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satellite73 Group Title

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

  • 2 years ago
  • 2 years ago

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  1. satellite73 Group Title
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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 years ago
  2. Callisto Group Title
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    Aren't you asking the same question in different form?

    • 2 years ago
  3. satellite73 Group Title
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    yes

    • 2 years ago
  4. satellite73 Group Title
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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

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

    • 2 years ago
  6. satellite73 Group Title
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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

    • 2 years ago
  7. satellite73 Group Title
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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

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

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

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

    • 2 years ago
  11. satellite73 Group Title
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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

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

    • 2 years ago
  13. satellite73 Group Title
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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

    • 2 years ago
  14. satellite73 Group Title
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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

    • 2 years ago
  15. satellite73 Group Title
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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

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

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

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

    • 2 years ago
  19. satellite73 Group Title
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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

    • 2 years ago
  20. Callisto Group Title
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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 :|

    • 2 years ago
  21. satellite73 Group Title
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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

    • 2 years ago
  22. KingGeorge Group Title
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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 \[\sqrt{5-\sqrt{21}}=\frac{\sqrt7-\sqrt3}{\sqrt2}\]

    • 2 years ago
  23. KingGeorge Group Title
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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)

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

    • 2 years ago
  25. satellite73 Group Title
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    this looks much snappier

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

    • 2 years ago
  27. KingGeorge Group Title
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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.

    • 2 years ago
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