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HorseCrazyGirlForever
Group Title
Simplify the following roots. Please click on my question to see the problems because I have to draw it out. Thanks.
 one year ago
 one year ago
HorseCrazyGirlForever Group Title
Simplify the following roots. Please click on my question to see the problems because I have to draw it out. Thanks.
 one year ago
 one year ago

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HorseCrazyGirlForever Group TitleBest ResponseYou've already chosen the best response.0
dw:1356140871952:dw
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
\[\huge \sqrt[3]{16}+5\;\sqrt[3]{54}\]Hmm so this is a bit tricky. We want to simplify the roots. So we need to find FACTORS of those numbers that are PERFECT CUBES. Examples of Perfect Cubes > 8, 27, 64, ...\[\huge 2 \cdot2\cdot2=8 \qquad \qquad 2^3=8\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Since neither of our numbers is larger than 64, we'll be trying to find factors of 27 or 8 under the roots.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
\[\huge \sqrt[3]{16} \quad \rightarrow \sqrt[3]{\color{orangered}{2}\cdot \color{orangered}{8}}\]Understand how I did the first part? :D
 one year ago

HorseCrazyGirlForever Group TitleBest ResponseYou've already chosen the best response.0
Yes.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
\[\huge \sqrt[3]{\color{orangered}{2}\cdot\color{orangered}{8}} \qquad = \qquad \sqrt[3]{\color{orangered}{2}}\cdot \sqrt[3]{\color{orangered}{8}} \qquad = \qquad \sqrt[3]{\color{orangered}{2}}\cdot 2\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Have any ideas for the second one? :)
 one year ago

HorseCrazyGirlForever Group TitleBest ResponseYou've already chosen the best response.0
So you are trying to break it down as much as possible?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Not AS MUCH AS POSSIBLE. We certainly could have broken the 8 down a little further, into 4*2, and then into another 2*2*2.. But that wouldn't have helped us. We're trying to break down the number into a perfect cube. 8 is a perfect cube, so we were able to break it down to 8 nicely.
 one year ago

HorseCrazyGirlForever Group TitleBest ResponseYou've already chosen the best response.0
Ohhh I see! So the factor of 54 would be 27?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Yah I think that works! 27*2 = 54
 one year ago

HorseCrazyGirlForever Group TitleBest ResponseYou've already chosen the best response.0
Awesomeness!
 one year ago

HorseCrazyGirlForever Group TitleBest ResponseYou've already chosen the best response.0
So it would just be written like that?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
\[\huge \sqrt[3]{54}\qquad =\qquad \sqrt[3]{\color{royalblue}{2}\cdot \color{royalblue}{27}}\qquad = \qquad \sqrt[3]{2}\cdot 3\]That part make sense? We have a little bit more work to do on this one.
 one year ago

HorseCrazyGirlForever Group TitleBest ResponseYou've already chosen the best response.0
I understand the first part, but I don't under stand how 2*27 turns into 2*3...
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
\[\huge \sqrt[3]{\color{royalblue}{2}\cdot \color{royalblue}{27}}\]From here, remember that you can split up the root if you want. So we'll write it like so,\[\huge \sqrt[3]{\color{royalblue}{2}}\cdot \sqrt[3]{\color{royalblue}{27}}\] From here, we just need to remember that 27 is a perfect square.\[\huge 3^3=27\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
So taking the cube root of 27 should give us 3.
 one year ago

HorseCrazyGirlForever Group TitleBest ResponseYou've already chosen the best response.0
Oh okay.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
So our problem has simplified to this,\[\huge \sqrt[3]{16}+5\;\sqrt[3]{54} \qquad \rightarrow \qquad 2\sqrt[3]{2}+5(3\sqrt[3]{2})\]
 one year ago

HorseCrazyGirlForever Group TitleBest ResponseYou've already chosen the best response.0
Okay I understand all of that. So how would I solve the rest?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Let's multiply the 3 and 5 out.\[\huge 2\sqrt[3]{2}+5(3\sqrt[3]{2}) \qquad =\qquad 2\;\sqrt[3]{2}+15\;\sqrt[3]{2}\] From here, you might notice that we have similar terms. Sure, they're kinda ugly but they should combine nonetheless.\[\huge 17\; \sqrt[3]{2}\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
If this part is confusing you, maybe think of the cuberoot 2 as something simpler, like x.\[\huge 2x+15x=17x\]
 one year ago

HorseCrazyGirlForever Group TitleBest ResponseYou've already chosen the best response.0
Awesome! Thank you so much! I have a few more, could you please help me with those to?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Close this thread, open a new one. In the description somewhere, type @zepdrix It will give me a little popup, so I can easily find your question. There are lots of good helpers on here. So if I can't get to your question I'm sure someone can. But yah I'll come take a look ^^ heh
 one year ago

HorseCrazyGirlForever Group TitleBest ResponseYou've already chosen the best response.0
Okay! Thank you so much!
 one year ago
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