Question

How can you write the expression with rationalized denominator?

Answer 1

@nooce

Answer 2

@zepdrix

Answer 3

Oh sorry I'm helping someone else, forgot about this one c: hehe
glad you pinged me XD

Answer 4

So we have an irrational number in the denominator. To fix that, let's ummm

Answer 5

\[\large \frac{1}{\sqrt2}\]Here's a quick example, when dealing with a SQUARE (2nd) root, we can multiply the top and bottom by \(\huge \frac{\sqrt2}{\sqrt2}\) and it will end up changing the denominator to a rational number, namely, 2.

But in this problem we have a CUBE (3rd) root, so we have to do something a lil bit fancier.

Answer 6

To get a 6 in the denominator we'll need to multiply the top and bottom by \(\huge \frac{6^{2/3}}{6^{2/3}}\)

Answer 7

okay....

Answer 8

\[\large \frac{2+\sqrt[3]{3}}{\sqrt[3]{6}} \qquad \rightarrow \qquad \frac{2+3^{1/3}}{6^{1/3}}\]Understand the fractional exponent notation ok? :)

Answer 9

ohhh okay (: i get ya

Answer 10

\[\large \frac{2+3^{1/3}}{6^{1/3}}\left(\frac{6^{2/3}}{6^{2/3}}\right)\]Understand how the bottom will simplify? c:

Answer 11

i believe so lol

Answer 12

now what?

Answer 13

When we MULTIPLY terms with similar bases, we ADD the exponents.
So in the bottom we'll get,\[\huge 6^{1/3}\cdot 6^{2/3}=6^{1/3+2/3}\]

Answer 14

\[\huge = \qquad 6^{3/3} \qquad = 6\]

Answer 15

so the bottom is 6 lol

Answer 16

Yah :3

Answer 17

so now the top?

Answer 18

Hmmm the top doesn't work out so nice... maybe there was a better way to do this. One sec lemme think :)

Answer 19

i could give u the possible answers..... they all have 6 under them

Answer 20

oh ok that might help ^^

Answer 21

okay hold on

Answer 22

Oh ok I see what they want us to do.

So now that we've taken care of the bottom, we'll write the \(\large 6^{2/3}\) like this \(\large \sqrt[3]{6^2}\) before we distribute it to the top.

Answer 23

\[\large (2+\sqrt[3]3)\sqrt[3]{6^2} \quad = \quad 2\sqrt[3]{6^2}+\sqrt[3]3\cdot\sqrt[3]{6^2}\]And I guess they want us to simplify a bit from here.

Answer 24

okay i see what ur sayying

Answer 25

\[\huge \sqrt[3]{3}\cdot\sqrt[3]{36}=\sqrt[3]{108}\]

Answer 26

This part is a little tricky. You want a FACTOR of 108 that is a PERFECT CUBE.

Answer 27

The following numbers are perfect cubes,
8, 27, 64, 125.

Because if you take the cube root of any of those numbers, they'll give you a nice clean value.\[\large 2^3=8, \qquad 3^3=27, \qquad ...\]

Answer 28

It turns out that 108 is divisible by 27!\[\huge \sqrt[3]{108}=\sqrt[3]{4\cdot27}\]

Answer 29

soooo

Answer 30

Ah sorry lost my connection D:

Answer 31

27 isnt a possible answer....

Answer 32

27 is a perfect cube! So let's take the cube root of 27.

Answer 33

\[\huge \sqrt[3]{4\cdot27}\quad =\quad \sqrt[3]{4}\cdot \sqrt[3]{27}=\sqrt[3]{4}\cdot 3\]

Answer 34

This is a really awful problem. Is this for algebra or something? D:

Answer 35

algabra 2 :/

Answer 36

So it looks like it isssss probablyyyyyyy .... d.
I'm like .. 65% sure :D lol

Answer 37

It's a really annoying problem. Easy to make a mistake on. But I think we did it correctly.

Answer 38

okay thankyou(:could u help me with another?

Answer 39

it's not like that one i promise lol

Answer 40

Close this thread.
Start a new one with your new question.

Type @zepdrix somewhere in the comments and I'll try to take a look if I have time ^^