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kandsbbf

How can you write the expression with rationalized denominator?

  • one year ago
  • one year ago

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  1. kandsbbf
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    @nooce

    • one year ago
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  2. kandsbbf
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    @zepdrix

    • one year ago
  3. zepdrix
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    Oh sorry I'm helping someone else, forgot about this one c: hehe glad you pinged me XD

    • one year ago
  4. zepdrix
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    So we have an irrational number in the denominator. To fix that, let's ummm

    • one year ago
  5. zepdrix
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    \[\large \frac{1}{\sqrt2}\]Here's a quick example, when dealing with a SQUARE (2nd) root, we can multiply the top and bottom by \(\huge \frac{\sqrt2}{\sqrt2}\) and it will end up changing the denominator to a rational number, namely, 2. But in this problem we have a CUBE (3rd) root, so we have to do something a lil bit fancier.

    • one year ago
  6. zepdrix
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    To get a 6 in the denominator we'll need to multiply the top and bottom by \(\huge \frac{6^{2/3}}{6^{2/3}}\)

    • one year ago
  7. kandsbbf
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    okay....

    • one year ago
  8. zepdrix
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    \[\large \frac{2+\sqrt[3]{3}}{\sqrt[3]{6}} \qquad \rightarrow \qquad \frac{2+3^{1/3}}{6^{1/3}}\]Understand the fractional exponent notation ok? :)

    • one year ago
  9. kandsbbf
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    ohhh okay (: i get ya

    • one year ago
  10. zepdrix
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    \[\large \frac{2+3^{1/3}}{6^{1/3}}\left(\frac{6^{2/3}}{6^{2/3}}\right)\]Understand how the bottom will simplify? c:

    • one year ago
  11. kandsbbf
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    i believe so lol

    • one year ago
  12. kandsbbf
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    now what?

    • one year ago
  13. zepdrix
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    When we MULTIPLY terms with similar bases, we ADD the exponents. So in the bottom we'll get,\[\huge 6^{1/3}\cdot 6^{2/3}=6^{1/3+2/3}\]

    • one year ago
  14. zepdrix
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    \[\huge = \qquad 6^{3/3} \qquad = 6\]

    • one year ago
  15. kandsbbf
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    so the bottom is 6 lol

    • one year ago
  16. zepdrix
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    Yah :3

    • one year ago
  17. kandsbbf
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    so now the top?

    • one year ago
  18. zepdrix
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    Hmmm the top doesn't work out so nice... maybe there was a better way to do this. One sec lemme think :)

    • one year ago
  19. kandsbbf
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    i could give u the possible answers..... they all have 6 under them

    • one year ago
  20. zepdrix
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    oh ok that might help ^^

    • one year ago
  21. kandsbbf
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    okay hold on

    • one year ago
  22. kandsbbf
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    • one year ago
  23. zepdrix
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    Oh ok I see what they want us to do. So now that we've taken care of the bottom, we'll write the \(\large 6^{2/3}\) like this \(\large \sqrt[3]{6^2}\) before we distribute it to the top.

    • one year ago
  24. zepdrix
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    \[\large (2+\sqrt[3]3)\sqrt[3]{6^2} \quad = \quad 2\sqrt[3]{6^2}+\sqrt[3]3\cdot\sqrt[3]{6^2}\]And I guess they want us to simplify a bit from here.

    • one year ago
  25. kandsbbf
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    okay i see what ur sayying

    • one year ago
  26. zepdrix
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    \[\huge \sqrt[3]{3}\cdot\sqrt[3]{36}=\sqrt[3]{108}\]

    • one year ago
  27. zepdrix
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    This part is a little tricky. You want a FACTOR of 108 that is a PERFECT CUBE.

    • one year ago
  28. zepdrix
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    The following numbers are perfect cubes, 8, 27, 64, 125. Because if you take the cube root of any of those numbers, they'll give you a nice clean value.\[\large 2^3=8, \qquad 3^3=27, \qquad ...\]

    • one year ago
  29. zepdrix
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    It turns out that 108 is divisible by 27!\[\huge \sqrt[3]{108}=\sqrt[3]{4\cdot27}\]

    • one year ago
  30. kandsbbf
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    soooo

    • one year ago
  31. zepdrix
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    Ah sorry lost my connection D:

    • one year ago
  32. kandsbbf
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    27 isnt a possible answer....

    • one year ago
  33. zepdrix
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    27 is a perfect cube! So let's take the cube root of 27.

    • one year ago
  34. zepdrix
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    \[\huge \sqrt[3]{4\cdot27}\quad =\quad \sqrt[3]{4}\cdot \sqrt[3]{27}=\sqrt[3]{4}\cdot 3\]

    • one year ago
  35. zepdrix
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    This is a really awful problem. Is this for algebra or something? D:

    • one year ago
  36. kandsbbf
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    algabra 2 :/

    • one year ago
  37. zepdrix
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    So it looks like it isssss probablyyyyyyy .... d. I'm like .. 65% sure :D lol

    • one year ago
  38. zepdrix
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    It's a really annoying problem. Easy to make a mistake on. But I think we did it correctly.

    • one year ago
  39. kandsbbf
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    okay thankyou(:could u help me with another?

    • one year ago
  40. kandsbbf
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    it's not like that one i promise lol

    • one year ago
  41. zepdrix
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    Close this thread. Start a new one with your new question. Type @zepdrix somewhere in the comments and I'll try to take a look if I have time ^^

    • one year ago
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