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avai70178

  • one year ago

I need some help rationalizing these denominators. Can anyone help?

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  1. avai70178
    • one year ago
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    \[\frac{ 4 }{ 3-3\sqrt{2} }\]

  2. anonymous
    • one year ago
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    Multiply by the conjugate of the denominator \(3+\sqrt2\) divided by itself \[\frac{ 4 }{ 3-3\sqrt{2} }\times \frac{ 3+3\sqrt2 }{ 3+3\sqrt2}\]

  3. avai70178
    • one year ago
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    wouldn't we multiply the top and bottom by \[-3\sqrt{3}\]

  4. anonymous
    • one year ago
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    no. you need to use the conjugate because it gives a difference of squares in the denominator and the radicals will cancel

  5. avai70178
    • one year ago
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    okay so would we end up with \[\frac{ 12\sqrt{3} }{ 3 }\] ?

  6. anonymous
    • one year ago
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    no. It looks like you multiplied by something with √3. |dw:1439590754304:dw|

  7. avai70178
    • one year ago
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    so what would we end up?

  8. anonymous
    • one year ago
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    you have to do the multiplication

  9. avai70178
    • one year ago
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    would the square root still be 2 in the numerator and be cancelled out in the denominator?

  10. anonymous
    • one year ago
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    yes. when you do FOIL on the denominator the middle terms will both have √2, but they'll have opposite signs and cancel.

  11. avai70178
    • one year ago
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    oh okay give me a second to multiply

  12. avai70178
    • one year ago
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    i feel like this is wrong but...\[\frac{ 12+12\sqrt{2} }{ -9 }\]

  13. anonymous
    • one year ago
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    That's right. Now all you have to do is simplify it. 12 and -9 are both divisible by 3, so you reduce by dividing all those by 3

  14. avai70178
    • one year ago
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    oh okay awesome so \[\frac{ 4+12\sqrt{2} }{ -3 }\]

  15. anonymous
    • one year ago
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    both 12's so \[\frac{ 4+4\sqrt{2} }{ -3 }\]

  16. avai70178
    • one year ago
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    oh my bad sorry

  17. avai70178
    • one year ago
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    But thank you so much!

  18. anonymous
    • one year ago
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    you're welcome

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