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kandsbbf

  • 2 years ago

How can you write the expression with rationalized denominator?

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  1. kandsbbf
    • 2 years ago
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    @nooce

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  2. kandsbbf
    • 2 years ago
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    @zepdrix

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

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

  5. zepdrix
    • 2 years ago
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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.

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

  7. kandsbbf
    • 2 years ago
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    okay....

  8. zepdrix
    • 2 years ago
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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? :)

  9. kandsbbf
    • 2 years ago
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    ohhh okay (: i get ya

  10. zepdrix
    • 2 years ago
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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:

  11. kandsbbf
    • 2 years ago
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    i believe so lol

  12. kandsbbf
    • 2 years ago
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    now what?

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

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

  15. kandsbbf
    • 2 years ago
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    so the bottom is 6 lol

  16. zepdrix
    • 2 years ago
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    Yah :3

  17. kandsbbf
    • 2 years ago
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    so now the top?

  18. zepdrix
    • 2 years ago
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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 :)

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

  20. zepdrix
    • 2 years ago
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    oh ok that might help ^^

  21. kandsbbf
    • 2 years ago
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    okay hold on

  22. kandsbbf
    • 2 years ago
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  23. zepdrix
    • 2 years ago
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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.

  24. zepdrix
    • 2 years ago
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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.

  25. kandsbbf
    • 2 years ago
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    okay i see what ur sayying

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

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

  28. zepdrix
    • 2 years ago
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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.\[\large 2^3=8, \qquad 3^3=27, \qquad ...\]

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

  30. kandsbbf
    • 2 years ago
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    soooo

  31. zepdrix
    • 2 years ago
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    Ah sorry lost my connection D:

  32. kandsbbf
    • 2 years ago
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    27 isnt a possible answer....

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

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

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

  36. kandsbbf
    • 2 years ago
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    algabra 2 :/

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

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

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

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

  41. zepdrix
    • 2 years ago
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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 ^^

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