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satellite73

  • 2 years ago

prove \(\sqrt{5-\sqrt{21}}=\frac{\sqrt{7}-\sqrt{3}}{\sqrt{2}}\)

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  1. satellite73
    • 2 years ago
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    any help ? i have seen something similar. here is what wofram says http://www.wolframalpha.com/input/?i=sqrt%285-sqrt%2821%29%29

  2. Callisto
    • 2 years ago
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    Aren't you asking the same question in different form?

  3. satellite73
    • 2 years ago
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    yes

  4. satellite73
    • 2 years ago
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    for two reasons. it might be easier if we know the answer and "simplify" doesn't mean anything in this context

  5. Callisto
    • 2 years ago
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    Sorry, but one simple question: in reality, is it possible to do it backward?

  6. satellite73
    • 2 years ago
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    this looks for all the world like one of those trig questions where you use one half angle formula and another half angle formula and you get two different looking but equal answers. then it is a pain to show they are the same, but it usually amounts to multiplying repeatedly

  7. satellite73
    • 2 years ago
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    ok here is a "well known formula from basic algebra" \[\sqrt{a+b\sqrt{c}}=\sqrt{\frac{1}{2}(a+\sqrt{q})}-\sqrt{\frac{1}{2}(a-\sqrt{q})}\] if \(q=a^2-b^2c\) is a perfect square

  8. satellite73
    • 2 years ago
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    i must have missed that day

  9. satellite73
    • 2 years ago
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    guess i have my homework cut out for me

  10. satellite73
    • 2 years ago
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    typo actually it should be \[\sqrt{a\pm b\sqrt{c}}=\sqrt{\frac{1}{2}(a+\sqrt{q})}\pm\sqrt{\frac{1}{2}(a-\sqrt{q})}\]

  11. satellite73
    • 2 years ago
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    given this formula we are done, since \(5^2-21=4\) a perfect square, so we get the answer immediately. how we know this i am not sure

  12. Callisto
    • 2 years ago
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    Sorry, I don't want to do so. but.. Please check this http://openstudy.com/study#/updates/4fda8ca0e4b0f2662fd103f2

  13. satellite73
    • 2 years ago
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    ok lets try this start with \[\sqrt{5-\sqrt{21}}\] and write the radicand as a perfect square. so \[\sqrt{5-\sqrt{21}}=x-y=\sqrt{(x-y)^2}=\sqrt{x^2+y^2-2xy}\] now we can put \(x^2+y^2=5, -2xy=\sqrt{21}, y=-\frac{\sqrt{21}}{2x}\) and so \(x^2+(\frac{\sqrt{21}}{2x})^2=5\) and so \[4x^4-20x^2+21=0\] solving gives \(x=\sqrt{\frac{7}{2}}\) i think

  14. satellite73
    • 2 years ago
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    i will have to look at this more the first "well known formula" comes directly from this paper, but no explanation, so not much help except in getting the answer

  15. satellite73
    • 2 years ago
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    the second method i tried to mimic what i read here http://gauravtiwari.org/2011/03/16/a-problem-on-ordinary-nested-radicals/ but the line "which on simplification yields..." was not so obvious to me

  16. satellite73
    • 2 years ago
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    ok, take a look at 4.5.2 here and the following theorem. this i think give it all

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  17. satellite73
    • 2 years ago
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    learn something new every day!

  18. satellite73
    • 2 years ago
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    today for example i learned where to get a really great hamburger

  19. satellite73
    • 2 years ago
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    your friend must have run in to someone who was either nuts ( a nutty math teacher, imagine) or just learned this and wanted to show off. i cannot imagine this in a basic algebra class

  20. Callisto
    • 2 years ago
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    *Disappointed* You're not learning HOW to get a great hamburger but you learnt WHERE to get it :( Thanks for all the notes and the 'well unknown formula' which I haven't heard of. Hmm.. Actually, I can solve it :|

  21. satellite73
    • 2 years ago
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    great. let me know how it goes. if they had had the internet when i was in school i would have 4 degrees by now

  22. KingGeorge
    • 2 years ago
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    This might be good enough. Look at \(\sqrt{5-\sqrt{21}}\). If you square it, you get \(5-\sqrt{21}\). We want this to be a perfect square of some binomial with square roots. This means, that we should want a square multiplied by two. Hence, multiply this by two. We get \[10-2\sqrt{21}\]If we want this to be a perfect square of some expression that looks like \(\sqrt x\pm\sqrt y\), we want to find an \(x, y\) such that \(x+y=10\) and \(xy=21\). Solving this, we get that \[10-2\sqrt{21}=(\sqrt7-\sqrt3)^2\]Now we work backward to get what we had at the start. Divide by 2, and then take the square root gives us that \[\sqrt{5-\sqrt{21}}=\frac{\sqrt7-\sqrt3}{\sqrt2}\]

  23. KingGeorge
    • 2 years ago
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    Typo in that first paragraph. I said "This means, that we should want a square multiplied by two." It should read "This means, that we should want a square root multiplied by two." (talking about the square root in \(5-\sqrt{21}\) of course)

  24. Callisto
    • 2 years ago
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    @KingGeorge I solved the question using this way. But no one here looks at it except myininaya

  25. satellite73
    • 2 years ago
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    this looks much snappier

  26. KingGeorge
    • 2 years ago
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    The two methods are very slightly different. Same trick, but slightly different process.

  27. KingGeorge
    • 2 years ago
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    Namely, I took things out of the square root, and you left them in. So I guess they really are just the same thing.

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