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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)

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

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!

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4 Answers 4

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Squaring the fraction gives

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

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  • $\begingroup$ Thank you! Nice visual representation ! $\endgroup$
    – Gaurang Tandon
    Mar 21, 2014 at 8:28
  • $\begingroup$ You're welcome. $\endgroup$
    – user63181
    Mar 21, 2014 at 8:29
  • $\begingroup$ Also, please answer the second part of my question. $\endgroup$
    – Gaurang Tandon
    Mar 21, 2014 at 8:29
  • $\begingroup$ @GaurangTandon The second part has no real answer. You know because you have done several similar calculations and because you try out several things, one of them then leads you to the right answer. $\endgroup$
    – 5xum
    Mar 21, 2014 at 8:31
  • $\begingroup$ Since there's nested roots and To get rid of one root the best idea is to square @GaurangTandon $\endgroup$
    – user63181
    Mar 21, 2014 at 8:33
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$$N^2 = \frac{(\sqrt5 + 2) + (\sqrt 5 - 2) + 2\sqrt{(\sqrt 5 + 2)(\sqrt 5 - 2)}}{\sqrt 5 + 1} =\\=\frac{2\sqrt 5 + 2\sqrt{5-4}}{\sqrt 5 + 1} =...$$ Can you see where this is going?

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  • $\begingroup$ Yes, I can ! Thanks! $\endgroup$
    – Gaurang Tandon
    Mar 21, 2014 at 8:28
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The next step would be to combine the square roots, $$\sqrt{\sqrt5 - 2} * \sqrt{\sqrt5 + 2} = \sqrt{(\sqrt5 - 2)(\sqrt5 + 2)}$$ Hope that helps.

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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.

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  • $\begingroup$ What is golden ratio? And what is that symbol? I am sorry but I can't understand it. Thanks though :) $\endgroup$
    – Gaurang Tandon
    Mar 21, 2014 at 8:49
  • $\begingroup$ @GaurangTandon Oops, I guess my hint won't help you much if you don't know what it is! You should look it up though, since it's just one of those special numbers from elementary geometry like $\pi$ and $\sqrt 2$ that you are expected to be familiar with after a certain point. I refer you to the wiki: en.wikipedia.org/wiki/Golden_ratio $\endgroup$
    – David H
    Mar 21, 2014 at 9:03
  • $\begingroup$ Ok, thanks. I will try and study more hard and learn this too :) $\endgroup$
    – Gaurang Tandon
    Mar 21, 2014 at 10:26

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