AngelaB97
  • AngelaB97
what is 6 --- ^3√4
Mathematics
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SOLVED
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chestercat
  • chestercat
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AngelaB97
  • AngelaB97
what is |dw:1441484588912:dw|
AngelaB97
  • AngelaB97
how do you rationalize it?
triciaal
  • triciaal
|dw:1441486622928:dw|

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More answers

triciaal
  • triciaal
to rationalize the denominator means to get rid of the radical. multiply by 1 expressed as a fraction
AngelaB97
  • AngelaB97
before you continue can i just ask is there some sort of rule for what you did when you crossed out those fractions. Meaning, could i be able to do that with any sort of radical that i want to rationalize?
triciaal
  • triciaal
|dw:1441486781234:dw|
triciaal
  • triciaal
rules for exponents the nth root is the same as a fractional exponent.
AngelaB97
  • AngelaB97
the answer is supposed to be |dw:1441486876177:dw|
Excalibur0126
  • Excalibur0126
Or in decimal form, 3.7797631...
triciaal
  • triciaal
|dw:1441486898347:dw|
triciaal
  • triciaal
give me a minute
AngelaB97
  • AngelaB97
sure
triciaal
  • triciaal
|dw:1441487116337:dw||dw:1441487355866:dw|
triciaal
  • triciaal
I think I messed up on something @zepdrix please correct
zepdrix
  • zepdrix
Hey Angela :) In order to rationalize this thing, you'd like the power on the 4 to be a 1. I like the first step that Tricia applied, writing the 4 with a rational exponent.\[\large\rm \frac{6}{\sqrt[3]{4}}=\frac{6}{4^{1/3}}\]So we don't want a 1/3 power, we want a 1 power down there. What do we have to add to 1/3 to get 1?
AngelaB97
  • AngelaB97
hey :)) we have to add 2/3
zepdrix
  • zepdrix
Good good good. We would like 2/3 exponent, we'll leave the base the same. So we actually want to multiply top and bottom by \(\large\rm 4^{2/3}\) Our rules of exponents will give us \(\large\rm 4^{1/3}\cdot4^{2/3}=4^{1/3+2/3}\) in the denominator!
triciaal
  • triciaal
@zepdrix thanks
zepdrix
  • zepdrix
\[\large\rm \frac{6}{4^{1/3}}\left(\frac{4^{2/3}}{4^{2/3}}\right)=\frac{6\cdot4^{2/3}}{4^{3/3}}\]Ok with that step Angela? :o
triciaal
  • triciaal
|dw:1441488139781:dw|
AngelaB97
  • AngelaB97
6 * 42/3 divided by 4
zepdrix
  • zepdrix
\[\large\rm \frac{\color{orangered}{6}\cdot 4^{2/3}}{\color{orangered}{4}}\]Mmmm k good. This can be simplified a little bit further since 6 and 4 share a common factor.
AngelaB97
  • AngelaB97
so on the bottom we get 2 and the 6 turns into a 3?
zepdrix
  • zepdrix
\[\large\rm \frac{3\cdot4^{2/3}}{2}\]Yayyyy good job \c:/
AngelaB97
  • AngelaB97
so we don't need to do any further simplifying? because the answer in my book shows |dw:1441488618998:dw| @zepdrix
zepdrix
  • zepdrix
OH interesting :O Ok sec I think about it
zepdrix
  • zepdrix
Ok maybe this is a better route to take then..\[\large\rm \frac{6}{4^{1/3}}=\frac{3\cdot2}{4^{1/3}}=\frac{3\cdot2}{(2^2)^{1/3}}=\frac{3\cdot2}{2^{2/3}}\]
zepdrix
  • zepdrix
I'm rewriting the 6 as 3*2. I'm rewriting 4 as 2 squared. and then applying exponent rule, when we have a power and a power like that, we multiply. So 2 times 1/3 gave me 2/3 for the power on the 2.
zepdrix
  • zepdrix
\[\large\rm =\frac{3\cdot2^1}{2^{2/3}}\]And now we can apply one of our exponent rules from here: \(\large\rm \frac{x^{a}}{x^b}=x^{a-b}\) Do you see how we can apply that to the 2's?
AngelaB97
  • AngelaB97
yes thank you very much
triciaal
  • triciaal
@AngelaB97 sorry if we confuse you but hopefully you understand more about the rules of exponents and how to rationalize it is very easy to make mistakes if not careful
AngelaB97
  • AngelaB97
@zepdrix @triciaal thank you both for your help!
zepdrix
  • zepdrix
yay team \c:/
triciaal
  • triciaal
welcome anytime

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