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eroshea Group Title

rationalize

  • 2 years ago
  • 2 years ago

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  1. eroshea Group Title
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    \[\frac{ 4-4\sqrt{3}i }{ -2\sqrt{3}+2i }\]

    • 2 years ago
  2. hartnn Group Title
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    multiply and divide by -2root3-2i

    • 2 years ago
  3. hartnn Group Title
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    or take -2i common from numerator. u get -2i(2i-2root 3) in numerator. so what gets cancelled?

    • 2 years ago
  4. eroshea Group Title
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    wait.. i'll try

    • 2 years ago
  5. eroshea Group Title
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    is there something that can be cancelled?

    • 2 years ago
  6. hartnn Group Title
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    see the factor in numerator and denominator!!

    • 2 years ago
  7. eroshea Group Title
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    owh.. -2i will remain??and the factor will cancel out the denominator?

    • 2 years ago
  8. hartnn Group Title
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    yup,-2i,got it ?

    • 2 years ago
  9. eroshea Group Title
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    but i think its wrong? i thnk this should be what in the numerator is , -2i(2i+2root3),, am i right??

    • 2 years ago
  10. hartnn Group Title
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    O.o

    • 2 years ago
  11. hartnn Group Title
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    options?

    • 2 years ago
  12. hartnn Group Title
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    and yes, u are right....

    • 2 years ago
  13. eroshea Group Title
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    i dont know what to do next

    • 2 years ago
  14. hartnn Group Title
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    multiply and divide by -2root3-2i

    • 2 years ago
  15. eroshea Group Title
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    i hate it when that "i" is in my solution .. :D i'll try solving it

    • 2 years ago
  16. eroshea Group Title
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    the denominator will be 8? but i have a difficulty in solving the numerator :((

    • 2 years ago
  17. hartnn Group Title
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    8 ?? or -16 ?

    • 2 years ago
  18. eroshea Group Title
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    my bad.. i believe the numerator should be 18?? 12-4i^2 = 12-4(-1) = 12+4= 16?

    • 2 years ago
  19. eroshea Group Title
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    i mean the DENOMINATOR

    • 2 years ago
  20. hartnn Group Title
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    *minus*

    • 2 years ago
  21. eroshea Group Title
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    y negative?

    • 2 years ago
  22. eroshea Group Title
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    \[(-2\sqrt{3}+2i)(-2\sqrt{3}-2i) = (-2\sqrt{3})^{2}-4i ^{2}=12-4(-1)=16\]

    • 2 years ago
  23. hartnn Group Title
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    my bad, sorry,i took - common from that....ok,so 16 numerator = ?

    • 2 years ago
  24. hartnn Group Title
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    and what are your options ?

    • 2 years ago
  25. eroshea Group Title
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    my final answer is -root3 - i

    • 2 years ago
  26. eroshea Group Title
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    if i simplified my numerator, -16root3 - 16i and divide that by 16 -root3 -i is my answer correct?

    • 2 years ago
  27. hartnn Group Title
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    hmm....its my silly mistake's day or i m getting i-root3

    • 2 years ago
  28. eroshea Group Title
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    owh @_@

    • 2 years ago
  29. eroshea Group Title
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    i'll try solving it again

    • 2 years ago
  30. eroshea Group Title
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    owh you're right!! hahaha... i had a mistake with my subtraction, 24i-8i should be 16i and not -16i.. thanks ! :))

    • 2 years ago
  31. hartnn Group Title
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    finally! welcome :)

    • 2 years ago
  32. eroshea Group Title
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    it's not your silly mistake day :))

    • 2 years ago
  33. hartnn Group Title
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    glad to know :)

    • 2 years ago
  34. eroshea Group Title
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    how long have you've been in this site?

    • 2 years ago
  35. hartnn Group Title
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    48 days around.....

    • 2 years ago
  36. eroshea Group Title
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    owh.. i just made my account yesterday :3 brb.. i'll just go out and eat first and then back in reviewing :)) i have my finals tomorrow for algebra and trigonometry

    • 2 years ago
  37. hartnn Group Title
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    best of luck.

    • 2 years ago
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