anonymous one year ago How would I find the least common denominator of these fractions (posted below)...

1. anonymous

$\frac{ \sqrt[3]{x} }{ 2\sqrt{x} } + \frac{ \sqrt{x}+3 }{ 3\sqrt[3]{x} }$

2. anonymous

yaaa.. no

3. anonymous

lol That's exactly what I thought.

4. anonymous

at first i thought you be like 1/18 and 1/4 or something like that and I be like I know EXACTLY how to do that :) but ya... no...

5. anonymous

srry xD

6. anonymous

@triciaal @geny55

7. anonymous

See that's my problem, and I even know the answer (thanks to calculators). Problem is I have to show work, and I can't do that since I have no idea how to. But yeah. Don't worry about it. :)

8. anonymous

@pooja195 @perl

9. anonymous

@Ashleyisakitty @west_coast

10. anonymous

i can help

11. anonymous

It would be appreciated. :)

12. anonymous

ok, paste both fractions in separate comments so i can see them clearer

13. anonymous

|dw:1442863551641:dw|

14. anonymous

Okay just a second. (and nice picture lol)

15. anonymous

TY TY :)

16. anonymous

$\frac{ \sqrt[3]{x} }{ 2\sqrt{x} }$

17. anonymous

ehhhhhhhh................. idk thiss

18. anonymous

$\frac{ \sqrt{x}+3 }{ 3\sqrt[3]{x} }$

19. anonymous

Eh. It's okay.

20. anonymous

@Nnesha @Preetha @kiamousekia @iGreen

21. anonymous

would you like help with this $\frac{ \sqrt[3]{x} }{ 2\sqrt{x} } + \frac{ \sqrt{x}+3 }{ 3\sqrt[3]{x} }$

22. anonymous

23. anonymous

$$\Large \frac{ \sqrt[3]{x} }{ 2\sqrt{x} }\cdot \frac{\sqrt x }{\sqrt x } + \frac{ \sqrt{x}+3 }{ 3\sqrt[3]{x} } \cdot \frac{\sqrt[3] {x^2} }{\sqrt[3] {x^2}}$$

24. anonymous

First i would rationalize the denominators

25. anonymous

Is what you posted above the rationalization?

26. phi

it might be easier to see if you know how to write the radicals as exponents $\frac{ \sqrt[3]{x} }{ 2\sqrt{x} } + \frac{ \sqrt{x}+3 }{ 3\sqrt[3]{x} } \\ \frac{ x^\frac{1}{3} }{ 2x^\frac{1}{2} } + \frac{ x^\frac{1}{2} +3 }{ 3x^\frac{1}{3} }$ the common denominator is the product of the two denominators you can also rationalize the denominators as jayz suggests. notice that the answer can be written in many different forms

27. anonymous

So $6x^{1/6}$ would be the common denominator ?

28. phi

you add exponents so 6 x^(5/6) you can remember this rule by remembering x *x = x^2 and x is x^1 in other words, $$x^1 \cdot x^1 = x^2$$

29. anonymous

Ah okay. Just to check, is this a problem that I have to multiply the numerators by the same values as the denominators?

30. anonymous

Because I have heard both yes and no from different people depending on the equations.

31. phi

once you know what the common denominator is, you know what you should multiply the "bottom" by (in the the first fraction, for example) we can't just multiply the bottom by something (that will change it) but if we multiply top and bottom by the same thing (for example x/x) that is the same as multiplying by 1 , so it does not change the fraction.

32. phi

$\frac{ x^\frac{1}{3} }{ 2x^\frac{1}{2} } + \frac{ x^\frac{1}{2} +3 }{ 3x^\frac{1}{3} }$ multiply the first fraction by 1 (in the form 3 x^(1/3) / (3 x^(1/3)) $\frac{ x^\frac{1}{3} }{ 2x^\frac{1}{2} } \cdot \frac{ 3x^\frac{1}{3} }{ 3x^\frac{1}{3} }+ \frac{ x^\frac{1}{2} +3 }{ 3x^\frac{1}{3} }$

33. phi

and multiply the second fraction by 1 in the form 2x^(1/2) / 2x^(1/2) $\frac{ x^\frac{1}{3} }{ 2x^\frac{1}{2} } \cdot \frac{ 3x^\frac{1}{3} }{ 3x^\frac{1}{3} }+ \frac{ (x^\frac{1}{2} +3 )}{ 3x^\frac{1}{3} }\cdot \frac{ 2x^\frac{1}{2} }{ 2x^\frac{1}{2} }$

34. phi

I assume you know how to add the exponents 1/2 + 1/3 that is the same as 3/6 + 2/6 = 5/6

35. phi

so you get 6 x^(5/6) as the common denominator

36. anonymous

Yes, I understand. Thank you. :) I am solving it out now.

37. anonymous

I got$\frac{ 3x^{2/3}+2\sqrt{x} (\sqrt{x}+3) }{ 6x^{5/6} }$

38. phi

yes, and that can be written as $\frac{ 3x^\frac{2}{3} + 2x + 6 x^\frac{1}{2}}{6x^\frac{5}{6}}$ and lots of other ways. for example: $\frac{1}{2x^\frac{1}{6}} + \frac{x^\frac{1}{6}}{3} +\frac{1}{x^\frac{1}{3}}$ (all ways look ugly!)

39. anonymous

And that would be the final answer correct?

40. phi

I think as good as any is $\frac{ 3x^\frac{2}{3} + 2x + 6 x^\frac{1}{2}}{6x^\frac{5}{6}}$ (that is your answer, but "distributing" the $$2 \sqrt{x}$$ on the other hand some people would rather have the denominator 6x (rather than 6 x^(5/6) ) so they would multiply top and bottom by x^(1/6)) But I would leave it the first way. Or show your teacher the different ways and ask if it matters.

41. anonymous

The way you did it makes sense to me, so I'm going with that one. Thank you so much for your help! :D