger question)}

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My approach:

After rationalizing the denominator, by multiplying fraction with $\frac{\sqrt{\sqrt{5}+1}}{\sqrt{\sqrt{5}+1}}$, I got:

$$\frac{(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2})*\sqrt{\sqrt{5}+1}}{\sqrt{5}+1}$$

Which is leading me nowhere. The textbook has simply asserted that $$N^2 = 2$$ ^{(Substituting the large fraction with $N$ to save space)}

I squared the numerator and got: $$2\sqrt{5}*2*(\sqrt{\sqrt{5}+2}*\sqrt{\sqrt{5}-2})$$

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My Question:

1. How would I further go with this to solve for $N$? or what would be the correct/easier way to find $N$ ?

2. How would I know when to rationalize denominators, and when to square the fractions, to get the answer ?

Thank you!

Answer 1

$$N^2 = \frac{(\sqrt5 + 2) + (\sqrt 5 - 2) + 2\sqrt{(\sqrt 5 + 2)(\sqrt 5 - 2)}}{\sqrt 5 + 1} =\\=\frac{2\sqrt 5 + 2\sqrt{5-4}}{\sqrt 5 + 1} =...$$ Can you see where this is going?

Answer 2

Squaring the fraction gives

$$\frac{2\sqrt5+2\sqrt{\underbrace{(\sqrt{5}+2)(\sqrt5-2)}_{=1}}}{\sqrt5+1}=\frac{2\sqrt5+2}{\sqrt5+1}=2$$

Answer 3

The next step would be to combine the square roots, $$\sqrt{\sqrt5 - 2} * \sqrt{\sqrt5 + 2} = \sqrt{(\sqrt5 - 2)(\sqrt5 + 2)}$$ Hope that helps.

Answer 4

Hint: I think a much more elegant way to go about this problem would be to capitalize on the convenient algebraic properties of the golden ratio. Start by converting all the roots of 5 to expressions involbing $\phi$ instead: e.g., $\phi=\frac{\sqrt{5}+1}{2}\implies \sqrt{5}+2=2\phi+1$, and so forth.