## avai70178 one year ago I need some help rationalizing these denominators. Can anyone help?

1. avai70178

$\frac{ 4 }{ 3-3\sqrt{2} }$

2. anonymous

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

wouldn't we multiply the top and bottom by $-3\sqrt{3}$

4. anonymous

no. you need to use the conjugate because it gives a difference of squares in the denominator and the radicals will cancel

5. avai70178

okay so would we end up with $\frac{ 12\sqrt{3} }{ 3 }$ ?

6. anonymous

no. It looks like you multiplied by something with √3. |dw:1439590754304:dw|

7. avai70178

so what would we end up?

8. anonymous

you have to do the multiplication

9. avai70178

would the square root still be 2 in the numerator and be cancelled out in the denominator?

10. anonymous

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

oh okay give me a second to multiply

12. avai70178

i feel like this is wrong but...$\frac{ 12+12\sqrt{2} }{ -9 }$

13. anonymous

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

oh okay awesome so $\frac{ 4+12\sqrt{2} }{ -3 }$

15. anonymous

both 12's so $\frac{ 4+4\sqrt{2} }{ -3 }$

16. avai70178

17. avai70178

But thank you so much!

18. anonymous

you're welcome