anonymous
  • anonymous
How can you write the expression with rationalized denominator?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
@nooce
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anonymous
  • anonymous
@zepdrix
zepdrix
  • zepdrix
Oh sorry I'm helping someone else, forgot about this one c: hehe glad you pinged me XD

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zepdrix
  • zepdrix
So we have an irrational number in the denominator. To fix that, let's ummm
zepdrix
  • zepdrix
\[\large \frac{1}{\sqrt2}\]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.
zepdrix
  • zepdrix
To get a 6 in the denominator we'll need to multiply the top and bottom by \(\huge \frac{6^{2/3}}{6^{2/3}}\)
anonymous
  • anonymous
okay....
zepdrix
  • zepdrix
\[\large \frac{2+\sqrt[3]{3}}{\sqrt[3]{6}} \qquad \rightarrow \qquad \frac{2+3^{1/3}}{6^{1/3}}\]Understand the fractional exponent notation ok? :)
anonymous
  • anonymous
ohhh okay (: i get ya
zepdrix
  • zepdrix
\[\large \frac{2+3^{1/3}}{6^{1/3}}\left(\frac{6^{2/3}}{6^{2/3}}\right)\]Understand how the bottom will simplify? c:
anonymous
  • anonymous
i believe so lol
anonymous
  • anonymous
now what?
zepdrix
  • zepdrix
When we MULTIPLY terms with similar bases, we ADD the exponents. So in the bottom we'll get,\[\huge 6^{1/3}\cdot 6^{2/3}=6^{1/3+2/3}\]
zepdrix
  • zepdrix
\[\huge = \qquad 6^{3/3} \qquad = 6\]
anonymous
  • anonymous
so the bottom is 6 lol
zepdrix
  • zepdrix
Yah :3
anonymous
  • anonymous
so now the top?
zepdrix
  • zepdrix
Hmmm the top doesn't work out so nice... maybe there was a better way to do this. One sec lemme think :)
anonymous
  • anonymous
i could give u the possible answers..... they all have 6 under them
zepdrix
  • zepdrix
oh ok that might help ^^
anonymous
  • anonymous
okay hold on
anonymous
  • anonymous
zepdrix
  • zepdrix
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.
zepdrix
  • zepdrix
\[\large (2+\sqrt[3]3)\sqrt[3]{6^2} \quad = \quad 2\sqrt[3]{6^2}+\sqrt[3]3\cdot\sqrt[3]{6^2}\]And I guess they want us to simplify a bit from here.
anonymous
  • anonymous
okay i see what ur sayying
zepdrix
  • zepdrix
\[\huge \sqrt[3]{3}\cdot\sqrt[3]{36}=\sqrt[3]{108}\]
zepdrix
  • zepdrix
This part is a little tricky. You want a FACTOR of 108 that is a PERFECT CUBE.
zepdrix
  • zepdrix
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.\[\large 2^3=8, \qquad 3^3=27, \qquad ...\]
zepdrix
  • zepdrix
It turns out that 108 is divisible by 27!\[\huge \sqrt[3]{108}=\sqrt[3]{4\cdot27}\]
anonymous
  • anonymous
soooo
zepdrix
  • zepdrix
Ah sorry lost my connection D:
anonymous
  • anonymous
27 isnt a possible answer....
zepdrix
  • zepdrix
27 is a perfect cube! So let's take the cube root of 27.
zepdrix
  • zepdrix
\[\huge \sqrt[3]{4\cdot27}\quad =\quad \sqrt[3]{4}\cdot \sqrt[3]{27}=\sqrt[3]{4}\cdot 3\]
zepdrix
  • zepdrix
This is a really awful problem. Is this for algebra or something? D:
anonymous
  • anonymous
algabra 2 :/
zepdrix
  • zepdrix
So it looks like it isssss probablyyyyyyy .... d. I'm like .. 65% sure :D lol
zepdrix
  • zepdrix
It's a really annoying problem. Easy to make a mistake on. But I think we did it correctly.
anonymous
  • anonymous
okay thankyou(:could u help me with another?
anonymous
  • anonymous
it's not like that one i promise lol
zepdrix
  • zepdrix
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 ^^

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