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

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

Mathematics
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\[\frac{ 4 }{ 3-3\sqrt{2} }\]
Multiply by the conjugate of the denominator \(3+\sqrt2\) divided by itself \[\frac{ 4 }{ 3-3\sqrt{2} }\times \frac{ 3+3\sqrt2 }{ 3+3\sqrt2}\]
wouldn't we multiply the top and bottom by \[-3\sqrt{3}\]

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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
okay so would we end up with \[\frac{ 12\sqrt{3} }{ 3 }\] ?
no. It looks like you multiplied by something with √3. |dw:1439590754304:dw|
so what would we end up?
you have to do the multiplication
would the square root still be 2 in the numerator and be cancelled out in the denominator?
yes. when you do FOIL on the denominator the middle terms will both have √2, but they'll have opposite signs and cancel.
oh okay give me a second to multiply
i feel like this is wrong but...\[\frac{ 12+12\sqrt{2} }{ -9 }\]
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
oh okay awesome so \[\frac{ 4+12\sqrt{2} }{ -3 }\]
both 12's so \[\frac{ 4+4\sqrt{2} }{ -3 }\]
oh my bad sorry
But thank you so much!
you're welcome

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