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Callisto Group Title

Tutor question #1 Simplify the following: \[\frac{1}{\sqrt2 + \sqrt3 + \sqrt5}\] (conditions: pen, paper, no calculator, done in 2 minutes) *

  • 2 years ago
  • 2 years ago

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  1. Callisto Group Title
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    I got \[ \frac{2\sqrt3 + 3\sqrt2 - \sqrt{30}}{12}\] Was that right?

    • 2 years ago
  2. RaphaelFilgueiras Group Title
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    yes

    • 2 years ago
  3. apoorvk Group Title
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    HUNDRED PERCENT!!!!!!!!!!!!!

    • 2 years ago
  4. Callisto Group Title
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    But it looks so messy...

    • 2 years ago
  5. bronzegoddess Group Title
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    i got something different: isn't it you cant have radicals under a fraction so you multiply by the conjugate?

    • 2 years ago
  6. Callisto Group Title
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    That's what rationalise means

    • 2 years ago
  7. apoorvk Group Title
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    Not messy at all, this is the only way to go about it.

    • 2 years ago
  8. bronzegoddess Group Title
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    yes but if you multiply 1 by the conjugate pair you get the conjugate pair.. your numerator seems wrong..

    • 2 years ago
  9. Callisto Group Title
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    Can you correct it?

    • 2 years ago
  10. bronzegoddess Group Title
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    otherwise can someone please show me a step by step to how you got that answer? and i'll show you what i did :)

    • 2 years ago
  11. Callisto Group Title
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    luckily, I haven't deleted that.. \[(\sqrt2 + \sqrt3 + \sqrt5)(\sqrt2 + \sqrt3 - \sqrt5)\]\[ = (\sqrt2 + \sqrt3)^2-5\]\[ = 2 + 3 + 2\sqrt6-5\]\[=2\sqrt6\] \[\frac{1}{\sqrt2 + \sqrt3 + \sqrt5}\]\[ = \frac{\sqrt2 + \sqrt3 - \sqrt5}{(\sqrt2 + \sqrt3 + \sqrt5)(\sqrt2 + \sqrt3 - \sqrt5)}\]\[ = \frac{\sqrt2 + \sqrt3 - \sqrt5}{2\sqrt6}\]\[ = \frac{2\sqrt3 + 3\sqrt2 - \sqrt{30}}{12}\]

    • 2 years ago
  12. RaphaelFilgueiras Group Title
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    \[1/(\sqrt{2}+\sqrt{3}+\sqrt{5})=1/(a+\sqrt{5})=(a-sqrt{5})/(a^{2}-5)\] \[(\sqrt{2}+\sqrt{3}-\sqrt{5})/(5-5+\sqrt{6})\] \[\sqrt{2}+\sqrt{3}-\sqrt{5}/2\sqrt{6}=2.\sqrt{3}+2.\sqrt{2}-\sqrt{30}/12\]

    • 2 years ago
  13. bronzegoddess Group Title
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    I ended at the second to last step... where did you get the 2 in front of the radicals if you are multiplying by 1?

    • 2 years ago
  14. Callisto Group Title
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    see the first part of it..

    • 2 years ago
  15. RaphaelFilgueiras Group Title
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    5-5+2sqrt(6)*

    • 2 years ago
  16. RaphaelFilgueiras Group Title
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    second line

    • 2 years ago
  17. Callisto Group Title
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    @RaphaelFilgueiras The second term of the numerator is 3\sqrt{2} ?! \sqrt{18} = \sqrt{9 times 2} = 3\sqrt2 ...

    • 2 years ago
  18. bronzegoddess Group Title
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    okay, so the you rationalized the second to last line because of the radical at the bottom?

    • 2 years ago
  19. Callisto Group Title
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    Of course! That's 'rationalise'

    • 2 years ago
  20. bronzegoddess Group Title
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    okay sorry, then you are right.. I just wanted to know where I went wrong...

    • 2 years ago
  21. RaphaelFilgueiras Group Title
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    no in my solution i put (5-5+sqrt(5))(WRONG) instead(5-5+2sqrt(5))

    • 2 years ago
  22. Callisto Group Title
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    Sorry, but I don't know what you're doing.. No offense.

    • 2 years ago
  23. Callisto Group Title
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    Do you mind re-type it again?

    • 2 years ago
  24. RaphaelFilgueiras Group Title
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    i did same as you,but call(a=sqrt(2)+sqrt(3))

    • 2 years ago
  25. Callisto Group Title
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    Okay, and the answer is the same?

    • 2 years ago
  26. RaphaelFilgueiras Group Title
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    yes

    • 2 years ago
  27. Callisto Group Title
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    Okay, thanks everyone!!!

    • 2 years ago
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